🔬 Model Lab

Problem 1:
We are asked about the parity of $o^2+no$ where $o$ is odd and $n$ is any whole number.

Factor:
$$o^2+no = o(o+n).$$

$o$ is odd, so $o^2 = o\cdot o$ is odd times odd = odd.

Hence parity is odd + $n\cdot o$.
$n\cdot o$ has the same parity as $n$, because $o$ is odd: if $n$ is even, $n o$ is even; if $n$ is odd, $n o$ is odd.

So:
- if $n$ even: odd + even = odd.
- if $n$ odd: odd + odd = even.

Thus the number is odd iff $n$ is even, and even iff $n$ is odd.

Check choices:
A. always odd - false, e.g. $o=1,n=1$ gives $1+1=2$ even.
B. always even - false, e.g. $o=1,n=2$ gives $1+2=3$ odd.
C. it is even only if $n$ is even - means even $\implies$ $n$ even. False, even happens when $n$ is odd.
D. it is odd only if $n$ is odd - means odd $\implies$ $n$ odd. False, odd happens when $n$ is even.
E. it is odd only if $n$ is even - means odd $\implies$ $n$ even. True by above.

ANSWER 1: E

Problem 2:
Isabella redeems on days $d, d+10, d+20, d+30, d+40, d+50$.
No date is a Sunday.

Modulo 7, $10\equiv 3$. So successive dates advance by 3 weekdays.

Starting from residue $r$, the residues are:
$$r,\ r+3,\ r+6,\ r+9=r+2,\ r+12=r+5,\ r+15=r+1 \pmod 7.$$
These 6 are all distinct, because adding 3 mod 7 cycles through all 7 residues. Continuing one more step would give $r+18=r+4$.

Thus the 6 dates occupy 6 distinct days of the week, missing exactly the 7th residue $r+4$.

Since none of the 6 is Sunday, the missing one must be Sunday. So:
$$r+4 \equiv \text{Sunday}.$$

Hence $r$ is 4 days before Sunday, i.e. Wednesday.

Check: Wed $\to$ Wed+3=Sat $\to$ +3=Tue $\to$ +3=Fri $\to$ +3=Mon $\to$ +3=Thu. That is Wed,Sat,Tue,Fri,Mon,Thu - no Sunday. Any other start would include Sunday.

ANSWER 2: C

Problem 3:
We interpret $3\ 2\cdot 4\ 3\cdots$ as fractions:
$$\frac32 \cdot \frac43 \cdot \frac54 \cdot \frac65 \cdots \frac{a}{b}=9.$$

The pattern is numerator $3,4,5,6,\dots$ and denominator $2,3,4,5,\dots$, i.e. term $k/(k-1)$. The last term is $a/b$, so we must have $b=a-1$ for the pattern to continue.

Then the product telescopes:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}
= \frac{a}{2},$$
because all intermediate $3,4,\dots,a-1$ cancel.

Set equal to 9:
$$\frac{a}{2}=9 \implies a=18.$$
Then $b=a-1=17$.

Sum $a+b=18+17=35$.

This rules out A,B,C,E.

ANSWER 3: D

Problem 4:
Let $x,y,z$ be numbers of 5-cent, 10-cent, 25-cent stamps used. $0\le x,y,z\le 20$ integers.
Value $710$ cents:
$$5x+10y+25z=710.$$
Divide by 5:
$$x+2y+5z=142. \tag{1}$$
Maximize $N=x+y+z$.

From (1), $x=142-2y-5z$, so $0\le x\le20$ gives:
$$122\le 2y+5z\le 142.$$
Also $N=x+y+z=142-y-4z$, so $y+4z=142-N$.

To use many stamps we want small denominations. Try $x=20,y=20$ fails to give integer $z$. Search systematically.

Since $y\le20$, to satisfy $2y+5z\ge122$ we need $z$ large. If $z\le16$, then $2y+5z\le40+80=120<122$, impossible. So $z\ge17$.

Try $z=17,y=19$: then $2y+5z=38+85=123$, so $x=142-123=19$, within $0-20$. This gives:
$$19\cdot5+19\cdot10+17\cdot25=95+190+425=710,$$
with $N=19+19+17=55$ stamps. So 55 is achievable.

Can we do 56 or more? Let $N\ge56$.
Then $y+4z=142-N\le86$ and $x=N-y-z\le20$ gives $y+z\ge N-20$.
Subtract: $3z=(y+4z)-(y+z)\le (142-N)-(N-20)=162-2N$.
So $z\le(162-2N)/3$.
Also $y=142-N-4z\le20$ gives $z\ge(122-N)/4$.

For $N=56$: $z\ge\lceil66/4\rceil=17$ and $z\le\lfloor50/3\rfloor=16$, contradiction. For $N>56$ the interval is even emptier; in general need $(122-N)/4\le(162-2N)/3$, i.e. $5N\le282$, so $N\le56.4$. Thus $N\ge57$ impossible, and $N=56$ impossible by the integer contradiction.

Hence 55 is greatest possible.

ANSWER 4: E

Problem 5:
We have $ABA\times CD = CDCD$, where $A,B,C,D$ are different digits, $A\ne0$, $C\ne0$.

Write $CD$ as the 2-digit number $10C+D$.
$CDCD$ as a 4-digit number is:
$$1000C+100D+10C+D =1010C+101D=101(10C+D)=101\cdot CD.$$

So:
$$ABA \times CD = 101\times CD.$$
Since $CD\ne0$, divide:
$$ABA =101.$$

Thus $A=1$, $B=0$, and $A+B=1$.

This is consistent with distinctness: we can choose e.g. $C=2,D=3$, all of $1,0,2,3$ different, and $101\times23=2323$. So valid.

Other sums are ruled out.

ANSWER 5: A

Problem 6:
Total 100: 52 boys, 48 girls. 40 from Jonas, 60 from Clay.

20 of the girls are from Jonas. So girls from Clay = total girls $-$ girls from Jonas:
$$48-20=28.$$

Clay total 60 = boys from Clay + girls from Clay, so:
$$\text{boys from Clay}=60-28=32.$$

Check: boys from Jonas $=52-32=20$, Jonas total $=20+20=40$. Works.

ANSWER 6: B

Problem 7:
Compute $(1901+\cdots+1993)-(101+\cdots+193)$.

Both sums have $1993-1901+1=93$ terms and $193-101+1=93$ terms.

Pair termwise: for $k=0,\dots,92$,
$$(1901+k)-(101+k)=1800.$$

So difference $=93\times1800$.
$93\times18=1674$, so $93\times1800=167{,}400$.

Other choices come from miscounting terms or arithmetic error.

ANSWER 7: A

Problem 8:
Find tens digit of $7^{2011}$, i.e. $7^{2011}\pmod{100}$.

Compute powers mod 100:
$7^1=7$,
$7^2=49$,
$7^3=343\equiv43$,
$7^4\equiv43\times7=301\equiv1$.

Since $7^4\equiv1\pmod{100}$, the last two digits repeat with period 4.

$2011 = 4\cdot502+3$, so:
$$7^{2011}=(7^4)^{502}\cdot7^3\equiv1^{502}\cdot43\equiv43\pmod{100}.$$

Thus the number ends $\dots43$, tens digit $4$.

ANSWER 8: D

Problem 9:
10 different positive whole numbers have mean 10, so sum $=10\times10=100$.

To make one number as large as possible, make the other nine as small as possible while distinct and positive.

Smallest 9 distinct positives are $1,2,3,4,5,6,7,8,9$, sum $=45$.

Then the tenth number $=100-45=55$, which is distinct from the others and valid. Any larger value would force the other nine to sum to $<45$, impossible with distinct positives.

So largest possible is 55.

ANSWER 9: C

Problem 10:
Count student-days (one student working one day).

Allen: $7$ students $\times$ $3$ days $=21$.
Balboa: $4\times5=20$.
Carver: $5\times9=45$.
Total $=21+20+45=86$ student-days.

Total pay $\$774$ for $86$ student-days at equal rate:
$$\text{rate}=774/86=9 \text{ dollars per student-day}.$$

Bal

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that $o$ is an odd whole number and $n$ is any whole number. We must determine which statement about $o^2 + no$ is always true.

**Step-by-Step Solution:**
1. Factor the expression:
   $$o^2 + no = o(o + n)$$
2. Since $o$ is an odd whole number, the parity of the product $o(o + n)$ is entirely determined by the parity of $(o + n)$:
   - If $n$ is **odd**, then $\text{odd} + \text{odd} = \text{even}$, so $o(o+n)$ is $\text{odd} \times \text{even} = \textbf{even}$.
   - If $n$ is **even**, then $\text{odd} + \text{even} = \text{odd}$, so $o(o+n)$ is $\text{odd} \times \text{odd} = \textbf{odd}$.
3. Therefore, the expression $o^2 + no$ is odd if and only if $n$ is even. 
4. This matches the statement: "it is odd only if $n$ is even."

ANSWER 1: E

---

### Problem 2:
**Restatement:** Isabella redeems 6 coupons, one every 10 days. Pete's is closed on Sundays, and none of the 6 dates falls on a Sunday. We need to find the day of the week she redeems her first coupon.

**Step-by-Step Solution:**
1. There are 7 days in a week. Moving forward by 10 days advances the day of the week by:
   $$10 \equiv 3 \pmod 7$$
2. Let $d$ represent the day of the week of the first coupon ($0 = \text{Sunday}, 1 = \text{Monday}, \dots, 6 = \text{Saturday}$).
3. The days of the week for the 6 coupons will be:
   - Coupon 1: $d$
   - Coupon 2: $d + 3$
   - Coupon 3: $d + 6$
   - Coupon 4: $d + 9 \equiv d + 2 \pmod 7$
   - Coupon 5: $d + 12 \equiv d + 5 \pmod 7$
   - Coupon 6: $d + 15 \equiv d + 1 \pmod 7$
4. The six days of the week visited are $d, d+1, d+2, d+3, d+5, d+6$. Notice that the only day of the week omitted from this list is $d+4$.
5. Since no circled date falls on a Sunday, Sunday must be the omitted day:
   $$d + 4 \equiv 0 \pmod 7 \implies d \equiv 3 \pmod 7$$
6. Day 3 corresponds to **Wednesday**.
   - Checking: Wed $\to$ Sat $\to$ Tue $\to$ Fri $\to$ Mon $\to$ Thu. None of these is Sunday.

ANSWER 2: C

---

### Problem 3:
**Restatement:** We are given the telescoping product $\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdots \frac{a}{b} = 9$. We must find the sum $a + b$.

**Step-by-Step Solution:**
1. Each fraction is of the form $\frac{k+1}{k}$, where the numerator of each term cancels with the denominator of the subsequent term:
   $$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{a}{b} = \frac{a}{2}$$
   where $b = a - 1$.
2. We are given that this product equals 9:
   $$\frac{a}{2} = 9 \implies a = 18$$
3. Since $b = a - 1$, we have $b = 17$.
4. The sum of $a$ and $b$ is:
   $$a + b = 18 + 17 = 35$$

ANSWER 3: D

---

### Problem 4:
**Restatement:** Nicolas has 20 stamps each of 5¢, 10¢, and 25¢. We want to find the greatest number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-Step Solution:**
1. Calculate the total value and count of all stamps:
   - $20 \times 5¢ = 100¢$
   - $20 \times 10¢ = 200¢$
   - $20 \times 25¢ = 500¢$
   - Total value $= 100 + 200 + 500 = 800¢$ using 60 stamps.
2. To use the **maximum** number of stamps to make 710¢, Nicolas must **minimize** the number of stamps removed to subtract:
   $$800¢ - 710¢ = 90¢$$
3. To remove the fewest stamps totaling 90¢, we try to use the largest denomination stamps (25¢):
   - With 3 stamps of 25¢ ($75¢$), the remaining amount needed is $15¢$.
   - Since Nicolas has only 5¢, 10¢, and 25¢ stamps, 15¢ cannot be made with a single stamp. It requires at least two stamps ($10¢ + 5¢$).
   - Thus, 90¢ can be formed with $3 \times 25¢ + 1 \times 10¢ + 1 \times 5¢$, which uses $3 + 1 + 1 = 5$ stamps.
   - Can 90¢ be formed using 4 stamps? The maximum value of 4 stamps less than or equal to 90¢ would be $3 \times 25¢ + 10¢ = 85¢ \ne 90¢$. Thus, 4 stamps is impossible.
4. The minimum number of stamps to remove is 5.
5. Therefore, the maximum number of stamps used is:
   $$60 - 5 = 55$$

ANSWER 4: E

---

### Problem 5:
**Restatement:** In the multiplication $ABA \times CD = CDCD$, $A, B, C, D$ are distinct digits. We need to find $A + B$.

**Step-by-Step Solution:**
1. Express $CDCD$ in terms of the two-digit number $CD$:
   $$CDCD = CD \times 100 + CD = CD \times 101$$
2. Substitute this into the equation:
   $$ABA \times CD = 101 \times CD$$
3. Since $CD$ is a non-zero two-digit number, divide both sides by $CD$:
   $$ABA = 101$$
4. Comparing digits, we have $A = 1$ and $B = 0$.
5. Thus:
   $$A + B = 1 + 0 = 1$$

ANSWER 5: A

---

### Problem 6:
**Restatement:** Out of 100 students at a camp, 52 are boys and 48 are girls. Also, 40 are from Jonas Middle School and 60 from Clay Middle School. Given that 20 of the girls are from Jonas, find how many boys are from Clay.

**Step-by-Step Solution:**
1. Find the number of girls from Clay Middle School:
   $$\text{Total girls} - \text{Girls from Jonas} = 48 - 20 = 28\text{ girls from Clay}$$
2. Find the number of boys from Clay Middle School:
   $$\text{Total students from Clay} - \text{Girls from Clay} = 60 - 28 = 32\text{ boys from Clay}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** Compute the value of:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-Step Solution:**
1. Pair each term in the first sum with the corresponding term in the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
2. Each paired difference is equal to:
   $$1900 + k - (100 + k) = 1800$$
3. Find the number of terms: from $k = 1$ to $k = 93$, there are:
   $$93 - 1 + 1 = 93\text{ terms}$$
4. Multiply the difference by the number of terms:
   $$93 \times 1800 = 93 \times 18 \times 100 = 1674 \times 100 = 167,400$$

ANSWER 7: A

---

### Problem 8:
**Restatement:** Find the tens digit of $7^{2011}$.

**Step-by-Step Solution:**
1. Find the pattern of the last two digits of powers of 7 by computing $7^n \pmod{100}$:
   - $7^1 = 07$
   - $7^2 = 49$
   - $7^3 = 343 \equiv 43 \pmod{100}$
   - $7^4 = 43

# Problem 1:
We must decide when $o^2 + no$ is odd or even, given $o$ is odd and $n$ is any whole number.

**Key facts:** odd × odd = odd, so $o^2$ is **odd**. Since $o$ is odd, multiplying by $o$ does not change $n$'s parity, so $no$ has the **same parity as $n$**.

**Case 1: $n$ even.** Then $no$ is even, so $o^2 + no = \text{odd} + \text{even} = \textbf{odd}$.
**Case 2: $n$ odd.** Then $no$ is odd, so $o^2 + no = \text{odd} + \text{odd} = \textbf{even}$.

So the number is odd exactly when $n$ is even. In particular, **if the number is odd, $n$ must be even**.

Ruling out traps: (A) fails since $o=1, n=1$ gives $1+1=2$, even; (B) fails since $n=2$ gives $1+2=3$, odd; (C) and (D) state the exact opposite of the truth. Only (E) matches.

ANSWER 1: E

# Problem 2:
The 6 redemption dates are 10 days apart, so their days of the week shift by $10 \bmod 7 = 3$ each time.

If the first date falls on weekday $W$, the six dates fall on (mod 7):
$$W,\; W+3,\; W+6,\; W+2,\; W+5,\; W+1$$
That is **six of the seven weekdays**, and the only weekday missed is $W+4$.

Since no circled date is a Sunday, the one missed weekday must be Sunday:
$$W + 4 = \text{Sunday} \implies W = \text{Wednesday}.$$

**Check:** Starting Wednesday, the dates fall on Wed, Sat, Tue, Fri, Mon, Thu — no Sunday. (Trap: starting Thursday gives Thu, Sun, ... — fails immediately.)

ANSWER 2: C

# Problem 3:
The product is a chain of fractions (the bars were lost in formatting):
$$\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdots\frac{a}{b} = 9.$$

Each numerator cancels the denominator before it (e.g., $\frac{3}{2}\cdot\frac{4}{3} = \frac{4}{2}$), and each fraction has the form $\frac{k+1}{k}$, so $b = a-1$. Everything telescopes:
$$\text{Product} = \frac{a}{2} = 9 \implies a = 18, \quad b = 17.$$
Then $a + b = 18 + 17 = 35$.

Trap check: $a+b=17$ would mean $a=9$, giving product $\tfrac92$, not $9$; only $35$ fits the pattern.

ANSWER 3: D

# Problem 4:
Work in units of 5¢: we need total value $710¢ = 142$ units, using stamps worth $1, 2, 5$ units (i.e., 5¢, 10¢, 25¢), at most 20 of each. To **maximize the count**, use as many cheap stamps as possible.

**Building a 55-stamp solution:** Take nineteen 1's and nineteen 2's: value $19 + 38 = 57$ units with 38 stamps. Remaining value $142 - 57 = 85 = 17\times 5$: seventeen 5's.
$$19\times 5¢ + 19\times 10¢ + 17\times 25¢ = 95 + 190 + 425 = 710¢ = \$7.10.\; ✓$$
That's $19+19+17 = 55$ stamps, all counts $\le 20$. ✓

**Why 56 is impossible:** with 56 stamps, at least $56 - 40 = 16$ must be 25¢ stamps (only 40 cheap ones exist). Let $z$ = number of 25¢ stamps:
- $z=16$: max value $= 100 + 200 + 400 = 700 < 710$. ✗
- $z=17$: need $5x+10y=285$, $x+y=39 \Rightarrow x=21>20$. ✗
- $z=18$: $x=24>20$. ✗  $z=19$: $x=27>20$. ✗  $z=20$: $x=30>20$. ✗

Traps: 54 comes from $20×5+16×10+18×25 = 710$; 51 from $20×5+11×10+20×25 = 710$ — both valid but not maximal.

ANSWER 4: E

# Problem 5:
Write $CDCD$ as a number:
$$CDCD = 1000C + 100D + 10C + D = 101(10C + D) = 101 \times CD.$$
The given equation is
$$ABA \times CD = CDCD = 101 \times CD.$$
Since $CD \neq 0$ (it's a two-digit number), we can divide both sides by $CD$:
$$ABA = 101.$$
So $A = 1$ and $B = 0$, giving $A + B = 1$.

**Consistency check:** $C, D$ just need to be distinct from $0, 1$ and each other, e.g., $101 \times 23 = 2323$. ✓ (Trap: guessing digits without using the factorization 101 wastes time or gives 9.)

ANSWER 5: A

# Problem 6:
Set up a two-way table with 100 total students.

- Girls: 48 total, and 20 girls are from Jonas $\Rightarrow$ girls from Clay $= 48 - 20 = 28$.
- Jonas has 40 students, of whom 20 are girls $\Rightarrow$ boys from Jonas $= 40 - 20 = 20$.
- Boys: 52 total $\Rightarrow$ boys from Clay $= 52 - 20 = \textbf{32}$.

**Check:** Jonas $20+20=40$ ✓; Clay $28+32=60$ ✓; boys $20+32=52$ ✓; girls $20+28=48$ ✓; total $100$ ✓. (Trap: 20 counts the boys from Jonas, not Clay.)

ANSWER 6: B

# Problem 7:
Both sums have the same number of terms: $1901$ to $1993$ has $1993-1901+1 = 93$ terms, and $101$ to $193$ has $193-101+1 = 93$ terms.

Pair them up: each term in the first sum is exactly $1800$ more than its partner ($1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$). So the difference is
$$93 \times 1800 = 167{,}400.$$

**Check via formulas:** $\frac{93(1901+1993)}{2} - \frac{93(101+193)}{2} = 181{,}071 - 13{,}671 = 167{,}400$. ✓ (Trap: 181,071 is the first sum alone — forgetting to subtract.)

ANSWER 7: A

# Problem 8:
Only the tens digit matters, so track the last two digits of $7^{k}$, which cycle with period 4:
$$7^1 = 07,\quad 7^2 = 49,\quad 7^3 = 343 \to 43,\quad 7^4 = 2401 \to 01,\quad 7^5 \to 07, \ldots$$

Find where $2011$ falls in the cycle:
$$2011 = 4 \times 502 + 3,$$
so $7^{2011}$ ends in the same two digits as $7^3$, namely $\ldots 43$.

The tens digit is $\textbf{4}$. (Trap: $7^{2012}$, a multiple of 4, ends in $01$, tens digit 0 — choice A; that would require the exponent to be divisible by 4.)

ANSWER 8: D

# Problem 9:
The ten **different** positive whole numbers have average 10, so their sum is $10 \times 10 = 100$.

To make one number as large as possible, make the other nine as small as possible. Since they must be **distinct positive** whole numbers, the smallest possible set is $1, 2, 3, \ldots, 9$, with sum
$$1+2+\cdots+9 = 45.$$
So the largest possible value is
$$100 - 45 = 55.$$

**Check:** $1,2,\ldots,9,55$ are ten different positive whole numbers with sum $100$ and average $10$. ✓ (Trap: 91 arises if you ignore "different" and use nine 1's; but the numbers must be distinct.)

ANSWER 9: C

# Problem 10:
Count student-days (the unit being paid):
- Allen: $7 \times 3 = 21$ student-days
- Balboa: $4 \times 5 = 20$ student-days
- Carver: $5 \times 9 = 45$ student-days

Total: $21 + 20 + 45 = 86$ student-days, paid $\$774$ altogether. So the daily rate is
$$774 \div 86 = 9 \text{ dollars per student-day}.$$
Balboa's students worked 20 student-days, so they earned
$$20 \times 9 = 18

Problem 1:  
We need the parity of \(o^2+no\), where \(o\) is odd and \(n\) is any whole number.

Factor:
\[
o^2+no=o(o+n).
\]

Since \(o\) is odd:

- If \(n\) is even, then \(o+n\) is odd, so \(o(o+n)\) is odd.
- If \(n\) is odd, then \(o+n\) is even, so \(o(o+n)\) is even.

Thus the expression is odd exactly when \(n\) is even.

Check the choices:
- A and B are false because the parity depends on \(n\).
- C says “even only if \(n\) is even,” but when the expression is even, \(n\) is actually odd.
- D says “odd only if \(n\) is odd,” but when the expression is odd, \(n\) is even.
- E says “odd only if \(n\) is even,” which is true.

ANSWER 1: E

Problem 2:  
Isabella redeems coupons 10 days apart. Since
\[
10\equiv 3 \pmod 7,
\]
the six circled days are the first day plus
\[
0,3,6,9,12,15
\]
days, which modulo 7 gives offsets
\[
0,3,6,2,5,1.
\]

So the six circled days are all weekdays except the one 4 days after the first day. Since none of the circled days is a Sunday, Sunday must be the un-circled day: 4 days after the first day.

Therefore the first coupon is 4 days before Sunday, which is Wednesday.

Check: Wednesday, Saturday, Tuesday, Friday, Monday, Thursday — no Sunday.

ANSWER 2: C

Problem 3:  
The product is the telescoping product of fractions:
\[
\frac32\cdot \frac43\cdot \frac54\cdot \frac65\cdots \frac{a}{b}=9.
\]

The numerators cancel the next denominators:
\[
\frac32\cdot \frac43=\frac42,
\]
\[
\frac42\cdot \frac54=\frac52,
\]
and continuing, the whole product simplifies to
\[
\frac{a}{2}.
\]

So
\[
\frac{a}{2}=9 \implies a=18.
\]

By the pattern, the denominator is one less than the numerator, so
\[
b=17.
\]

Thus
\[
a+b=18+17=35.
\]

ANSWER 3: D

Problem 4:  
Nicolas has 20 of each type of stamp, so he has
\[
20+20+20=60
\]
stamps total. Their total value is
\[
20(5+10+25)=20(40)=800
\]
cents.

He wants postage of
\[
\$7.10=710
\]
cents, so the stamps he does **not** use must total
\[
800-710=90
\]
cents.

To maximize the number of stamps he uses, we minimize the number of stamps he leaves unused.

Can 4 unused stamps total 90 cents? Let \(x,y,z\) be the numbers of 25-, 10-, and 5-cent unused stamps. Then
\[
x+y+z=4
\]
and
\[
25x+10y+5z=90.
\]

Divide by 5:
\[
5x+2y+z=18.
\]

Subtract \(x+y+z=4\):
\[
4x+y=14.
\]

Testing possible \(x=0,1,2,3,4\):

- \(x=0\Rightarrow y=14\), impossible.
- \(x=1\Rightarrow y=10\), impossible.
- \(x=2\Rightarrow y=6\), impossible.
- \(x=3\Rightarrow y=2\), then \(z=4-3-2=-1\), impossible.
- \(x=4\Rightarrow y=-2\), impossible.

So 4 unused stamps cannot total 90 cents.

Can 5 unused stamps total 90 cents? Yes:
\[
25+25+25+10+5=90.
\]

Then he uses
\[
60-5=55
\]
stamps. For example, he could use 19 five-cent stamps, 19 ten-cent stamps, and 17 twenty-five-cent stamps:
\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So the greatest possible number of stamps is 55. Choice D, 54, is possible but not greatest.

ANSWER 4: E

Problem 5:  
We are told
\[
ABA\times CD=CDCD,
\]
where \(A,B,C,D\) are different digits.

Write the numbers in place-value form:
\[
ABA=101A+10B,
\]
\[
CD=10C+D,
\]
\[
CDCD=1000C+100D+10C+D=101(10C+D).
\]

So
\[
(101A+10B)(10C+D)=101(10C+D).
\]

Since \(CD\) is a two-digit number, \(10C+D\neq 0\), so we can divide both sides by \(10C+D\):
\[
101A+10B=101.
\]

The only digit solution is
\[
A=1,\quad B=0.
\]

Therefore
\[
A+B=1+0=1.
\]

ANSWER 5: A

Problem 6:  
There are 48 girls total, and 20 of the girls are from Jonas Middle School. So the number of girls from Clay Middle School is
\[
48-20=28.
\]

There are 60 students total from Clay Middle School, so the number of boys from Clay is
\[
60-28=32.
\]

ANSWER 6: B

Problem 7:  
Pair the terms:
\[
(1901+1902+\cdots+1993)-(101+102+\cdots+193)
\]
\[
=(1901-101)+(1902-102)+\cdots+(1993-193).
\]

Each pair is
\[
1800.
\]

Now count the number of terms:
\[
193-101+1=93.
\]

So the value is
\[
93\cdot 1800=167400.
\]

ANSWER 7: A

Problem 8:  
We need the tens digit of \(7^{2011}\), so we need the last two digits.

Powers of 7 modulo 100 cycle:
\[
7^1=07,
\]
\[
7^2=49,
\]
\[
7^3=343\equiv 43,
\]
\[
7^4=2401\equiv 01.
\]

Then the cycle repeats:
\[
07,\ 49,\ 43,\ 01.
\]

Since
\[
2011\equiv 3 \pmod 4,
\]
we have
\[
7^{2011}\equiv 7^3\equiv 43 \pmod{100}.
\]

The last two digits are 43, so the tens digit is 4.

ANSWER 8: D

Problem 9:  
The average of 10 different positive whole numbers is 10, so their sum is
\[
10\cdot 10=100.
\]

To make one number as large as possible, make the other 9 numbers as small as possible while still being different positive whole numbers. The smallest 9 distinct positive whole numbers are
\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is
\[
1+2+\cdots+9=45.
\]

So the largest number can be
\[
100-45=55.
\]

This is possible: \(1,2,3,4,5,6,7,8,9,55\) are all different positive whole numbers.

ANSWER 9: C

Problem 10:  
First count total student-days worked:

- Allen: \(7\cdot 3=21\) student-days
- Balboa: \(4\cdot 5=20\) student-days
- Carver: \(5\cdot 9=45\) student-days

Total:
\[
21+20+45=86
\]
student-days.

The total payment is \$774, so the daily rate is
\[
774\div 86=9
\]
dollars per student-day.

Balboa students worked 20 student-days, so they earned
\[
20\cdot 9=180
\]
dollars.

ANSWER 10: C

# Problem 1:
We need a statement always true about o² + no, where o is odd.

Factor: o² + no = o(o + n).

Since o is odd, the parity of the product is the parity of (o + n).
- If n is even, o + n is odd → product = odd·odd = **odd**.
- If n is odd, o + n is even → product = odd·even = **even**.

So the expression is odd exactly when n is even.

Ruling out others: A fails (n odd gives even), B fails (n even gives odd), C says "even only if n is even" — but it's even when n is *odd*, false. D says "odd only if n is odd" — false, it's odd when n is even. E: "odd only if n is even" — true.

ANSWER 1: E

# Problem 2:
The redemption days are 0, 10, 20, 30, 40, 50 days after the first.

Reduce mod 7 (days of week shift): 0, 3, 6, 2, 5, 1.

These residues are {0, 1, 2, 3, 5, 6} — every residue except **4**. So the only day of the week never hit is the day 4 days after the starting day. For no coupon to land on Sunday, Sunday must be that missed day:

start + 4 ≡ Sunday.

Counting back 4 days from Sunday: Saturday, Friday, Thursday, **Wednesday**.

Check: Wed → +10 Sat → +20 Tue → +30 Fri → +40 Mon → +50 Thu. No Sundays. ✓

ANSWER 2: C

# Problem 3:
The product is (3/2)·(4/3)·(5/4)·(6/5)·…·(a/b), where each numerator is one more than its denominator, so b = a − 1.

Telescoping: everything cancels except the last numerator and first denominator:
a/2 = 9 → a = 18, so b = 17.

a + b = 18 + 17 = 35.

(Choice E, 37, would come from mistakenly taking a = 19.)

ANSWER 3: D

# Problem 4:
Total value of all 60 stamps: 20(5 + 10 + 25) = 20·40 = 800 cents.

We need 710 cents, so we must *remove* stamps worth exactly 800 − 710 = 90 cents, using as **few** stamps as possible (to keep the most stamps).

Can 90 be made with 4 or fewer stamps? Max with 4 stamps is 100; possible 4-stamp totals near 90: 100, 85, 80, 75, 70… none equal 90. Three or fewer is at most 75. So 4 or fewer is impossible.

With 5 stamps: 25 + 25 + 25 + 10 + 5 = 90 ✓

So remove 5 stamps: 60 − 5 = 55 stamps.

Check: 17(25) + 19(10) + 19(5) = 425 + 190 + 95 = 710 cents, using 17 + 19 + 19 = 55 stamps.

ANSWER 4: E

# Problem 5:
CDCD as a number = CD·100 + CD = CD·101.

So ABA × CD = 101 × CD, and since CD ≠ 0, we get ABA = 101.

Thus A = 1, B = 0 (and indeed A, B, C, D distinct works, e.g. 101 × 23 = 2323).

A + B = 1 + 0 = 1.

ANSWER 5: A

# Problem 6:
Jonas total = 40, and 20 of those were girls, so boys from Jonas = 40 − 20 = 20.

Total boys = 52, so boys from Clay = 52 − 20 = 32.

(Check: girls from Clay = 48 − 20 = 28; 32 + 28 = 60 = Clay total ✓.)

Choice C (40) is the Jonas total, and E (52) is all boys — traps.

ANSWER 6: B

# Problem 7:
Pair each term: 1901 − 101 = 1800, 1902 − 102 = 1800, …, each pair differs by 1800.

Number of terms: 1993 − 1901 + 1 = 93.

Total = 93 × 1800 = 167,400.

ANSWER 7: A

# Problem 8:
Look at 7ⁿ mod 100:
- 7¹ = 07
- 7² = 49
- 7³ = 343 → 43
- 7⁴ = 2401 → 01
- then the pattern repeats with period 4.

2011 ÷ 4 = 502 remainder 3, so 7²⁰¹¹ ≡ 7³ ≡ 43 (mod 100).

Last two digits are 43, so the tens digit is 4.

ANSWER 8: D

# Problem 9:
Sum of the 10 numbers = 10 × 10 = 100.

To make one number as large as possible, make the other nine as small as possible while distinct and positive: 1 + 2 + … + 9 = 45.

Largest = 100 − 45 = 55.

(90 or 91 would require repeats or a zero, which isn't allowed.)

ANSWER 9: C

# Problem 10:
Total student-days:
- Allen: 7 × 3 = 21
- Balboa: 4 × 5 = 20
- Carver: 5 × 9 = 45

Total = 86 student-days.

Daily wage = 774 ÷ 86 = $9.00 per student-day.

Balboa earnings = 20 × 9 = $180.00.

(A is one day's pay for one student; E is Carver's total.)

ANSWER 10: C

Problem 1:

We need determine the parity of

\[
o^2+no.
\]

Factor out \(o\):

\[
o^2+no=o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on \(o+n\).

- If \(n\) is even, then odd \(+\) even \(=\) odd, so \(o+n\) is odd. Thus odd \(\cdot\) odd is odd.
- If \(n\) is odd, then odd \(+\) odd \(=\) even, so the product is even.

Therefore, the expression is odd only when \(n\) is even.

- A is wrong because the expression is not always odd.
- B is wrong because it is odd when \(n\) is even.
- C reverses the condition for being even.
- D reverses the condition for being odd.

ANSWER 1: E

Problem 2:

Isabella uses a coupon every 10 days. Since

\[
10\equiv 3\pmod 7,
\]

each coupon date moves forward by 3 days of the week.

If the first coupon is redeemed on day \(d\), the six redemption days are:

\[
d,\ d+3,\ d+6,\ d+2,\ d+5,\ d+1.
\]

These are six different days of the week. The only day missing is

\[
d+4.
\]

Since none of the six coupon dates is Sunday, the missing day must be Sunday. Thus,

\[
d+4=\text{Sunday}.
\]

Four days before Sunday is Wednesday.

Indeed, beginning on Wednesday gives:

\[
\text{Wednesday, Saturday, Tuesday, Friday, Monday, Thursday}.
\]

Sunday is the only missing day.

ANSWER 2: C

Problem 3:

The product is

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9.
\]

Almost every numerator cancels with a denominator:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab
=\frac a2.
\]

So

\[
\frac a2=9.
\]

Therefore,

\[
a=18.
\]

The final fraction must be

\[
\frac{18}{17},
\]

so \(b=17\). Therefore,

\[
a+b=18+17=35.
\]

Choices 11, 13, and 17 are too small, while 37 would correspond to a different ending fraction.

ANSWER 3: D

Problem 4:

Let

- \(x\) = number of 5-cent stamps,
- \(y\) = number of 10-cent stamps,
- \(z\) = number of 25-cent stamps.

In units of 5 cents, the postage total is

\[
\frac{710}{5}=142.
\]

Thus,

\[
x+2y+5z=142.
\]

We want to maximize

\[
x+y+z.
\]

Since there are at most 20 five-cent stamps, \(x\leq 20\). Therefore,

\[
2y+5z=142-x\geq 142-20=122.
\]

Since \(y\leq20\), if \(z\leq16\), then

\[
2y+5z\leq 2(20)+5(16)=40+80=120,
\]

which is not enough. So \(z\geq17\).

If \(z=17\), then

\[
2y+5(17)\geq122,
\]

so

\[
2y+85\geq122,
\]

\[
2y\geq37,
\]

and hence \(y\geq19\).

Now use

\[
x+y+z=142-y-4z.
\]

Since \(y\geq19\) and \(z\geq17\),

\[
x+y+z\leq142-19-4(17)
=142-19-68
=55.
\]

This maximum is attainable with

\[
x=19,\qquad y=19,\qquad z=17.
\]

Check:

\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So 55 stamps can be used. Choices 45, 46, 51, and 54 are possible totals to consider, but 55 is the greatest possible.

ANSWER 4: E

Problem 5:

We are told

\[
ABA\times CD=CDCD.
\]

The number \(ABA\) is

\[
100A+10B+A=101A+10B.
\]

The number \(CDCD\) is

\[
1000C+100D+10C+D.
\]

Group the \(C\)'s and \(D\)'s:

\[
CDCD=1010C+101D=101(10C+D)=101\cdot CD.
\]

Thus the equation becomes

\[
ABA\cdot CD=101\cdot CD.
\]

Since \(CD\) is a two-digit number and is not zero, divide both sides by \(CD\):

\[
ABA=101.
\]

Therefore,

\[
A=1,\qquad B=0.
\]

So

\[
A+B=1+0=1.
\]

For example, \(101\times23=2323\), and the digits \(1,0,2,3\) are all different.

ANSWER 5: A

Problem 6:

There are 40 students from Jonas Middle School, and 20 of those are girls.

Therefore the number of boys from Jonas is

\[
40-20=20.
\]

There are 52 boys altogether. Therefore the number of boys from Clay Middle School is

\[
52-20=32.
\]

The other choices do not correctly account for the 20 Jonas girls and the total of 52 boys.

ANSWER 6: B

Problem 7:

There are

\[
1993-1901+1=93
\]

terms in the first sum.

There are also

\[
193-101+1=93
\]

terms in the second sum.

Pair corresponding terms:

\[
1901-101=1800,
\]

\[
1902-102=1800,
\]

and so on, through

\[
1993-193=1800.
\]

Thus there are 93 differences of 1800:

\[
93\cdot1800=167400.
\]

Therefore the value is

\[
167,\!400.
\]

ANSWER 7: A

Problem 8:

We need the tens digit of

\[
7^{2011}.
\]

It is enough to examine powers of 7 modulo 100:

\[
7^1=7,
\]

\[
7^2=49,
\]

\[
7^3=343,
\]

whose last two digits are 43, and

\[
7^4=2401,
\]

whose last two digits are 01.

Thus the last two digits repeat every 4 powers because

\[
7^4\equiv1\pmod{100}.
\]

Now,

\[
2011\div4
\]

leaves remainder 3, since

\[
2011=4(502)+3.
\]

Therefore,

\[
7^{2011}
\]

has the same last two digits as

\[
7^3=343,
\]

which are 43. The tens digit is 4.

ANSWER 8: D

Problem 9:

The average of 10 different positive whole numbers is 10, so their total sum is

\[
10\cdot10=100.
\]

To make one number as large as possible, make the other nine numbers as small as possible.

The nine smallest different positive whole numbers are

\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is

\[
1+2+3+4+5+6+7+8+9=45.
\]

So the largest possible tenth number is

\[
100-45=55.
\]

This works because the ten numbers can be

\[
1,2,3,4,5,6,7,8,9,55,
\]

which are all different and have total 100.

A value such as 90 or 91 would leave too small a total for nine distinct positive whole numbers.

ANSWER 9: C

Problem 10:

Compute the total number of student-days worked.

Allen students worked:

\[
7\cdot3=21
\]

student-days.

Balboa students worked:

\[
4\cdot5=20
\]

student-days.

Carver students worked:

\[
5\cdot9=45
\]

student-days.

So the total number of student-days is

\[
21+20+45=86.
\]

The total pay was \$774, so the pay per student per day was

\[
\frac{774}{86}=9.
\]

Balboa students worked 20 student-days, so they earned

\[
20\cdot9=180.
\]

Thus the Balboa students earned \$180.00 altogether.

ANSWER 10: C

Problem 1:
We are asked about the parity of $o^2+no$ where $o$ is odd and $n$ is any whole number.

Factor:
$$o^2+no = o(o+n).$$

$o$ is odd, so $o^2 = o\cdot o$ is odd times odd = odd.

Hence parity is odd + $n\cdot o$.
$n\cdot o$ has the same parity as $n$, because $o$ is odd: if $n$ is even, $n o$ is even; if $n$ is odd, $n o$ is odd.

So:
- if $n$ even: odd + even = odd.
- if $n$ odd: odd + odd = even.

Thus the number is odd iff $n$ is even, and even iff $n$ is odd.

Check choices:
A. always odd - false, e.g. $o=1,n=1$ gives $1+1=2$ even.
B. always even - false, e.g. $o=1,n=2$ gives $1+2=3$ odd.
C. it is even only if $n$ is even - means even $\implies$ $n$ even. False, even happens when $n$ is odd.
D. it is odd only if $n$ is odd - means odd $\implies$ $n$ odd. False, odd happens when $n$ is even.
E. it is odd only if $n$ is even - means odd $\implies$ $n$ even. True by above.

ANSWER 1: E

Problem 2:
Isabella redeems on days $d, d+10, d+20, d+30, d+40, d+50$.
No date is a Sunday.

Modulo 7, $10\equiv 3$. So successive dates advance by 3 weekdays.

Starting from residue $r$, the residues are:
$$r,\ r+3,\ r+6,\ r+9=r+2,\ r+12=r+5,\ r+15=r+1 \pmod 7.$$
These 6 are all distinct, because adding 3 mod 7 cycles through all 7 residues. Continuing one more step would give $r+18=r+4$.

Thus the 6 dates occupy 6 distinct days of the week, missing exactly the 7th residue $r+4$.

Since none of the 6 is Sunday, the missing one must be Sunday. So:
$$r+4 \equiv \text{Sunday}.$$

Hence $r$ is 4 days before Sunday, i.e. Wednesday.

Check: Wed $\to$ Wed+3=Sat $\to$ +3=Tue $\to$ +3=Fri $\to$ +3=Mon $\to$ +3=Thu. That is Wed,Sat,Tue,Fri,Mon,Thu - no Sunday. Any other start would include Sunday.

ANSWER 2: C

Problem 3:
We interpret $3\ 2\cdot 4\ 3\cdots$ as fractions:
$$\frac32 \cdot \frac43 \cdot \frac54 \cdot \frac65 \cdots \frac{a}{b}=9.$$

The pattern is numerator $3,4,5,6,\dots$ and denominator $2,3,4,5,\dots$, i.e. term $k/(k-1)$. The last term is $a/b$, so we must have $b=a-1$ for the pattern to continue.

Then the product telescopes:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}
= \frac{a}{2},$$
because all intermediate $3,4,\dots,a-1$ cancel.

Set equal to 9:
$$\frac{a}{2}=9 \implies a=18.$$
Then $b=a-1=17$.

Sum $a+b=18+17=35$.

This rules out A,B,C,E.

ANSWER 3: D

Problem 4:
Let $x,y,z$ be numbers of 5-cent, 10-cent, 25-cent stamps used. $0\le x,y,z\le 20$ integers.
Value $710$ cents:
$$5x+10y+25z=710.$$
Divide by 5:
$$x+2y+5z=142. \tag{1}$$
Maximize $N=x+y+z$.

From (1), $x=142-2y-5z$, so $0\le x\le20$ gives:
$$122\le 2y+5z\le 142.$$
Also $N=x+y+z=142-y-4z$, so $y+4z=142-N$.

To use many stamps we want small denominations. Try $x=20,y=20$ fails to give integer $z$. Search systematically.

Since $y\le20$, to satisfy $2y+5z\ge122$ we need $z$ large. If $z\le16$, then $2y+5z\le40+80=120<122$, impossible. So $z\ge17$.

Try $z=17,y=19$: then $2y+5z=38+85=123$, so $x=142-123=19$, within $0-20$. This gives:
$$19\cdot5+19\cdot10+17\cdot25=95+190+425=710,$$
with $N=19+19+17=55$ stamps. So 55 is achievable.

Can we do 56 or more? Let $N\ge56$.
Then $y+4z=142-N\le86$ and $x=N-y-z\le20$ gives $y+z\ge N-20$.
Subtract: $3z=(y+4z)-(y+z)\le (142-N)-(N-20)=162-2N$.
So $z\le(162-2N)/3$.
Also $y=142-N-4z\le20$ gives $z\ge(122-N)/4$.

For $N=56$: $z\ge\lceil66/4\rceil=17$ and $z\le\lfloor50/3\rfloor=16$, contradiction. For $N>56$ the interval is even emptier; in general need $(122-N)/4\le(162-2N)/3$, i.e. $5N\le282$, so $N\le56.4$. Thus $N\ge57$ impossible, and $N=56$ impossible by the integer contradiction.

Hence 55 is greatest possible.

ANSWER 4: E

Problem 5:
We have $ABA\times CD = CDCD$, where $A,B,C,D$ are different digits, $A\ne0$, $C\ne0$.

Write $CD$ as the 2-digit number $10C+D$.
$CDCD$ as a 4-digit number is:
$$1000C+100D+10C+D =1010C+101D=101(10C+D)=101\cdot CD.$$

So:
$$ABA \times CD = 101\times CD.$$
Since $CD\ne0$, divide:
$$ABA =101.$$

Thus $A=1$, $B=0$, and $A+B=1$.

This is consistent with distinctness: we can choose e.g. $C=2,D=3$, all of $1,0,2,3$ different, and $101\times23=2323$. So valid.

Other sums are ruled out.

ANSWER 5: A

Problem 6:
Total 100: 52 boys, 48 girls. 40 from Jonas, 60 from Clay.

20 of the girls are from Jonas. So girls from Clay = total girls $-$ girls from Jonas:
$$48-20=28.$$

Clay total 60 = boys from Clay + girls from Clay, so:
$$\text{boys from Clay}=60-28=32.$$

Check: boys from Jonas $=52-32=20$, Jonas total $=20+20=40$. Works.

ANSWER 6: B

Problem 7:
Compute $(1901+\cdots+1993)-(101+\cdots+193)$.

Both sums have $1993-1901+1=93$ terms and $193-101+1=93$ terms.

Pair termwise: for $k=0,\dots,92$,
$$(1901+k)-(101+k)=1800.$$

So difference $=93\times1800$.
$93\times18=1674$, so $93\times1800=167{,}400$.

Other choices come from miscounting terms or arithmetic error.

ANSWER 7: A

Problem 8:
Find tens digit of $7^{2011}$, i.e. $7^{2011}\pmod{100}$.

Compute powers mod 100:
$7^1=7$,
$7^2=49$,
$7^3=343\equiv43$,
$7^4\equiv43\times7=301\equiv1$.

Since $7^4\equiv1\pmod{100}$, the last two digits repeat with period 4.

$2011 = 4\cdot502+3$, so:
$$7^{2011}=(7^4)^{502}\cdot7^3\equiv1^{502}\cdot43\equiv43\pmod{100}.$$

Thus the number ends $\dots43$, tens digit $4$.

ANSWER 8: D

Problem 9:
10 different positive whole numbers have mean 10, so sum $=10\times10=100$.

To make one number as large as possible, make the other nine as small as possible while distinct and positive.

Smallest 9 distinct positives are $1,2,3,4,5,6,7,8,9$, sum $=45$.

Then the tenth number $=100-45=55$, which is distinct from the others and valid. Any larger value would force the other nine to sum to $<45$, impossible with distinct positives.

So largest possible is 55.

ANSWER 9: C

Problem 10:
Count student-days (one student working one day).

Allen: $7$ students $\times$ $3$ days $=21$.
Balboa: $4\times5=20$.
Carver: $5\times9=45$.
Total $=21+20+45=86$ student-days.

Total pay $\$774$ for $86$ student-days at equal rate:
$$\text{rate}=774/86=9 \text{ dollars per student-day}.$$

Bal

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that $o$ is an odd whole number and $n$ is any whole number. We must determine which statement about $o^2 + no$ is always true.

**Step-by-Step Solution:**
1. Factor the expression:
   $$o^2 + no = o(o + n)$$
2. Since $o$ is an odd whole number, the parity of the product $o(o + n)$ is entirely determined by the parity of $(o + n)$:
   - If $n$ is **odd**, then $\text{odd} + \text{odd} = \text{even}$, so $o(o+n)$ is $\text{odd} \times \text{even} = \textbf{even}$.
   - If $n$ is **even**, then $\text{odd} + \text{even} = \text{odd}$, so $o(o+n)$ is $\text{odd} \times \text{odd} = \textbf{odd}$.
3. Therefore, the expression $o^2 + no$ is odd if and only if $n$ is even. 
4. This matches the statement: "it is odd only if $n$ is even."

ANSWER 1: E

---

### Problem 2:
**Restatement:** Isabella redeems 6 coupons, one every 10 days. Pete's is closed on Sundays, and none of the 6 dates falls on a Sunday. We need to find the day of the week she redeems her first coupon.

**Step-by-Step Solution:**
1. There are 7 days in a week. Moving forward by 10 days advances the day of the week by:
   $$10 \equiv 3 \pmod 7$$
2. Let $d$ represent the day of the week of the first coupon ($0 = \text{Sunday}, 1 = \text{Monday}, \dots, 6 = \text{Saturday}$).
3. The days of the week for the 6 coupons will be:
   - Coupon 1: $d$
   - Coupon 2: $d + 3$
   - Coupon 3: $d + 6$
   - Coupon 4: $d + 9 \equiv d + 2 \pmod 7$
   - Coupon 5: $d + 12 \equiv d + 5 \pmod 7$
   - Coupon 6: $d + 15 \equiv d + 1 \pmod 7$
4. The six days of the week visited are $d, d+1, d+2, d+3, d+5, d+6$. Notice that the only day of the week omitted from this list is $d+4$.
5. Since no circled date falls on a Sunday, Sunday must be the omitted day:
   $$d + 4 \equiv 0 \pmod 7 \implies d \equiv 3 \pmod 7$$
6. Day 3 corresponds to **Wednesday**.
   - Checking: Wed $\to$ Sat $\to$ Tue $\to$ Fri $\to$ Mon $\to$ Thu. None of these is Sunday.

ANSWER 2: C

---

### Problem 3:
**Restatement:** We are given the telescoping product $\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdots \frac{a}{b} = 9$. We must find the sum $a + b$.

**Step-by-Step Solution:**
1. Each fraction is of the form $\frac{k+1}{k}$, where the numerator of each term cancels with the denominator of the subsequent term:
   $$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{a}{b} = \frac{a}{2}$$
   where $b = a - 1$.
2. We are given that this product equals 9:
   $$\frac{a}{2} = 9 \implies a = 18$$
3. Since $b = a - 1$, we have $b = 17$.
4. The sum of $a$ and $b$ is:
   $$a + b = 18 + 17 = 35$$

ANSWER 3: D

---

### Problem 4:
**Restatement:** Nicolas has 20 stamps each of 5¢, 10¢, and 25¢. We want to find the greatest number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-Step Solution:**
1. Calculate the total value and count of all stamps:
   - $20 \times 5¢ = 100¢$
   - $20 \times 10¢ = 200¢$
   - $20 \times 25¢ = 500¢$
   - Total value $= 100 + 200 + 500 = 800¢$ using 60 stamps.
2. To use the **maximum** number of stamps to make 710¢, Nicolas must **minimize** the number of stamps removed to subtract:
   $$800¢ - 710¢ = 90¢$$
3. To remove the fewest stamps totaling 90¢, we try to use the largest denomination stamps (25¢):
   - With 3 stamps of 25¢ ($75¢$), the remaining amount needed is $15¢$.
   - Since Nicolas has only 5¢, 10¢, and 25¢ stamps, 15¢ cannot be made with a single stamp. It requires at least two stamps ($10¢ + 5¢$).
   - Thus, 90¢ can be formed with $3 \times 25¢ + 1 \times 10¢ + 1 \times 5¢$, which uses $3 + 1 + 1 = 5$ stamps.
   - Can 90¢ be formed using 4 stamps? The maximum value of 4 stamps less than or equal to 90¢ would be $3 \times 25¢ + 10¢ = 85¢ \ne 90¢$. Thus, 4 stamps is impossible.
4. The minimum number of stamps to remove is 5.
5. Therefore, the maximum number of stamps used is:
   $$60 - 5 = 55$$

ANSWER 4: E

---

### Problem 5:
**Restatement:** In the multiplication $ABA \times CD = CDCD$, $A, B, C, D$ are distinct digits. We need to find $A + B$.

**Step-by-Step Solution:**
1. Express $CDCD$ in terms of the two-digit number $CD$:
   $$CDCD = CD \times 100 + CD = CD \times 101$$
2. Substitute this into the equation:
   $$ABA \times CD = 101 \times CD$$
3. Since $CD$ is a non-zero two-digit number, divide both sides by $CD$:
   $$ABA = 101$$
4. Comparing digits, we have $A = 1$ and $B = 0$.
5. Thus:
   $$A + B = 1 + 0 = 1$$

ANSWER 5: A

---

### Problem 6:
**Restatement:** Out of 100 students at a camp, 52 are boys and 48 are girls. Also, 40 are from Jonas Middle School and 60 from Clay Middle School. Given that 20 of the girls are from Jonas, find how many boys are from Clay.

**Step-by-Step Solution:**
1. Find the number of girls from Clay Middle School:
   $$\text{Total girls} - \text{Girls from Jonas} = 48 - 20 = 28\text{ girls from Clay}$$
2. Find the number of boys from Clay Middle School:
   $$\text{Total students from Clay} - \text{Girls from Clay} = 60 - 28 = 32\text{ boys from Clay}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** Compute the value of:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-Step Solution:**
1. Pair each term in the first sum with the corresponding term in the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
2. Each paired difference is equal to:
   $$1900 + k - (100 + k) = 1800$$
3. Find the number of terms: from $k = 1$ to $k = 93$, there are:
   $$93 - 1 + 1 = 93\text{ terms}$$
4. Multiply the difference by the number of terms:
   $$93 \times 1800 = 93 \times 18 \times 100 = 1674 \times 100 = 167,400$$

ANSWER 7: A

---

### Problem 8:
**Restatement:** Find the tens digit of $7^{2011}$.

**Step-by-Step Solution:**
1. Find the pattern of the last two digits of powers of 7 by computing $7^n \pmod{100}$:
   - $7^1 = 07$
   - $7^2 = 49$
   - $7^3 = 343 \equiv 43 \pmod{100}$
   - $7^4 = 43

# Problem 1:
We must decide when $o^2 + no$ is odd or even, given $o$ is odd and $n$ is any whole number.

**Key facts:** odd × odd = odd, so $o^2$ is **odd**. Since $o$ is odd, multiplying by $o$ does not change $n$'s parity, so $no$ has the **same parity as $n$**.

**Case 1: $n$ even.** Then $no$ is even, so $o^2 + no = \text{odd} + \text{even} = \textbf{odd}$.
**Case 2: $n$ odd.** Then $no$ is odd, so $o^2 + no = \text{odd} + \text{odd} = \textbf{even}$.

So the number is odd exactly when $n$ is even. In particular, **if the number is odd, $n$ must be even**.

Ruling out traps: (A) fails since $o=1, n=1$ gives $1+1=2$, even; (B) fails since $n=2$ gives $1+2=3$, odd; (C) and (D) state the exact opposite of the truth. Only (E) matches.

ANSWER 1: E

# Problem 2:
The 6 redemption dates are 10 days apart, so their days of the week shift by $10 \bmod 7 = 3$ each time.

If the first date falls on weekday $W$, the six dates fall on (mod 7):
$$W,\; W+3,\; W+6,\; W+2,\; W+5,\; W+1$$
That is **six of the seven weekdays**, and the only weekday missed is $W+4$.

Since no circled date is a Sunday, the one missed weekday must be Sunday:
$$W + 4 = \text{Sunday} \implies W = \text{Wednesday}.$$

**Check:** Starting Wednesday, the dates fall on Wed, Sat, Tue, Fri, Mon, Thu — no Sunday. (Trap: starting Thursday gives Thu, Sun, ... — fails immediately.)

ANSWER 2: C

# Problem 3:
The product is a chain of fractions (the bars were lost in formatting):
$$\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdots\frac{a}{b} = 9.$$

Each numerator cancels the denominator before it (e.g., $\frac{3}{2}\cdot\frac{4}{3} = \frac{4}{2}$), and each fraction has the form $\frac{k+1}{k}$, so $b = a-1$. Everything telescopes:
$$\text{Product} = \frac{a}{2} = 9 \implies a = 18, \quad b = 17.$$
Then $a + b = 18 + 17 = 35$.

Trap check: $a+b=17$ would mean $a=9$, giving product $\tfrac92$, not $9$; only $35$ fits the pattern.

ANSWER 3: D

# Problem 4:
Work in units of 5¢: we need total value $710¢ = 142$ units, using stamps worth $1, 2, 5$ units (i.e., 5¢, 10¢, 25¢), at most 20 of each. To **maximize the count**, use as many cheap stamps as possible.

**Building a 55-stamp solution:** Take nineteen 1's and nineteen 2's: value $19 + 38 = 57$ units with 38 stamps. Remaining value $142 - 57 = 85 = 17\times 5$: seventeen 5's.
$$19\times 5¢ + 19\times 10¢ + 17\times 25¢ = 95 + 190 + 425 = 710¢ = \$7.10.\; ✓$$
That's $19+19+17 = 55$ stamps, all counts $\le 20$. ✓

**Why 56 is impossible:** with 56 stamps, at least $56 - 40 = 16$ must be 25¢ stamps (only 40 cheap ones exist). Let $z$ = number of 25¢ stamps:
- $z=16$: max value $= 100 + 200 + 400 = 700 < 710$. ✗
- $z=17$: need $5x+10y=285$, $x+y=39 \Rightarrow x=21>20$. ✗
- $z=18$: $x=24>20$. ✗  $z=19$: $x=27>20$. ✗  $z=20$: $x=30>20$. ✗

Traps: 54 comes from $20×5+16×10+18×25 = 710$; 51 from $20×5+11×10+20×25 = 710$ — both valid but not maximal.

ANSWER 4: E

# Problem 5:
Write $CDCD$ as a number:
$$CDCD = 1000C + 100D + 10C + D = 101(10C + D) = 101 \times CD.$$
The given equation is
$$ABA \times CD = CDCD = 101 \times CD.$$
Since $CD \neq 0$ (it's a two-digit number), we can divide both sides by $CD$:
$$ABA = 101.$$
So $A = 1$ and $B = 0$, giving $A + B = 1$.

**Consistency check:** $C, D$ just need to be distinct from $0, 1$ and each other, e.g., $101 \times 23 = 2323$. ✓ (Trap: guessing digits without using the factorization 101 wastes time or gives 9.)

ANSWER 5: A

# Problem 6:
Set up a two-way table with 100 total students.

- Girls: 48 total, and 20 girls are from Jonas $\Rightarrow$ girls from Clay $= 48 - 20 = 28$.
- Jonas has 40 students, of whom 20 are girls $\Rightarrow$ boys from Jonas $= 40 - 20 = 20$.
- Boys: 52 total $\Rightarrow$ boys from Clay $= 52 - 20 = \textbf{32}$.

**Check:** Jonas $20+20=40$ ✓; Clay $28+32=60$ ✓; boys $20+32=52$ ✓; girls $20+28=48$ ✓; total $100$ ✓. (Trap: 20 counts the boys from Jonas, not Clay.)

ANSWER 6: B

# Problem 7:
Both sums have the same number of terms: $1901$ to $1993$ has $1993-1901+1 = 93$ terms, and $101$ to $193$ has $193-101+1 = 93$ terms.

Pair them up: each term in the first sum is exactly $1800$ more than its partner ($1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$). So the difference is
$$93 \times 1800 = 167{,}400.$$

**Check via formulas:** $\frac{93(1901+1993)}{2} - \frac{93(101+193)}{2} = 181{,}071 - 13{,}671 = 167{,}400$. ✓ (Trap: 181,071 is the first sum alone — forgetting to subtract.)

ANSWER 7: A

# Problem 8:
Only the tens digit matters, so track the last two digits of $7^{k}$, which cycle with period 4:
$$7^1 = 07,\quad 7^2 = 49,\quad 7^3 = 343 \to 43,\quad 7^4 = 2401 \to 01,\quad 7^5 \to 07, \ldots$$

Find where $2011$ falls in the cycle:
$$2011 = 4 \times 502 + 3,$$
so $7^{2011}$ ends in the same two digits as $7^3$, namely $\ldots 43$.

The tens digit is $\textbf{4}$. (Trap: $7^{2012}$, a multiple of 4, ends in $01$, tens digit 0 — choice A; that would require the exponent to be divisible by 4.)

ANSWER 8: D

# Problem 9:
The ten **different** positive whole numbers have average 10, so their sum is $10 \times 10 = 100$.

To make one number as large as possible, make the other nine as small as possible. Since they must be **distinct positive** whole numbers, the smallest possible set is $1, 2, 3, \ldots, 9$, with sum
$$1+2+\cdots+9 = 45.$$
So the largest possible value is
$$100 - 45 = 55.$$

**Check:** $1,2,\ldots,9,55$ are ten different positive whole numbers with sum $100$ and average $10$. ✓ (Trap: 91 arises if you ignore "different" and use nine 1's; but the numbers must be distinct.)

ANSWER 9: C

# Problem 10:
Count student-days (the unit being paid):
- Allen: $7 \times 3 = 21$ student-days
- Balboa: $4 \times 5 = 20$ student-days
- Carver: $5 \times 9 = 45$ student-days

Total: $21 + 20 + 45 = 86$ student-days, paid $\$774$ altogether. So the daily rate is
$$774 \div 86 = 9 \text{ dollars per student-day}.$$
Balboa's students worked 20 student-days, so they earned
$$20 \times 9 = 18

Problem 1:  
We need the parity of \(o^2+no\), where \(o\) is odd and \(n\) is any whole number.

Factor:
\[
o^2+no=o(o+n).
\]

Since \(o\) is odd:

- If \(n\) is even, then \(o+n\) is odd, so \(o(o+n)\) is odd.
- If \(n\) is odd, then \(o+n\) is even, so \(o(o+n)\) is even.

Thus the expression is odd exactly when \(n\) is even.

Check the choices:
- A and B are false because the parity depends on \(n\).
- C says “even only if \(n\) is even,” but when the expression is even, \(n\) is actually odd.
- D says “odd only if \(n\) is odd,” but when the expression is odd, \(n\) is even.
- E says “odd only if \(n\) is even,” which is true.

ANSWER 1: E

Problem 2:  
Isabella redeems coupons 10 days apart. Since
\[
10\equiv 3 \pmod 7,
\]
the six circled days are the first day plus
\[
0,3,6,9,12,15
\]
days, which modulo 7 gives offsets
\[
0,3,6,2,5,1.
\]

So the six circled days are all weekdays except the one 4 days after the first day. Since none of the circled days is a Sunday, Sunday must be the un-circled day: 4 days after the first day.

Therefore the first coupon is 4 days before Sunday, which is Wednesday.

Check: Wednesday, Saturday, Tuesday, Friday, Monday, Thursday — no Sunday.

ANSWER 2: C

Problem 3:  
The product is the telescoping product of fractions:
\[
\frac32\cdot \frac43\cdot \frac54\cdot \frac65\cdots \frac{a}{b}=9.
\]

The numerators cancel the next denominators:
\[
\frac32\cdot \frac43=\frac42,
\]
\[
\frac42\cdot \frac54=\frac52,
\]
and continuing, the whole product simplifies to
\[
\frac{a}{2}.
\]

So
\[
\frac{a}{2}=9 \implies a=18.
\]

By the pattern, the denominator is one less than the numerator, so
\[
b=17.
\]

Thus
\[
a+b=18+17=35.
\]

ANSWER 3: D

Problem 4:  
Nicolas has 20 of each type of stamp, so he has
\[
20+20+20=60
\]
stamps total. Their total value is
\[
20(5+10+25)=20(40)=800
\]
cents.

He wants postage of
\[
\$7.10=710
\]
cents, so the stamps he does **not** use must total
\[
800-710=90
\]
cents.

To maximize the number of stamps he uses, we minimize the number of stamps he leaves unused.

Can 4 unused stamps total 90 cents? Let \(x,y,z\) be the numbers of 25-, 10-, and 5-cent unused stamps. Then
\[
x+y+z=4
\]
and
\[
25x+10y+5z=90.
\]

Divide by 5:
\[
5x+2y+z=18.
\]

Subtract \(x+y+z=4\):
\[
4x+y=14.
\]

Testing possible \(x=0,1,2,3,4\):

- \(x=0\Rightarrow y=14\), impossible.
- \(x=1\Rightarrow y=10\), impossible.
- \(x=2\Rightarrow y=6\), impossible.
- \(x=3\Rightarrow y=2\), then \(z=4-3-2=-1\), impossible.
- \(x=4\Rightarrow y=-2\), impossible.

So 4 unused stamps cannot total 90 cents.

Can 5 unused stamps total 90 cents? Yes:
\[
25+25+25+10+5=90.
\]

Then he uses
\[
60-5=55
\]
stamps. For example, he could use 19 five-cent stamps, 19 ten-cent stamps, and 17 twenty-five-cent stamps:
\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So the greatest possible number of stamps is 55. Choice D, 54, is possible but not greatest.

ANSWER 4: E

Problem 5:  
We are told
\[
ABA\times CD=CDCD,
\]
where \(A,B,C,D\) are different digits.

Write the numbers in place-value form:
\[
ABA=101A+10B,
\]
\[
CD=10C+D,
\]
\[
CDCD=1000C+100D+10C+D=101(10C+D).
\]

So
\[
(101A+10B)(10C+D)=101(10C+D).
\]

Since \(CD\) is a two-digit number, \(10C+D\neq 0\), so we can divide both sides by \(10C+D\):
\[
101A+10B=101.
\]

The only digit solution is
\[
A=1,\quad B=0.
\]

Therefore
\[
A+B=1+0=1.
\]

ANSWER 5: A

Problem 6:  
There are 48 girls total, and 20 of the girls are from Jonas Middle School. So the number of girls from Clay Middle School is
\[
48-20=28.
\]

There are 60 students total from Clay Middle School, so the number of boys from Clay is
\[
60-28=32.
\]

ANSWER 6: B

Problem 7:  
Pair the terms:
\[
(1901+1902+\cdots+1993)-(101+102+\cdots+193)
\]
\[
=(1901-101)+(1902-102)+\cdots+(1993-193).
\]

Each pair is
\[
1800.
\]

Now count the number of terms:
\[
193-101+1=93.
\]

So the value is
\[
93\cdot 1800=167400.
\]

ANSWER 7: A

Problem 8:  
We need the tens digit of \(7^{2011}\), so we need the last two digits.

Powers of 7 modulo 100 cycle:
\[
7^1=07,
\]
\[
7^2=49,
\]
\[
7^3=343\equiv 43,
\]
\[
7^4=2401\equiv 01.
\]

Then the cycle repeats:
\[
07,\ 49,\ 43,\ 01.
\]

Since
\[
2011\equiv 3 \pmod 4,
\]
we have
\[
7^{2011}\equiv 7^3\equiv 43 \pmod{100}.
\]

The last two digits are 43, so the tens digit is 4.

ANSWER 8: D

Problem 9:  
The average of 10 different positive whole numbers is 10, so their sum is
\[
10\cdot 10=100.
\]

To make one number as large as possible, make the other 9 numbers as small as possible while still being different positive whole numbers. The smallest 9 distinct positive whole numbers are
\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is
\[
1+2+\cdots+9=45.
\]

So the largest number can be
\[
100-45=55.
\]

This is possible: \(1,2,3,4,5,6,7,8,9,55\) are all different positive whole numbers.

ANSWER 9: C

Problem 10:  
First count total student-days worked:

- Allen: \(7\cdot 3=21\) student-days
- Balboa: \(4\cdot 5=20\) student-days
- Carver: \(5\cdot 9=45\) student-days

Total:
\[
21+20+45=86
\]
student-days.

The total payment is \$774, so the daily rate is
\[
774\div 86=9
\]
dollars per student-day.

Balboa students worked 20 student-days, so they earned
\[
20\cdot 9=180
\]
dollars.

ANSWER 10: C

# Problem 1:
We need a statement always true about o² + no, where o is odd.

Factor: o² + no = o(o + n).

Since o is odd, the parity of the product is the parity of (o + n).
- If n is even, o + n is odd → product = odd·odd = **odd**.
- If n is odd, o + n is even → product = odd·even = **even**.

So the expression is odd exactly when n is even.

Ruling out others: A fails (n odd gives even), B fails (n even gives odd), C says "even only if n is even" — but it's even when n is *odd*, false. D says "odd only if n is odd" — false, it's odd when n is even. E: "odd only if n is even" — true.

ANSWER 1: E

# Problem 2:
The redemption days are 0, 10, 20, 30, 40, 50 days after the first.

Reduce mod 7 (days of week shift): 0, 3, 6, 2, 5, 1.

These residues are {0, 1, 2, 3, 5, 6} — every residue except **4**. So the only day of the week never hit is the day 4 days after the starting day. For no coupon to land on Sunday, Sunday must be that missed day:

start + 4 ≡ Sunday.

Counting back 4 days from Sunday: Saturday, Friday, Thursday, **Wednesday**.

Check: Wed → +10 Sat → +20 Tue → +30 Fri → +40 Mon → +50 Thu. No Sundays. ✓

ANSWER 2: C

# Problem 3:
The product is (3/2)·(4/3)·(5/4)·(6/5)·…·(a/b), where each numerator is one more than its denominator, so b = a − 1.

Telescoping: everything cancels except the last numerator and first denominator:
a/2 = 9 → a = 18, so b = 17.

a + b = 18 + 17 = 35.

(Choice E, 37, would come from mistakenly taking a = 19.)

ANSWER 3: D

# Problem 4:
Total value of all 60 stamps: 20(5 + 10 + 25) = 20·40 = 800 cents.

We need 710 cents, so we must *remove* stamps worth exactly 800 − 710 = 90 cents, using as **few** stamps as possible (to keep the most stamps).

Can 90 be made with 4 or fewer stamps? Max with 4 stamps is 100; possible 4-stamp totals near 90: 100, 85, 80, 75, 70… none equal 90. Three or fewer is at most 75. So 4 or fewer is impossible.

With 5 stamps: 25 + 25 + 25 + 10 + 5 = 90 ✓

So remove 5 stamps: 60 − 5 = 55 stamps.

Check: 17(25) + 19(10) + 19(5) = 425 + 190 + 95 = 710 cents, using 17 + 19 + 19 = 55 stamps.

ANSWER 4: E

# Problem 5:
CDCD as a number = CD·100 + CD = CD·101.

So ABA × CD = 101 × CD, and since CD ≠ 0, we get ABA = 101.

Thus A = 1, B = 0 (and indeed A, B, C, D distinct works, e.g. 101 × 23 = 2323).

A + B = 1 + 0 = 1.

ANSWER 5: A

# Problem 6:
Jonas total = 40, and 20 of those were girls, so boys from Jonas = 40 − 20 = 20.

Total boys = 52, so boys from Clay = 52 − 20 = 32.

(Check: girls from Clay = 48 − 20 = 28; 32 + 28 = 60 = Clay total ✓.)

Choice C (40) is the Jonas total, and E (52) is all boys — traps.

ANSWER 6: B

# Problem 7:
Pair each term: 1901 − 101 = 1800, 1902 − 102 = 1800, …, each pair differs by 1800.

Number of terms: 1993 − 1901 + 1 = 93.

Total = 93 × 1800 = 167,400.

ANSWER 7: A

# Problem 8:
Look at 7ⁿ mod 100:
- 7¹ = 07
- 7² = 49
- 7³ = 343 → 43
- 7⁴ = 2401 → 01
- then the pattern repeats with period 4.

2011 ÷ 4 = 502 remainder 3, so 7²⁰¹¹ ≡ 7³ ≡ 43 (mod 100).

Last two digits are 43, so the tens digit is 4.

ANSWER 8: D

# Problem 9:
Sum of the 10 numbers = 10 × 10 = 100.

To make one number as large as possible, make the other nine as small as possible while distinct and positive: 1 + 2 + … + 9 = 45.

Largest = 100 − 45 = 55.

(90 or 91 would require repeats or a zero, which isn't allowed.)

ANSWER 9: C

# Problem 10:
Total student-days:
- Allen: 7 × 3 = 21
- Balboa: 4 × 5 = 20
- Carver: 5 × 9 = 45

Total = 86 student-days.

Daily wage = 774 ÷ 86 = $9.00 per student-day.

Balboa earnings = 20 × 9 = $180.00.

(A is one day's pay for one student; E is Carver's total.)

ANSWER 10: C

Problem 1:

We need determine the parity of

\[
o^2+no.
\]

Factor out \(o\):

\[
o^2+no=o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on \(o+n\).

- If \(n\) is even, then odd \(+\) even \(=\) odd, so \(o+n\) is odd. Thus odd \(\cdot\) odd is odd.
- If \(n\) is odd, then odd \(+\) odd \(=\) even, so the product is even.

Therefore, the expression is odd only when \(n\) is even.

- A is wrong because the expression is not always odd.
- B is wrong because it is odd when \(n\) is even.
- C reverses the condition for being even.
- D reverses the condition for being odd.

ANSWER 1: E

Problem 2:

Isabella uses a coupon every 10 days. Since

\[
10\equiv 3\pmod 7,
\]

each coupon date moves forward by 3 days of the week.

If the first coupon is redeemed on day \(d\), the six redemption days are:

\[
d,\ d+3,\ d+6,\ d+2,\ d+5,\ d+1.
\]

These are six different days of the week. The only day missing is

\[
d+4.
\]

Since none of the six coupon dates is Sunday, the missing day must be Sunday. Thus,

\[
d+4=\text{Sunday}.
\]

Four days before Sunday is Wednesday.

Indeed, beginning on Wednesday gives:

\[
\text{Wednesday, Saturday, Tuesday, Friday, Monday, Thursday}.
\]

Sunday is the only missing day.

ANSWER 2: C

Problem 3:

The product is

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9.
\]

Almost every numerator cancels with a denominator:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab
=\frac a2.
\]

So

\[
\frac a2=9.
\]

Therefore,

\[
a=18.
\]

The final fraction must be

\[
\frac{18}{17},
\]

so \(b=17\). Therefore,

\[
a+b=18+17=35.
\]

Choices 11, 13, and 17 are too small, while 37 would correspond to a different ending fraction.

ANSWER 3: D

Problem 4:

Let

- \(x\) = number of 5-cent stamps,
- \(y\) = number of 10-cent stamps,
- \(z\) = number of 25-cent stamps.

In units of 5 cents, the postage total is

\[
\frac{710}{5}=142.
\]

Thus,

\[
x+2y+5z=142.
\]

We want to maximize

\[
x+y+z.
\]

Since there are at most 20 five-cent stamps, \(x\leq 20\). Therefore,

\[
2y+5z=142-x\geq 142-20=122.
\]

Since \(y\leq20\), if \(z\leq16\), then

\[
2y+5z\leq 2(20)+5(16)=40+80=120,
\]

which is not enough. So \(z\geq17\).

If \(z=17\), then

\[
2y+5(17)\geq122,
\]

so

\[
2y+85\geq122,
\]

\[
2y\geq37,
\]

and hence \(y\geq19\).

Now use

\[
x+y+z=142-y-4z.
\]

Since \(y\geq19\) and \(z\geq17\),

\[
x+y+z\leq142-19-4(17)
=142-19-68
=55.
\]

This maximum is attainable with

\[
x=19,\qquad y=19,\qquad z=17.
\]

Check:

\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So 55 stamps can be used. Choices 45, 46, 51, and 54 are possible totals to consider, but 55 is the greatest possible.

ANSWER 4: E

Problem 5:

We are told

\[
ABA\times CD=CDCD.
\]

The number \(ABA\) is

\[
100A+10B+A=101A+10B.
\]

The number \(CDCD\) is

\[
1000C+100D+10C+D.
\]

Group the \(C\)'s and \(D\)'s:

\[
CDCD=1010C+101D=101(10C+D)=101\cdot CD.
\]

Thus the equation becomes

\[
ABA\cdot CD=101\cdot CD.
\]

Since \(CD\) is a two-digit number and is not zero, divide both sides by \(CD\):

\[
ABA=101.
\]

Therefore,

\[
A=1,\qquad B=0.
\]

So

\[
A+B=1+0=1.
\]

For example, \(101\times23=2323\), and the digits \(1,0,2,3\) are all different.

ANSWER 5: A

Problem 6:

There are 40 students from Jonas Middle School, and 20 of those are girls.

Therefore the number of boys from Jonas is

\[
40-20=20.
\]

There are 52 boys altogether. Therefore the number of boys from Clay Middle School is

\[
52-20=32.
\]

The other choices do not correctly account for the 20 Jonas girls and the total of 52 boys.

ANSWER 6: B

Problem 7:

There are

\[
1993-1901+1=93
\]

terms in the first sum.

There are also

\[
193-101+1=93
\]

terms in the second sum.

Pair corresponding terms:

\[
1901-101=1800,
\]

\[
1902-102=1800,
\]

and so on, through

\[
1993-193=1800.
\]

Thus there are 93 differences of 1800:

\[
93\cdot1800=167400.
\]

Therefore the value is

\[
167,\!400.
\]

ANSWER 7: A

Problem 8:

We need the tens digit of

\[
7^{2011}.
\]

It is enough to examine powers of 7 modulo 100:

\[
7^1=7,
\]

\[
7^2=49,
\]

\[
7^3=343,
\]

whose last two digits are 43, and

\[
7^4=2401,
\]

whose last two digits are 01.

Thus the last two digits repeat every 4 powers because

\[
7^4\equiv1\pmod{100}.
\]

Now,

\[
2011\div4
\]

leaves remainder 3, since

\[
2011=4(502)+3.
\]

Therefore,

\[
7^{2011}
\]

has the same last two digits as

\[
7^3=343,
\]

which are 43. The tens digit is 4.

ANSWER 8: D

Problem 9:

The average of 10 different positive whole numbers is 10, so their total sum is

\[
10\cdot10=100.
\]

To make one number as large as possible, make the other nine numbers as small as possible.

The nine smallest different positive whole numbers are

\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is

\[
1+2+3+4+5+6+7+8+9=45.
\]

So the largest possible tenth number is

\[
100-45=55.
\]

This works because the ten numbers can be

\[
1,2,3,4,5,6,7,8,9,55,
\]

which are all different and have total 100.

A value such as 90 or 91 would leave too small a total for nine distinct positive whole numbers.

ANSWER 9: C

Problem 10:

Compute the total number of student-days worked.

Allen students worked:

\[
7\cdot3=21
\]

student-days.

Balboa students worked:

\[
4\cdot5=20
\]

student-days.

Carver students worked:

\[
5\cdot9=45
\]

student-days.

So the total number of student-days is

\[
21+20+45=86.
\]

The total pay was \$774, so the pay per student per day was

\[
\frac{774}{86}=9.
\]

Balboa students worked 20 student-days, so they earned

\[
20\cdot9=180.
\]

Thus the Balboa students earned \$180.00 altogether.

ANSWER 10: C

Problem 1:
We are asked about the parity of $o^2+no$ where $o$ is odd and $n$ is any whole number.

Factor:
$$o^2+no = o(o+n).$$

$o$ is odd, so $o^2 = o\cdot o$ is odd times odd = odd.

Hence parity is odd + $n\cdot o$.
$n\cdot o$ has the same parity as $n$, because $o$ is odd: if $n$ is even, $n o$ is even; if $n$ is odd, $n o$ is odd.

So:
- if $n$ even: odd + even = odd.
- if $n$ odd: odd + odd = even.

Thus the number is odd iff $n$ is even, and even iff $n$ is odd.

Check choices:
A. always odd - false, e.g. $o=1,n=1$ gives $1+1=2$ even.
B. always even - false, e.g. $o=1,n=2$ gives $1+2=3$ odd.
C. it is even only if $n$ is even - means even $\implies$ $n$ even. False, even happens when $n$ is odd.
D. it is odd only if $n$ is odd - means odd $\implies$ $n$ odd. False, odd happens when $n$ is even.
E. it is odd only if $n$ is even - means odd $\implies$ $n$ even. True by above.

ANSWER 1: E

Problem 2:
Isabella redeems on days $d, d+10, d+20, d+30, d+40, d+50$.
No date is a Sunday.

Modulo 7, $10\equiv 3$. So successive dates advance by 3 weekdays.

Starting from residue $r$, the residues are:
$$r,\ r+3,\ r+6,\ r+9=r+2,\ r+12=r+5,\ r+15=r+1 \pmod 7.$$
These 6 are all distinct, because adding 3 mod 7 cycles through all 7 residues. Continuing one more step would give $r+18=r+4$.

Thus the 6 dates occupy 6 distinct days of the week, missing exactly the 7th residue $r+4$.

Since none of the 6 is Sunday, the missing one must be Sunday. So:
$$r+4 \equiv \text{Sunday}.$$

Hence $r$ is 4 days before Sunday, i.e. Wednesday.

Check: Wed $\to$ Wed+3=Sat $\to$ +3=Tue $\to$ +3=Fri $\to$ +3=Mon $\to$ +3=Thu. That is Wed,Sat,Tue,Fri,Mon,Thu - no Sunday. Any other start would include Sunday.

ANSWER 2: C

Problem 3:
We interpret $3\ 2\cdot 4\ 3\cdots$ as fractions:
$$\frac32 \cdot \frac43 \cdot \frac54 \cdot \frac65 \cdots \frac{a}{b}=9.$$

The pattern is numerator $3,4,5,6,\dots$ and denominator $2,3,4,5,\dots$, i.e. term $k/(k-1)$. The last term is $a/b$, so we must have $b=a-1$ for the pattern to continue.

Then the product telescopes:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}
= \frac{a}{2},$$
because all intermediate $3,4,\dots,a-1$ cancel.

Set equal to 9:
$$\frac{a}{2}=9 \implies a=18.$$
Then $b=a-1=17$.

Sum $a+b=18+17=35$.

This rules out A,B,C,E.

ANSWER 3: D

Problem 4:
Let $x,y,z$ be numbers of 5-cent, 10-cent, 25-cent stamps used. $0\le x,y,z\le 20$ integers.
Value $710$ cents:
$$5x+10y+25z=710.$$
Divide by 5:
$$x+2y+5z=142. \tag{1}$$
Maximize $N=x+y+z$.

From (1), $x=142-2y-5z$, so $0\le x\le20$ gives:
$$122\le 2y+5z\le 142.$$
Also $N=x+y+z=142-y-4z$, so $y+4z=142-N$.

To use many stamps we want small denominations. Try $x=20,y=20$ fails to give integer $z$. Search systematically.

Since $y\le20$, to satisfy $2y+5z\ge122$ we need $z$ large. If $z\le16$, then $2y+5z\le40+80=120<122$, impossible. So $z\ge17$.

Try $z=17,y=19$: then $2y+5z=38+85=123$, so $x=142-123=19$, within $0-20$. This gives:
$$19\cdot5+19\cdot10+17\cdot25=95+190+425=710,$$
with $N=19+19+17=55$ stamps. So 55 is achievable.

Can we do 56 or more? Let $N\ge56$.
Then $y+4z=142-N\le86$ and $x=N-y-z\le20$ gives $y+z\ge N-20$.
Subtract: $3z=(y+4z)-(y+z)\le (142-N)-(N-20)=162-2N$.
So $z\le(162-2N)/3$.
Also $y=142-N-4z\le20$ gives $z\ge(122-N)/4$.

For $N=56$: $z\ge\lceil66/4\rceil=17$ and $z\le\lfloor50/3\rfloor=16$, contradiction. For $N>56$ the interval is even emptier; in general need $(122-N)/4\le(162-2N)/3$, i.e. $5N\le282$, so $N\le56.4$. Thus $N\ge57$ impossible, and $N=56$ impossible by the integer contradiction.

Hence 55 is greatest possible.

ANSWER 4: E

Problem 5:
We have $ABA\times CD = CDCD$, where $A,B,C,D$ are different digits, $A\ne0$, $C\ne0$.

Write $CD$ as the 2-digit number $10C+D$.
$CDCD$ as a 4-digit number is:
$$1000C+100D+10C+D =1010C+101D=101(10C+D)=101\cdot CD.$$

So:
$$ABA \times CD = 101\times CD.$$
Since $CD\ne0$, divide:
$$ABA =101.$$

Thus $A=1$, $B=0$, and $A+B=1$.

This is consistent with distinctness: we can choose e.g. $C=2,D=3$, all of $1,0,2,3$ different, and $101\times23=2323$. So valid.

Other sums are ruled out.

ANSWER 5: A

Problem 6:
Total 100: 52 boys, 48 girls. 40 from Jonas, 60 from Clay.

20 of the girls are from Jonas. So girls from Clay = total girls $-$ girls from Jonas:
$$48-20=28.$$

Clay total 60 = boys from Clay + girls from Clay, so:
$$\text{boys from Clay}=60-28=32.$$

Check: boys from Jonas $=52-32=20$, Jonas total $=20+20=40$. Works.

ANSWER 6: B

Problem 7:
Compute $(1901+\cdots+1993)-(101+\cdots+193)$.

Both sums have $1993-1901+1=93$ terms and $193-101+1=93$ terms.

Pair termwise: for $k=0,\dots,92$,
$$(1901+k)-(101+k)=1800.$$

So difference $=93\times1800$.
$93\times18=1674$, so $93\times1800=167{,}400$.

Other choices come from miscounting terms or arithmetic error.

ANSWER 7: A

Problem 8:
Find tens digit of $7^{2011}$, i.e. $7^{2011}\pmod{100}$.

Compute powers mod 100:
$7^1=7$,
$7^2=49$,
$7^3=343\equiv43$,
$7^4\equiv43\times7=301\equiv1$.

Since $7^4\equiv1\pmod{100}$, the last two digits repeat with period 4.

$2011 = 4\cdot502+3$, so:
$$7^{2011}=(7^4)^{502}\cdot7^3\equiv1^{502}\cdot43\equiv43\pmod{100}.$$

Thus the number ends $\dots43$, tens digit $4$.

ANSWER 8: D

Problem 9:
10 different positive whole numbers have mean 10, so sum $=10\times10=100$.

To make one number as large as possible, make the other nine as small as possible while distinct and positive.

Smallest 9 distinct positives are $1,2,3,4,5,6,7,8,9$, sum $=45$.

Then the tenth number $=100-45=55$, which is distinct from the others and valid. Any larger value would force the other nine to sum to $<45$, impossible with distinct positives.

So largest possible is 55.

ANSWER 9: C

Problem 10:
Count student-days (one student working one day).

Allen: $7$ students $\times$ $3$ days $=21$.
Balboa: $4\times5=20$.
Carver: $5\times9=45$.
Total $=21+20+45=86$ student-days.

Total pay $\$774$ for $86$ student-days at equal rate:
$$\text{rate}=774/86=9 \text{ dollars per student-day}.$$

Bal

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that $o$ is an odd whole number and $n$ is any whole number. We must determine which statement about $o^2 + no$ is always true.

**Step-by-Step Solution:**
1. Factor the expression:
   $$o^2 + no = o(o + n)$$
2. Since $o$ is an odd whole number, the parity of the product $o(o + n)$ is entirely determined by the parity of $(o + n)$:
   - If $n$ is **odd**, then $\text{odd} + \text{odd} = \text{even}$, so $o(o+n)$ is $\text{odd} \times \text{even} = \textbf{even}$.
   - If $n$ is **even**, then $\text{odd} + \text{even} = \text{odd}$, so $o(o+n)$ is $\text{odd} \times \text{odd} = \textbf{odd}$.
3. Therefore, the expression $o^2 + no$ is odd if and only if $n$ is even. 
4. This matches the statement: "it is odd only if $n$ is even."

ANSWER 1: E

---

### Problem 2:
**Restatement:** Isabella redeems 6 coupons, one every 10 days. Pete's is closed on Sundays, and none of the 6 dates falls on a Sunday. We need to find the day of the week she redeems her first coupon.

**Step-by-Step Solution:**
1. There are 7 days in a week. Moving forward by 10 days advances the day of the week by:
   $$10 \equiv 3 \pmod 7$$
2. Let $d$ represent the day of the week of the first coupon ($0 = \text{Sunday}, 1 = \text{Monday}, \dots, 6 = \text{Saturday}$).
3. The days of the week for the 6 coupons will be:
   - Coupon 1: $d$
   - Coupon 2: $d + 3$
   - Coupon 3: $d + 6$
   - Coupon 4: $d + 9 \equiv d + 2 \pmod 7$
   - Coupon 5: $d + 12 \equiv d + 5 \pmod 7$
   - Coupon 6: $d + 15 \equiv d + 1 \pmod 7$
4. The six days of the week visited are $d, d+1, d+2, d+3, d+5, d+6$. Notice that the only day of the week omitted from this list is $d+4$.
5. Since no circled date falls on a Sunday, Sunday must be the omitted day:
   $$d + 4 \equiv 0 \pmod 7 \implies d \equiv 3 \pmod 7$$
6. Day 3 corresponds to **Wednesday**.
   - Checking: Wed $\to$ Sat $\to$ Tue $\to$ Fri $\to$ Mon $\to$ Thu. None of these is Sunday.

ANSWER 2: C

---

### Problem 3:
**Restatement:** We are given the telescoping product $\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdots \frac{a}{b} = 9$. We must find the sum $a + b$.

**Step-by-Step Solution:**
1. Each fraction is of the form $\frac{k+1}{k}$, where the numerator of each term cancels with the denominator of the subsequent term:
   $$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{a}{b} = \frac{a}{2}$$
   where $b = a - 1$.
2. We are given that this product equals 9:
   $$\frac{a}{2} = 9 \implies a = 18$$
3. Since $b = a - 1$, we have $b = 17$.
4. The sum of $a$ and $b$ is:
   $$a + b = 18 + 17 = 35$$

ANSWER 3: D

---

### Problem 4:
**Restatement:** Nicolas has 20 stamps each of 5¢, 10¢, and 25¢. We want to find the greatest number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-Step Solution:**
1. Calculate the total value and count of all stamps:
   - $20 \times 5¢ = 100¢$
   - $20 \times 10¢ = 200¢$
   - $20 \times 25¢ = 500¢$
   - Total value $= 100 + 200 + 500 = 800¢$ using 60 stamps.
2. To use the **maximum** number of stamps to make 710¢, Nicolas must **minimize** the number of stamps removed to subtract:
   $$800¢ - 710¢ = 90¢$$
3. To remove the fewest stamps totaling 90¢, we try to use the largest denomination stamps (25¢):
   - With 3 stamps of 25¢ ($75¢$), the remaining amount needed is $15¢$.
   - Since Nicolas has only 5¢, 10¢, and 25¢ stamps, 15¢ cannot be made with a single stamp. It requires at least two stamps ($10¢ + 5¢$).
   - Thus, 90¢ can be formed with $3 \times 25¢ + 1 \times 10¢ + 1 \times 5¢$, which uses $3 + 1 + 1 = 5$ stamps.
   - Can 90¢ be formed using 4 stamps? The maximum value of 4 stamps less than or equal to 90¢ would be $3 \times 25¢ + 10¢ = 85¢ \ne 90¢$. Thus, 4 stamps is impossible.
4. The minimum number of stamps to remove is 5.
5. Therefore, the maximum number of stamps used is:
   $$60 - 5 = 55$$

ANSWER 4: E

---

### Problem 5:
**Restatement:** In the multiplication $ABA \times CD = CDCD$, $A, B, C, D$ are distinct digits. We need to find $A + B$.

**Step-by-Step Solution:**
1. Express $CDCD$ in terms of the two-digit number $CD$:
   $$CDCD = CD \times 100 + CD = CD \times 101$$
2. Substitute this into the equation:
   $$ABA \times CD = 101 \times CD$$
3. Since $CD$ is a non-zero two-digit number, divide both sides by $CD$:
   $$ABA = 101$$
4. Comparing digits, we have $A = 1$ and $B = 0$.
5. Thus:
   $$A + B = 1 + 0 = 1$$

ANSWER 5: A

---

### Problem 6:
**Restatement:** Out of 100 students at a camp, 52 are boys and 48 are girls. Also, 40 are from Jonas Middle School and 60 from Clay Middle School. Given that 20 of the girls are from Jonas, find how many boys are from Clay.

**Step-by-Step Solution:**
1. Find the number of girls from Clay Middle School:
   $$\text{Total girls} - \text{Girls from Jonas} = 48 - 20 = 28\text{ girls from Clay}$$
2. Find the number of boys from Clay Middle School:
   $$\text{Total students from Clay} - \text{Girls from Clay} = 60 - 28 = 32\text{ boys from Clay}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** Compute the value of:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-Step Solution:**
1. Pair each term in the first sum with the corresponding term in the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
2. Each paired difference is equal to:
   $$1900 + k - (100 + k) = 1800$$
3. Find the number of terms: from $k = 1$ to $k = 93$, there are:
   $$93 - 1 + 1 = 93\text{ terms}$$
4. Multiply the difference by the number of terms:
   $$93 \times 1800 = 93 \times 18 \times 100 = 1674 \times 100 = 167,400$$

ANSWER 7: A

---

### Problem 8:
**Restatement:** Find the tens digit of $7^{2011}$.

**Step-by-Step Solution:**
1. Find the pattern of the last two digits of powers of 7 by computing $7^n \pmod{100}$:
   - $7^1 = 07$
   - $7^2 = 49$
   - $7^3 = 343 \equiv 43 \pmod{100}$
   - $7^4 = 43

# Problem 1:
We must decide when $o^2 + no$ is odd or even, given $o$ is odd and $n$ is any whole number.

**Key facts:** odd × odd = odd, so $o^2$ is **odd**. Since $o$ is odd, multiplying by $o$ does not change $n$'s parity, so $no$ has the **same parity as $n$**.

**Case 1: $n$ even.** Then $no$ is even, so $o^2 + no = \text{odd} + \text{even} = \textbf{odd}$.
**Case 2: $n$ odd.** Then $no$ is odd, so $o^2 + no = \text{odd} + \text{odd} = \textbf{even}$.

So the number is odd exactly when $n$ is even. In particular, **if the number is odd, $n$ must be even**.

Ruling out traps: (A) fails since $o=1, n=1$ gives $1+1=2$, even; (B) fails since $n=2$ gives $1+2=3$, odd; (C) and (D) state the exact opposite of the truth. Only (E) matches.

ANSWER 1: E

# Problem 2:
The 6 redemption dates are 10 days apart, so their days of the week shift by $10 \bmod 7 = 3$ each time.

If the first date falls on weekday $W$, the six dates fall on (mod 7):
$$W,\; W+3,\; W+6,\; W+2,\; W+5,\; W+1$$
That is **six of the seven weekdays**, and the only weekday missed is $W+4$.

Since no circled date is a Sunday, the one missed weekday must be Sunday:
$$W + 4 = \text{Sunday} \implies W = \text{Wednesday}.$$

**Check:** Starting Wednesday, the dates fall on Wed, Sat, Tue, Fri, Mon, Thu — no Sunday. (Trap: starting Thursday gives Thu, Sun, ... — fails immediately.)

ANSWER 2: C

# Problem 3:
The product is a chain of fractions (the bars were lost in formatting):
$$\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdots\frac{a}{b} = 9.$$

Each numerator cancels the denominator before it (e.g., $\frac{3}{2}\cdot\frac{4}{3} = \frac{4}{2}$), and each fraction has the form $\frac{k+1}{k}$, so $b = a-1$. Everything telescopes:
$$\text{Product} = \frac{a}{2} = 9 \implies a = 18, \quad b = 17.$$
Then $a + b = 18 + 17 = 35$.

Trap check: $a+b=17$ would mean $a=9$, giving product $\tfrac92$, not $9$; only $35$ fits the pattern.

ANSWER 3: D

# Problem 4:
Work in units of 5¢: we need total value $710¢ = 142$ units, using stamps worth $1, 2, 5$ units (i.e., 5¢, 10¢, 25¢), at most 20 of each. To **maximize the count**, use as many cheap stamps as possible.

**Building a 55-stamp solution:** Take nineteen 1's and nineteen 2's: value $19 + 38 = 57$ units with 38 stamps. Remaining value $142 - 57 = 85 = 17\times 5$: seventeen 5's.
$$19\times 5¢ + 19\times 10¢ + 17\times 25¢ = 95 + 190 + 425 = 710¢ = \$7.10.\; ✓$$
That's $19+19+17 = 55$ stamps, all counts $\le 20$. ✓

**Why 56 is impossible:** with 56 stamps, at least $56 - 40 = 16$ must be 25¢ stamps (only 40 cheap ones exist). Let $z$ = number of 25¢ stamps:
- $z=16$: max value $= 100 + 200 + 400 = 700 < 710$. ✗
- $z=17$: need $5x+10y=285$, $x+y=39 \Rightarrow x=21>20$. ✗
- $z=18$: $x=24>20$. ✗  $z=19$: $x=27>20$. ✗  $z=20$: $x=30>20$. ✗

Traps: 54 comes from $20×5+16×10+18×25 = 710$; 51 from $20×5+11×10+20×25 = 710$ — both valid but not maximal.

ANSWER 4: E

# Problem 5:
Write $CDCD$ as a number:
$$CDCD = 1000C + 100D + 10C + D = 101(10C + D) = 101 \times CD.$$
The given equation is
$$ABA \times CD = CDCD = 101 \times CD.$$
Since $CD \neq 0$ (it's a two-digit number), we can divide both sides by $CD$:
$$ABA = 101.$$
So $A = 1$ and $B = 0$, giving $A + B = 1$.

**Consistency check:** $C, D$ just need to be distinct from $0, 1$ and each other, e.g., $101 \times 23 = 2323$. ✓ (Trap: guessing digits without using the factorization 101 wastes time or gives 9.)

ANSWER 5: A

# Problem 6:
Set up a two-way table with 100 total students.

- Girls: 48 total, and 20 girls are from Jonas $\Rightarrow$ girls from Clay $= 48 - 20 = 28$.
- Jonas has 40 students, of whom 20 are girls $\Rightarrow$ boys from Jonas $= 40 - 20 = 20$.
- Boys: 52 total $\Rightarrow$ boys from Clay $= 52 - 20 = \textbf{32}$.

**Check:** Jonas $20+20=40$ ✓; Clay $28+32=60$ ✓; boys $20+32=52$ ✓; girls $20+28=48$ ✓; total $100$ ✓. (Trap: 20 counts the boys from Jonas, not Clay.)

ANSWER 6: B

# Problem 7:
Both sums have the same number of terms: $1901$ to $1993$ has $1993-1901+1 = 93$ terms, and $101$ to $193$ has $193-101+1 = 93$ terms.

Pair them up: each term in the first sum is exactly $1800$ more than its partner ($1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$). So the difference is
$$93 \times 1800 = 167{,}400.$$

**Check via formulas:** $\frac{93(1901+1993)}{2} - \frac{93(101+193)}{2} = 181{,}071 - 13{,}671 = 167{,}400$. ✓ (Trap: 181,071 is the first sum alone — forgetting to subtract.)

ANSWER 7: A

# Problem 8:
Only the tens digit matters, so track the last two digits of $7^{k}$, which cycle with period 4:
$$7^1 = 07,\quad 7^2 = 49,\quad 7^3 = 343 \to 43,\quad 7^4 = 2401 \to 01,\quad 7^5 \to 07, \ldots$$

Find where $2011$ falls in the cycle:
$$2011 = 4 \times 502 + 3,$$
so $7^{2011}$ ends in the same two digits as $7^3$, namely $\ldots 43$.

The tens digit is $\textbf{4}$. (Trap: $7^{2012}$, a multiple of 4, ends in $01$, tens digit 0 — choice A; that would require the exponent to be divisible by 4.)

ANSWER 8: D

# Problem 9:
The ten **different** positive whole numbers have average 10, so their sum is $10 \times 10 = 100$.

To make one number as large as possible, make the other nine as small as possible. Since they must be **distinct positive** whole numbers, the smallest possible set is $1, 2, 3, \ldots, 9$, with sum
$$1+2+\cdots+9 = 45.$$
So the largest possible value is
$$100 - 45 = 55.$$

**Check:** $1,2,\ldots,9,55$ are ten different positive whole numbers with sum $100$ and average $10$. ✓ (Trap: 91 arises if you ignore "different" and use nine 1's; but the numbers must be distinct.)

ANSWER 9: C

# Problem 10:
Count student-days (the unit being paid):
- Allen: $7 \times 3 = 21$ student-days
- Balboa: $4 \times 5 = 20$ student-days
- Carver: $5 \times 9 = 45$ student-days

Total: $21 + 20 + 45 = 86$ student-days, paid $\$774$ altogether. So the daily rate is
$$774 \div 86 = 9 \text{ dollars per student-day}.$$
Balboa's students worked 20 student-days, so they earned
$$20 \times 9 = 18

Problem 1:  
We need the parity of \(o^2+no\), where \(o\) is odd and \(n\) is any whole number.

Factor:
\[
o^2+no=o(o+n).
\]

Since \(o\) is odd:

- If \(n\) is even, then \(o+n\) is odd, so \(o(o+n)\) is odd.
- If \(n\) is odd, then \(o+n\) is even, so \(o(o+n)\) is even.

Thus the expression is odd exactly when \(n\) is even.

Check the choices:
- A and B are false because the parity depends on \(n\).
- C says “even only if \(n\) is even,” but when the expression is even, \(n\) is actually odd.
- D says “odd only if \(n\) is odd,” but when the expression is odd, \(n\) is even.
- E says “odd only if \(n\) is even,” which is true.

ANSWER 1: E

Problem 2:  
Isabella redeems coupons 10 days apart. Since
\[
10\equiv 3 \pmod 7,
\]
the six circled days are the first day plus
\[
0,3,6,9,12,15
\]
days, which modulo 7 gives offsets
\[
0,3,6,2,5,1.
\]

So the six circled days are all weekdays except the one 4 days after the first day. Since none of the circled days is a Sunday, Sunday must be the un-circled day: 4 days after the first day.

Therefore the first coupon is 4 days before Sunday, which is Wednesday.

Check: Wednesday, Saturday, Tuesday, Friday, Monday, Thursday — no Sunday.

ANSWER 2: C

Problem 3:  
The product is the telescoping product of fractions:
\[
\frac32\cdot \frac43\cdot \frac54\cdot \frac65\cdots \frac{a}{b}=9.
\]

The numerators cancel the next denominators:
\[
\frac32\cdot \frac43=\frac42,
\]
\[
\frac42\cdot \frac54=\frac52,
\]
and continuing, the whole product simplifies to
\[
\frac{a}{2}.
\]

So
\[
\frac{a}{2}=9 \implies a=18.
\]

By the pattern, the denominator is one less than the numerator, so
\[
b=17.
\]

Thus
\[
a+b=18+17=35.
\]

ANSWER 3: D

Problem 4:  
Nicolas has 20 of each type of stamp, so he has
\[
20+20+20=60
\]
stamps total. Their total value is
\[
20(5+10+25)=20(40)=800
\]
cents.

He wants postage of
\[
\$7.10=710
\]
cents, so the stamps he does **not** use must total
\[
800-710=90
\]
cents.

To maximize the number of stamps he uses, we minimize the number of stamps he leaves unused.

Can 4 unused stamps total 90 cents? Let \(x,y,z\) be the numbers of 25-, 10-, and 5-cent unused stamps. Then
\[
x+y+z=4
\]
and
\[
25x+10y+5z=90.
\]

Divide by 5:
\[
5x+2y+z=18.
\]

Subtract \(x+y+z=4\):
\[
4x+y=14.
\]

Testing possible \(x=0,1,2,3,4\):

- \(x=0\Rightarrow y=14\), impossible.
- \(x=1\Rightarrow y=10\), impossible.
- \(x=2\Rightarrow y=6\), impossible.
- \(x=3\Rightarrow y=2\), then \(z=4-3-2=-1\), impossible.
- \(x=4\Rightarrow y=-2\), impossible.

So 4 unused stamps cannot total 90 cents.

Can 5 unused stamps total 90 cents? Yes:
\[
25+25+25+10+5=90.
\]

Then he uses
\[
60-5=55
\]
stamps. For example, he could use 19 five-cent stamps, 19 ten-cent stamps, and 17 twenty-five-cent stamps:
\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So the greatest possible number of stamps is 55. Choice D, 54, is possible but not greatest.

ANSWER 4: E

Problem 5:  
We are told
\[
ABA\times CD=CDCD,
\]
where \(A,B,C,D\) are different digits.

Write the numbers in place-value form:
\[
ABA=101A+10B,
\]
\[
CD=10C+D,
\]
\[
CDCD=1000C+100D+10C+D=101(10C+D).
\]

So
\[
(101A+10B)(10C+D)=101(10C+D).
\]

Since \(CD\) is a two-digit number, \(10C+D\neq 0\), so we can divide both sides by \(10C+D\):
\[
101A+10B=101.
\]

The only digit solution is
\[
A=1,\quad B=0.
\]

Therefore
\[
A+B=1+0=1.
\]

ANSWER 5: A

Problem 6:  
There are 48 girls total, and 20 of the girls are from Jonas Middle School. So the number of girls from Clay Middle School is
\[
48-20=28.
\]

There are 60 students total from Clay Middle School, so the number of boys from Clay is
\[
60-28=32.
\]

ANSWER 6: B

Problem 7:  
Pair the terms:
\[
(1901+1902+\cdots+1993)-(101+102+\cdots+193)
\]
\[
=(1901-101)+(1902-102)+\cdots+(1993-193).
\]

Each pair is
\[
1800.
\]

Now count the number of terms:
\[
193-101+1=93.
\]

So the value is
\[
93\cdot 1800=167400.
\]

ANSWER 7: A

Problem 8:  
We need the tens digit of \(7^{2011}\), so we need the last two digits.

Powers of 7 modulo 100 cycle:
\[
7^1=07,
\]
\[
7^2=49,
\]
\[
7^3=343\equiv 43,
\]
\[
7^4=2401\equiv 01.
\]

Then the cycle repeats:
\[
07,\ 49,\ 43,\ 01.
\]

Since
\[
2011\equiv 3 \pmod 4,
\]
we have
\[
7^{2011}\equiv 7^3\equiv 43 \pmod{100}.
\]

The last two digits are 43, so the tens digit is 4.

ANSWER 8: D

Problem 9:  
The average of 10 different positive whole numbers is 10, so their sum is
\[
10\cdot 10=100.
\]

To make one number as large as possible, make the other 9 numbers as small as possible while still being different positive whole numbers. The smallest 9 distinct positive whole numbers are
\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is
\[
1+2+\cdots+9=45.
\]

So the largest number can be
\[
100-45=55.
\]

This is possible: \(1,2,3,4,5,6,7,8,9,55\) are all different positive whole numbers.

ANSWER 9: C

Problem 10:  
First count total student-days worked:

- Allen: \(7\cdot 3=21\) student-days
- Balboa: \(4\cdot 5=20\) student-days
- Carver: \(5\cdot 9=45\) student-days

Total:
\[
21+20+45=86
\]
student-days.

The total payment is \$774, so the daily rate is
\[
774\div 86=9
\]
dollars per student-day.

Balboa students worked 20 student-days, so they earned
\[
20\cdot 9=180
\]
dollars.

ANSWER 10: C

# Problem 1:
We need a statement always true about o² + no, where o is odd.

Factor: o² + no = o(o + n).

Since o is odd, the parity of the product is the parity of (o + n).
- If n is even, o + n is odd → product = odd·odd = **odd**.
- If n is odd, o + n is even → product = odd·even = **even**.

So the expression is odd exactly when n is even.

Ruling out others: A fails (n odd gives even), B fails (n even gives odd), C says "even only if n is even" — but it's even when n is *odd*, false. D says "odd only if n is odd" — false, it's odd when n is even. E: "odd only if n is even" — true.

ANSWER 1: E

# Problem 2:
The redemption days are 0, 10, 20, 30, 40, 50 days after the first.

Reduce mod 7 (days of week shift): 0, 3, 6, 2, 5, 1.

These residues are {0, 1, 2, 3, 5, 6} — every residue except **4**. So the only day of the week never hit is the day 4 days after the starting day. For no coupon to land on Sunday, Sunday must be that missed day:

start + 4 ≡ Sunday.

Counting back 4 days from Sunday: Saturday, Friday, Thursday, **Wednesday**.

Check: Wed → +10 Sat → +20 Tue → +30 Fri → +40 Mon → +50 Thu. No Sundays. ✓

ANSWER 2: C

# Problem 3:
The product is (3/2)·(4/3)·(5/4)·(6/5)·…·(a/b), where each numerator is one more than its denominator, so b = a − 1.

Telescoping: everything cancels except the last numerator and first denominator:
a/2 = 9 → a = 18, so b = 17.

a + b = 18 + 17 = 35.

(Choice E, 37, would come from mistakenly taking a = 19.)

ANSWER 3: D

# Problem 4:
Total value of all 60 stamps: 20(5 + 10 + 25) = 20·40 = 800 cents.

We need 710 cents, so we must *remove* stamps worth exactly 800 − 710 = 90 cents, using as **few** stamps as possible (to keep the most stamps).

Can 90 be made with 4 or fewer stamps? Max with 4 stamps is 100; possible 4-stamp totals near 90: 100, 85, 80, 75, 70… none equal 90. Three or fewer is at most 75. So 4 or fewer is impossible.

With 5 stamps: 25 + 25 + 25 + 10 + 5 = 90 ✓

So remove 5 stamps: 60 − 5 = 55 stamps.

Check: 17(25) + 19(10) + 19(5) = 425 + 190 + 95 = 710 cents, using 17 + 19 + 19 = 55 stamps.

ANSWER 4: E

# Problem 5:
CDCD as a number = CD·100 + CD = CD·101.

So ABA × CD = 101 × CD, and since CD ≠ 0, we get ABA = 101.

Thus A = 1, B = 0 (and indeed A, B, C, D distinct works, e.g. 101 × 23 = 2323).

A + B = 1 + 0 = 1.

ANSWER 5: A

# Problem 6:
Jonas total = 40, and 20 of those were girls, so boys from Jonas = 40 − 20 = 20.

Total boys = 52, so boys from Clay = 52 − 20 = 32.

(Check: girls from Clay = 48 − 20 = 28; 32 + 28 = 60 = Clay total ✓.)

Choice C (40) is the Jonas total, and E (52) is all boys — traps.

ANSWER 6: B

# Problem 7:
Pair each term: 1901 − 101 = 1800, 1902 − 102 = 1800, …, each pair differs by 1800.

Number of terms: 1993 − 1901 + 1 = 93.

Total = 93 × 1800 = 167,400.

ANSWER 7: A

# Problem 8:
Look at 7ⁿ mod 100:
- 7¹ = 07
- 7² = 49
- 7³ = 343 → 43
- 7⁴ = 2401 → 01
- then the pattern repeats with period 4.

2011 ÷ 4 = 502 remainder 3, so 7²⁰¹¹ ≡ 7³ ≡ 43 (mod 100).

Last two digits are 43, so the tens digit is 4.

ANSWER 8: D

# Problem 9:
Sum of the 10 numbers = 10 × 10 = 100.

To make one number as large as possible, make the other nine as small as possible while distinct and positive: 1 + 2 + … + 9 = 45.

Largest = 100 − 45 = 55.

(90 or 91 would require repeats or a zero, which isn't allowed.)

ANSWER 9: C

# Problem 10:
Total student-days:
- Allen: 7 × 3 = 21
- Balboa: 4 × 5 = 20
- Carver: 5 × 9 = 45

Total = 86 student-days.

Daily wage = 774 ÷ 86 = $9.00 per student-day.

Balboa earnings = 20 × 9 = $180.00.

(A is one day's pay for one student; E is Carver's total.)

ANSWER 10: C

Problem 1:

We need determine the parity of

\[
o^2+no.
\]

Factor out \(o\):

\[
o^2+no=o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on \(o+n\).

- If \(n\) is even, then odd \(+\) even \(=\) odd, so \(o+n\) is odd. Thus odd \(\cdot\) odd is odd.
- If \(n\) is odd, then odd \(+\) odd \(=\) even, so the product is even.

Therefore, the expression is odd only when \(n\) is even.

- A is wrong because the expression is not always odd.
- B is wrong because it is odd when \(n\) is even.
- C reverses the condition for being even.
- D reverses the condition for being odd.

ANSWER 1: E

Problem 2:

Isabella uses a coupon every 10 days. Since

\[
10\equiv 3\pmod 7,
\]

each coupon date moves forward by 3 days of the week.

If the first coupon is redeemed on day \(d\), the six redemption days are:

\[
d,\ d+3,\ d+6,\ d+2,\ d+5,\ d+1.
\]

These are six different days of the week. The only day missing is

\[
d+4.
\]

Since none of the six coupon dates is Sunday, the missing day must be Sunday. Thus,

\[
d+4=\text{Sunday}.
\]

Four days before Sunday is Wednesday.

Indeed, beginning on Wednesday gives:

\[
\text{Wednesday, Saturday, Tuesday, Friday, Monday, Thursday}.
\]

Sunday is the only missing day.

ANSWER 2: C

Problem 3:

The product is

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9.
\]

Almost every numerator cancels with a denominator:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab
=\frac a2.
\]

So

\[
\frac a2=9.
\]

Therefore,

\[
a=18.
\]

The final fraction must be

\[
\frac{18}{17},
\]

so \(b=17\). Therefore,

\[
a+b=18+17=35.
\]

Choices 11, 13, and 17 are too small, while 37 would correspond to a different ending fraction.

ANSWER 3: D

Problem 4:

Let

- \(x\) = number of 5-cent stamps,
- \(y\) = number of 10-cent stamps,
- \(z\) = number of 25-cent stamps.

In units of 5 cents, the postage total is

\[
\frac{710}{5}=142.
\]

Thus,

\[
x+2y+5z=142.
\]

We want to maximize

\[
x+y+z.
\]

Since there are at most 20 five-cent stamps, \(x\leq 20\). Therefore,

\[
2y+5z=142-x\geq 142-20=122.
\]

Since \(y\leq20\), if \(z\leq16\), then

\[
2y+5z\leq 2(20)+5(16)=40+80=120,
\]

which is not enough. So \(z\geq17\).

If \(z=17\), then

\[
2y+5(17)\geq122,
\]

so

\[
2y+85\geq122,
\]

\[
2y\geq37,
\]

and hence \(y\geq19\).

Now use

\[
x+y+z=142-y-4z.
\]

Since \(y\geq19\) and \(z\geq17\),

\[
x+y+z\leq142-19-4(17)
=142-19-68
=55.
\]

This maximum is attainable with

\[
x=19,\qquad y=19,\qquad z=17.
\]

Check:

\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So 55 stamps can be used. Choices 45, 46, 51, and 54 are possible totals to consider, but 55 is the greatest possible.

ANSWER 4: E

Problem 5:

We are told

\[
ABA\times CD=CDCD.
\]

The number \(ABA\) is

\[
100A+10B+A=101A+10B.
\]

The number \(CDCD\) is

\[
1000C+100D+10C+D.
\]

Group the \(C\)'s and \(D\)'s:

\[
CDCD=1010C+101D=101(10C+D)=101\cdot CD.
\]

Thus the equation becomes

\[
ABA\cdot CD=101\cdot CD.
\]

Since \(CD\) is a two-digit number and is not zero, divide both sides by \(CD\):

\[
ABA=101.
\]

Therefore,

\[
A=1,\qquad B=0.
\]

So

\[
A+B=1+0=1.
\]

For example, \(101\times23=2323\), and the digits \(1,0,2,3\) are all different.

ANSWER 5: A

Problem 6:

There are 40 students from Jonas Middle School, and 20 of those are girls.

Therefore the number of boys from Jonas is

\[
40-20=20.
\]

There are 52 boys altogether. Therefore the number of boys from Clay Middle School is

\[
52-20=32.
\]

The other choices do not correctly account for the 20 Jonas girls and the total of 52 boys.

ANSWER 6: B

Problem 7:

There are

\[
1993-1901+1=93
\]

terms in the first sum.

There are also

\[
193-101+1=93
\]

terms in the second sum.

Pair corresponding terms:

\[
1901-101=1800,
\]

\[
1902-102=1800,
\]

and so on, through

\[
1993-193=1800.
\]

Thus there are 93 differences of 1800:

\[
93\cdot1800=167400.
\]

Therefore the value is

\[
167,\!400.
\]

ANSWER 7: A

Problem 8:

We need the tens digit of

\[
7^{2011}.
\]

It is enough to examine powers of 7 modulo 100:

\[
7^1=7,
\]

\[
7^2=49,
\]

\[
7^3=343,
\]

whose last two digits are 43, and

\[
7^4=2401,
\]

whose last two digits are 01.

Thus the last two digits repeat every 4 powers because

\[
7^4\equiv1\pmod{100}.
\]

Now,

\[
2011\div4
\]

leaves remainder 3, since

\[
2011=4(502)+3.
\]

Therefore,

\[
7^{2011}
\]

has the same last two digits as

\[
7^3=343,
\]

which are 43. The tens digit is 4.

ANSWER 8: D

Problem 9:

The average of 10 different positive whole numbers is 10, so their total sum is

\[
10\cdot10=100.
\]

To make one number as large as possible, make the other nine numbers as small as possible.

The nine smallest different positive whole numbers are

\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is

\[
1+2+3+4+5+6+7+8+9=45.
\]

So the largest possible tenth number is

\[
100-45=55.
\]

This works because the ten numbers can be

\[
1,2,3,4,5,6,7,8,9,55,
\]

which are all different and have total 100.

A value such as 90 or 91 would leave too small a total for nine distinct positive whole numbers.

ANSWER 9: C

Problem 10:

Compute the total number of student-days worked.

Allen students worked:

\[
7\cdot3=21
\]

student-days.

Balboa students worked:

\[
4\cdot5=20
\]

student-days.

Carver students worked:

\[
5\cdot9=45
\]

student-days.

So the total number of student-days is

\[
21+20+45=86.
\]

The total pay was \$774, so the pay per student per day was

\[
\frac{774}{86}=9.
\]

Balboa students worked 20 student-days, so they earned

\[
20\cdot9=180.
\]

Thus the Balboa students earned \$180.00 altogether.

ANSWER 10: C

Problem 1:
We are asked about the parity of $o^2+no$ where $o$ is odd and $n$ is any whole number.

Factor:
$$o^2+no = o(o+n).$$

$o$ is odd, so $o^2 = o\cdot o$ is odd times odd = odd.

Hence parity is odd + $n\cdot o$.
$n\cdot o$ has the same parity as $n$, because $o$ is odd: if $n$ is even, $n o$ is even; if $n$ is odd, $n o$ is odd.

So:
- if $n$ even: odd + even = odd.
- if $n$ odd: odd + odd = even.

Thus the number is odd iff $n$ is even, and even iff $n$ is odd.

Check choices:
A. always odd - false, e.g. $o=1,n=1$ gives $1+1=2$ even.
B. always even - false, e.g. $o=1,n=2$ gives $1+2=3$ odd.
C. it is even only if $n$ is even - means even $\implies$ $n$ even. False, even happens when $n$ is odd.
D. it is odd only if $n$ is odd - means odd $\implies$ $n$ odd. False, odd happens when $n$ is even.
E. it is odd only if $n$ is even - means odd $\implies$ $n$ even. True by above.

ANSWER 1: E

Problem 2:
Isabella redeems on days $d, d+10, d+20, d+30, d+40, d+50$.
No date is a Sunday.

Modulo 7, $10\equiv 3$. So successive dates advance by 3 weekdays.

Starting from residue $r$, the residues are:
$$r,\ r+3,\ r+6,\ r+9=r+2,\ r+12=r+5,\ r+15=r+1 \pmod 7.$$
These 6 are all distinct, because adding 3 mod 7 cycles through all 7 residues. Continuing one more step would give $r+18=r+4$.

Thus the 6 dates occupy 6 distinct days of the week, missing exactly the 7th residue $r+4$.

Since none of the 6 is Sunday, the missing one must be Sunday. So:
$$r+4 \equiv \text{Sunday}.$$

Hence $r$ is 4 days before Sunday, i.e. Wednesday.

Check: Wed $\to$ Wed+3=Sat $\to$ +3=Tue $\to$ +3=Fri $\to$ +3=Mon $\to$ +3=Thu. That is Wed,Sat,Tue,Fri,Mon,Thu - no Sunday. Any other start would include Sunday.

ANSWER 2: C

Problem 3:
We interpret $3\ 2\cdot 4\ 3\cdots$ as fractions:
$$\frac32 \cdot \frac43 \cdot \frac54 \cdot \frac65 \cdots \frac{a}{b}=9.$$

The pattern is numerator $3,4,5,6,\dots$ and denominator $2,3,4,5,\dots$, i.e. term $k/(k-1)$. The last term is $a/b$, so we must have $b=a-1$ for the pattern to continue.

Then the product telescopes:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}
= \frac{a}{2},$$
because all intermediate $3,4,\dots,a-1$ cancel.

Set equal to 9:
$$\frac{a}{2}=9 \implies a=18.$$
Then $b=a-1=17$.

Sum $a+b=18+17=35$.

This rules out A,B,C,E.

ANSWER 3: D

Problem 4:
Let $x,y,z$ be numbers of 5-cent, 10-cent, 25-cent stamps used. $0\le x,y,z\le 20$ integers.
Value $710$ cents:
$$5x+10y+25z=710.$$
Divide by 5:
$$x+2y+5z=142. \tag{1}$$
Maximize $N=x+y+z$.

From (1), $x=142-2y-5z$, so $0\le x\le20$ gives:
$$122\le 2y+5z\le 142.$$
Also $N=x+y+z=142-y-4z$, so $y+4z=142-N$.

To use many stamps we want small denominations. Try $x=20,y=20$ fails to give integer $z$. Search systematically.

Since $y\le20$, to satisfy $2y+5z\ge122$ we need $z$ large. If $z\le16$, then $2y+5z\le40+80=120<122$, impossible. So $z\ge17$.

Try $z=17,y=19$: then $2y+5z=38+85=123$, so $x=142-123=19$, within $0-20$. This gives:
$$19\cdot5+19\cdot10+17\cdot25=95+190+425=710,$$
with $N=19+19+17=55$ stamps. So 55 is achievable.

Can we do 56 or more? Let $N\ge56$.
Then $y+4z=142-N\le86$ and $x=N-y-z\le20$ gives $y+z\ge N-20$.
Subtract: $3z=(y+4z)-(y+z)\le (142-N)-(N-20)=162-2N$.
So $z\le(162-2N)/3$.
Also $y=142-N-4z\le20$ gives $z\ge(122-N)/4$.

For $N=56$: $z\ge\lceil66/4\rceil=17$ and $z\le\lfloor50/3\rfloor=16$, contradiction. For $N>56$ the interval is even emptier; in general need $(122-N)/4\le(162-2N)/3$, i.e. $5N\le282$, so $N\le56.4$. Thus $N\ge57$ impossible, and $N=56$ impossible by the integer contradiction.

Hence 55 is greatest possible.

ANSWER 4: E

Problem 5:
We have $ABA\times CD = CDCD$, where $A,B,C,D$ are different digits, $A\ne0$, $C\ne0$.

Write $CD$ as the 2-digit number $10C+D$.
$CDCD$ as a 4-digit number is:
$$1000C+100D+10C+D =1010C+101D=101(10C+D)=101\cdot CD.$$

So:
$$ABA \times CD = 101\times CD.$$
Since $CD\ne0$, divide:
$$ABA =101.$$

Thus $A=1$, $B=0$, and $A+B=1$.

This is consistent with distinctness: we can choose e.g. $C=2,D=3$, all of $1,0,2,3$ different, and $101\times23=2323$. So valid.

Other sums are ruled out.

ANSWER 5: A

Problem 6:
Total 100: 52 boys, 48 girls. 40 from Jonas, 60 from Clay.

20 of the girls are from Jonas. So girls from Clay = total girls $-$ girls from Jonas:
$$48-20=28.$$

Clay total 60 = boys from Clay + girls from Clay, so:
$$\text{boys from Clay}=60-28=32.$$

Check: boys from Jonas $=52-32=20$, Jonas total $=20+20=40$. Works.

ANSWER 6: B

Problem 7:
Compute $(1901+\cdots+1993)-(101+\cdots+193)$.

Both sums have $1993-1901+1=93$ terms and $193-101+1=93$ terms.

Pair termwise: for $k=0,\dots,92$,
$$(1901+k)-(101+k)=1800.$$

So difference $=93\times1800$.
$93\times18=1674$, so $93\times1800=167{,}400$.

Other choices come from miscounting terms or arithmetic error.

ANSWER 7: A

Problem 8:
Find tens digit of $7^{2011}$, i.e. $7^{2011}\pmod{100}$.

Compute powers mod 100:
$7^1=7$,
$7^2=49$,
$7^3=343\equiv43$,
$7^4\equiv43\times7=301\equiv1$.

Since $7^4\equiv1\pmod{100}$, the last two digits repeat with period 4.

$2011 = 4\cdot502+3$, so:
$$7^{2011}=(7^4)^{502}\cdot7^3\equiv1^{502}\cdot43\equiv43\pmod{100}.$$

Thus the number ends $\dots43$, tens digit $4$.

ANSWER 8: D

Problem 9:
10 different positive whole numbers have mean 10, so sum $=10\times10=100$.

To make one number as large as possible, make the other nine as small as possible while distinct and positive.

Smallest 9 distinct positives are $1,2,3,4,5,6,7,8,9$, sum $=45$.

Then the tenth number $=100-45=55$, which is distinct from the others and valid. Any larger value would force the other nine to sum to $<45$, impossible with distinct positives.

So largest possible is 55.

ANSWER 9: C

Problem 10:
Count student-days (one student working one day).

Allen: $7$ students $\times$ $3$ days $=21$.
Balboa: $4\times5=20$.
Carver: $5\times9=45$.
Total $=21+20+45=86$ student-days.

Total pay $\$774$ for $86$ student-days at equal rate:
$$\text{rate}=774/86=9 \text{ dollars per student-day}.$$

Bal

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that $o$ is an odd whole number and $n$ is any whole number. We must determine which statement about $o^2 + no$ is always true.

**Step-by-Step Solution:**
1. Factor the expression:
   $$o^2 + no = o(o + n)$$
2. Since $o$ is an odd whole number, the parity of the product $o(o + n)$ is entirely determined by the parity of $(o + n)$:
   - If $n$ is **odd**, then $\text{odd} + \text{odd} = \text{even}$, so $o(o+n)$ is $\text{odd} \times \text{even} = \textbf{even}$.
   - If $n$ is **even**, then $\text{odd} + \text{even} = \text{odd}$, so $o(o+n)$ is $\text{odd} \times \text{odd} = \textbf{odd}$.
3. Therefore, the expression $o^2 + no$ is odd if and only if $n$ is even. 
4. This matches the statement: "it is odd only if $n$ is even."

ANSWER 1: E

---

### Problem 2:
**Restatement:** Isabella redeems 6 coupons, one every 10 days. Pete's is closed on Sundays, and none of the 6 dates falls on a Sunday. We need to find the day of the week she redeems her first coupon.

**Step-by-Step Solution:**
1. There are 7 days in a week. Moving forward by 10 days advances the day of the week by:
   $$10 \equiv 3 \pmod 7$$
2. Let $d$ represent the day of the week of the first coupon ($0 = \text{Sunday}, 1 = \text{Monday}, \dots, 6 = \text{Saturday}$).
3. The days of the week for the 6 coupons will be:
   - Coupon 1: $d$
   - Coupon 2: $d + 3$
   - Coupon 3: $d + 6$
   - Coupon 4: $d + 9 \equiv d + 2 \pmod 7$
   - Coupon 5: $d + 12 \equiv d + 5 \pmod 7$
   - Coupon 6: $d + 15 \equiv d + 1 \pmod 7$
4. The six days of the week visited are $d, d+1, d+2, d+3, d+5, d+6$. Notice that the only day of the week omitted from this list is $d+4$.
5. Since no circled date falls on a Sunday, Sunday must be the omitted day:
   $$d + 4 \equiv 0 \pmod 7 \implies d \equiv 3 \pmod 7$$
6. Day 3 corresponds to **Wednesday**.
   - Checking: Wed $\to$ Sat $\to$ Tue $\to$ Fri $\to$ Mon $\to$ Thu. None of these is Sunday.

ANSWER 2: C

---

### Problem 3:
**Restatement:** We are given the telescoping product $\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdots \frac{a}{b} = 9$. We must find the sum $a + b$.

**Step-by-Step Solution:**
1. Each fraction is of the form $\frac{k+1}{k}$, where the numerator of each term cancels with the denominator of the subsequent term:
   $$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{a}{b} = \frac{a}{2}$$
   where $b = a - 1$.
2. We are given that this product equals 9:
   $$\frac{a}{2} = 9 \implies a = 18$$
3. Since $b = a - 1$, we have $b = 17$.
4. The sum of $a$ and $b$ is:
   $$a + b = 18 + 17 = 35$$

ANSWER 3: D

---

### Problem 4:
**Restatement:** Nicolas has 20 stamps each of 5¢, 10¢, and 25¢. We want to find the greatest number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-Step Solution:**
1. Calculate the total value and count of all stamps:
   - $20 \times 5¢ = 100¢$
   - $20 \times 10¢ = 200¢$
   - $20 \times 25¢ = 500¢$
   - Total value $= 100 + 200 + 500 = 800¢$ using 60 stamps.
2. To use the **maximum** number of stamps to make 710¢, Nicolas must **minimize** the number of stamps removed to subtract:
   $$800¢ - 710¢ = 90¢$$
3. To remove the fewest stamps totaling 90¢, we try to use the largest denomination stamps (25¢):
   - With 3 stamps of 25¢ ($75¢$), the remaining amount needed is $15¢$.
   - Since Nicolas has only 5¢, 10¢, and 25¢ stamps, 15¢ cannot be made with a single stamp. It requires at least two stamps ($10¢ + 5¢$).
   - Thus, 90¢ can be formed with $3 \times 25¢ + 1 \times 10¢ + 1 \times 5¢$, which uses $3 + 1 + 1 = 5$ stamps.
   - Can 90¢ be formed using 4 stamps? The maximum value of 4 stamps less than or equal to 90¢ would be $3 \times 25¢ + 10¢ = 85¢ \ne 90¢$. Thus, 4 stamps is impossible.
4. The minimum number of stamps to remove is 5.
5. Therefore, the maximum number of stamps used is:
   $$60 - 5 = 55$$

ANSWER 4: E

---

### Problem 5:
**Restatement:** In the multiplication $ABA \times CD = CDCD$, $A, B, C, D$ are distinct digits. We need to find $A + B$.

**Step-by-Step Solution:**
1. Express $CDCD$ in terms of the two-digit number $CD$:
   $$CDCD = CD \times 100 + CD = CD \times 101$$
2. Substitute this into the equation:
   $$ABA \times CD = 101 \times CD$$
3. Since $CD$ is a non-zero two-digit number, divide both sides by $CD$:
   $$ABA = 101$$
4. Comparing digits, we have $A = 1$ and $B = 0$.
5. Thus:
   $$A + B = 1 + 0 = 1$$

ANSWER 5: A

---

### Problem 6:
**Restatement:** Out of 100 students at a camp, 52 are boys and 48 are girls. Also, 40 are from Jonas Middle School and 60 from Clay Middle School. Given that 20 of the girls are from Jonas, find how many boys are from Clay.

**Step-by-Step Solution:**
1. Find the number of girls from Clay Middle School:
   $$\text{Total girls} - \text{Girls from Jonas} = 48 - 20 = 28\text{ girls from Clay}$$
2. Find the number of boys from Clay Middle School:
   $$\text{Total students from Clay} - \text{Girls from Clay} = 60 - 28 = 32\text{ boys from Clay}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** Compute the value of:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-Step Solution:**
1. Pair each term in the first sum with the corresponding term in the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
2. Each paired difference is equal to:
   $$1900 + k - (100 + k) = 1800$$
3. Find the number of terms: from $k = 1$ to $k = 93$, there are:
   $$93 - 1 + 1 = 93\text{ terms}$$
4. Multiply the difference by the number of terms:
   $$93 \times 1800 = 93 \times 18 \times 100 = 1674 \times 100 = 167,400$$

ANSWER 7: A

---

### Problem 8:
**Restatement:** Find the tens digit of $7^{2011}$.

**Step-by-Step Solution:**
1. Find the pattern of the last two digits of powers of 7 by computing $7^n \pmod{100}$:
   - $7^1 = 07$
   - $7^2 = 49$
   - $7^3 = 343 \equiv 43 \pmod{100}$
   - $7^4 = 43

# Problem 1:
We must decide when $o^2 + no$ is odd or even, given $o$ is odd and $n$ is any whole number.

**Key facts:** odd × odd = odd, so $o^2$ is **odd**. Since $o$ is odd, multiplying by $o$ does not change $n$'s parity, so $no$ has the **same parity as $n$**.

**Case 1: $n$ even.** Then $no$ is even, so $o^2 + no = \text{odd} + \text{even} = \textbf{odd}$.
**Case 2: $n$ odd.** Then $no$ is odd, so $o^2 + no = \text{odd} + \text{odd} = \textbf{even}$.

So the number is odd exactly when $n$ is even. In particular, **if the number is odd, $n$ must be even**.

Ruling out traps: (A) fails since $o=1, n=1$ gives $1+1=2$, even; (B) fails since $n=2$ gives $1+2=3$, odd; (C) and (D) state the exact opposite of the truth. Only (E) matches.

ANSWER 1: E

# Problem 2:
The 6 redemption dates are 10 days apart, so their days of the week shift by $10 \bmod 7 = 3$ each time.

If the first date falls on weekday $W$, the six dates fall on (mod 7):
$$W,\; W+3,\; W+6,\; W+2,\; W+5,\; W+1$$
That is **six of the seven weekdays**, and the only weekday missed is $W+4$.

Since no circled date is a Sunday, the one missed weekday must be Sunday:
$$W + 4 = \text{Sunday} \implies W = \text{Wednesday}.$$

**Check:** Starting Wednesday, the dates fall on Wed, Sat, Tue, Fri, Mon, Thu — no Sunday. (Trap: starting Thursday gives Thu, Sun, ... — fails immediately.)

ANSWER 2: C

# Problem 3:
The product is a chain of fractions (the bars were lost in formatting):
$$\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdots\frac{a}{b} = 9.$$

Each numerator cancels the denominator before it (e.g., $\frac{3}{2}\cdot\frac{4}{3} = \frac{4}{2}$), and each fraction has the form $\frac{k+1}{k}$, so $b = a-1$. Everything telescopes:
$$\text{Product} = \frac{a}{2} = 9 \implies a = 18, \quad b = 17.$$
Then $a + b = 18 + 17 = 35$.

Trap check: $a+b=17$ would mean $a=9$, giving product $\tfrac92$, not $9$; only $35$ fits the pattern.

ANSWER 3: D

# Problem 4:
Work in units of 5¢: we need total value $710¢ = 142$ units, using stamps worth $1, 2, 5$ units (i.e., 5¢, 10¢, 25¢), at most 20 of each. To **maximize the count**, use as many cheap stamps as possible.

**Building a 55-stamp solution:** Take nineteen 1's and nineteen 2's: value $19 + 38 = 57$ units with 38 stamps. Remaining value $142 - 57 = 85 = 17\times 5$: seventeen 5's.
$$19\times 5¢ + 19\times 10¢ + 17\times 25¢ = 95 + 190 + 425 = 710¢ = \$7.10.\; ✓$$
That's $19+19+17 = 55$ stamps, all counts $\le 20$. ✓

**Why 56 is impossible:** with 56 stamps, at least $56 - 40 = 16$ must be 25¢ stamps (only 40 cheap ones exist). Let $z$ = number of 25¢ stamps:
- $z=16$: max value $= 100 + 200 + 400 = 700 < 710$. ✗
- $z=17$: need $5x+10y=285$, $x+y=39 \Rightarrow x=21>20$. ✗
- $z=18$: $x=24>20$. ✗  $z=19$: $x=27>20$. ✗  $z=20$: $x=30>20$. ✗

Traps: 54 comes from $20×5+16×10+18×25 = 710$; 51 from $20×5+11×10+20×25 = 710$ — both valid but not maximal.

ANSWER 4: E

# Problem 5:
Write $CDCD$ as a number:
$$CDCD = 1000C + 100D + 10C + D = 101(10C + D) = 101 \times CD.$$
The given equation is
$$ABA \times CD = CDCD = 101 \times CD.$$
Since $CD \neq 0$ (it's a two-digit number), we can divide both sides by $CD$:
$$ABA = 101.$$
So $A = 1$ and $B = 0$, giving $A + B = 1$.

**Consistency check:** $C, D$ just need to be distinct from $0, 1$ and each other, e.g., $101 \times 23 = 2323$. ✓ (Trap: guessing digits without using the factorization 101 wastes time or gives 9.)

ANSWER 5: A

# Problem 6:
Set up a two-way table with 100 total students.

- Girls: 48 total, and 20 girls are from Jonas $\Rightarrow$ girls from Clay $= 48 - 20 = 28$.
- Jonas has 40 students, of whom 20 are girls $\Rightarrow$ boys from Jonas $= 40 - 20 = 20$.
- Boys: 52 total $\Rightarrow$ boys from Clay $= 52 - 20 = \textbf{32}$.

**Check:** Jonas $20+20=40$ ✓; Clay $28+32=60$ ✓; boys $20+32=52$ ✓; girls $20+28=48$ ✓; total $100$ ✓. (Trap: 20 counts the boys from Jonas, not Clay.)

ANSWER 6: B

# Problem 7:
Both sums have the same number of terms: $1901$ to $1993$ has $1993-1901+1 = 93$ terms, and $101$ to $193$ has $193-101+1 = 93$ terms.

Pair them up: each term in the first sum is exactly $1800$ more than its partner ($1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$). So the difference is
$$93 \times 1800 = 167{,}400.$$

**Check via formulas:** $\frac{93(1901+1993)}{2} - \frac{93(101+193)}{2} = 181{,}071 - 13{,}671 = 167{,}400$. ✓ (Trap: 181,071 is the first sum alone — forgetting to subtract.)

ANSWER 7: A

# Problem 8:
Only the tens digit matters, so track the last two digits of $7^{k}$, which cycle with period 4:
$$7^1 = 07,\quad 7^2 = 49,\quad 7^3 = 343 \to 43,\quad 7^4 = 2401 \to 01,\quad 7^5 \to 07, \ldots$$

Find where $2011$ falls in the cycle:
$$2011 = 4 \times 502 + 3,$$
so $7^{2011}$ ends in the same two digits as $7^3$, namely $\ldots 43$.

The tens digit is $\textbf{4}$. (Trap: $7^{2012}$, a multiple of 4, ends in $01$, tens digit 0 — choice A; that would require the exponent to be divisible by 4.)

ANSWER 8: D

# Problem 9:
The ten **different** positive whole numbers have average 10, so their sum is $10 \times 10 = 100$.

To make one number as large as possible, make the other nine as small as possible. Since they must be **distinct positive** whole numbers, the smallest possible set is $1, 2, 3, \ldots, 9$, with sum
$$1+2+\cdots+9 = 45.$$
So the largest possible value is
$$100 - 45 = 55.$$

**Check:** $1,2,\ldots,9,55$ are ten different positive whole numbers with sum $100$ and average $10$. ✓ (Trap: 91 arises if you ignore "different" and use nine 1's; but the numbers must be distinct.)

ANSWER 9: C

# Problem 10:
Count student-days (the unit being paid):
- Allen: $7 \times 3 = 21$ student-days
- Balboa: $4 \times 5 = 20$ student-days
- Carver: $5 \times 9 = 45$ student-days

Total: $21 + 20 + 45 = 86$ student-days, paid $\$774$ altogether. So the daily rate is
$$774 \div 86 = 9 \text{ dollars per student-day}.$$
Balboa's students worked 20 student-days, so they earned
$$20 \times 9 = 18

Problem 1:  
We need the parity of \(o^2+no\), where \(o\) is odd and \(n\) is any whole number.

Factor:
\[
o^2+no=o(o+n).
\]

Since \(o\) is odd:

- If \(n\) is even, then \(o+n\) is odd, so \(o(o+n)\) is odd.
- If \(n\) is odd, then \(o+n\) is even, so \(o(o+n)\) is even.

Thus the expression is odd exactly when \(n\) is even.

Check the choices:
- A and B are false because the parity depends on \(n\).
- C says “even only if \(n\) is even,” but when the expression is even, \(n\) is actually odd.
- D says “odd only if \(n\) is odd,” but when the expression is odd, \(n\) is even.
- E says “odd only if \(n\) is even,” which is true.

ANSWER 1: E

Problem 2:  
Isabella redeems coupons 10 days apart. Since
\[
10\equiv 3 \pmod 7,
\]
the six circled days are the first day plus
\[
0,3,6,9,12,15
\]
days, which modulo 7 gives offsets
\[
0,3,6,2,5,1.
\]

So the six circled days are all weekdays except the one 4 days after the first day. Since none of the circled days is a Sunday, Sunday must be the un-circled day: 4 days after the first day.

Therefore the first coupon is 4 days before Sunday, which is Wednesday.

Check: Wednesday, Saturday, Tuesday, Friday, Monday, Thursday — no Sunday.

ANSWER 2: C

Problem 3:  
The product is the telescoping product of fractions:
\[
\frac32\cdot \frac43\cdot \frac54\cdot \frac65\cdots \frac{a}{b}=9.
\]

The numerators cancel the next denominators:
\[
\frac32\cdot \frac43=\frac42,
\]
\[
\frac42\cdot \frac54=\frac52,
\]
and continuing, the whole product simplifies to
\[
\frac{a}{2}.
\]

So
\[
\frac{a}{2}=9 \implies a=18.
\]

By the pattern, the denominator is one less than the numerator, so
\[
b=17.
\]

Thus
\[
a+b=18+17=35.
\]

ANSWER 3: D

Problem 4:  
Nicolas has 20 of each type of stamp, so he has
\[
20+20+20=60
\]
stamps total. Their total value is
\[
20(5+10+25)=20(40)=800
\]
cents.

He wants postage of
\[
\$7.10=710
\]
cents, so the stamps he does **not** use must total
\[
800-710=90
\]
cents.

To maximize the number of stamps he uses, we minimize the number of stamps he leaves unused.

Can 4 unused stamps total 90 cents? Let \(x,y,z\) be the numbers of 25-, 10-, and 5-cent unused stamps. Then
\[
x+y+z=4
\]
and
\[
25x+10y+5z=90.
\]

Divide by 5:
\[
5x+2y+z=18.
\]

Subtract \(x+y+z=4\):
\[
4x+y=14.
\]

Testing possible \(x=0,1,2,3,4\):

- \(x=0\Rightarrow y=14\), impossible.
- \(x=1\Rightarrow y=10\), impossible.
- \(x=2\Rightarrow y=6\), impossible.
- \(x=3\Rightarrow y=2\), then \(z=4-3-2=-1\), impossible.
- \(x=4\Rightarrow y=-2\), impossible.

So 4 unused stamps cannot total 90 cents.

Can 5 unused stamps total 90 cents? Yes:
\[
25+25+25+10+5=90.
\]

Then he uses
\[
60-5=55
\]
stamps. For example, he could use 19 five-cent stamps, 19 ten-cent stamps, and 17 twenty-five-cent stamps:
\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So the greatest possible number of stamps is 55. Choice D, 54, is possible but not greatest.

ANSWER 4: E

Problem 5:  
We are told
\[
ABA\times CD=CDCD,
\]
where \(A,B,C,D\) are different digits.

Write the numbers in place-value form:
\[
ABA=101A+10B,
\]
\[
CD=10C+D,
\]
\[
CDCD=1000C+100D+10C+D=101(10C+D).
\]

So
\[
(101A+10B)(10C+D)=101(10C+D).
\]

Since \(CD\) is a two-digit number, \(10C+D\neq 0\), so we can divide both sides by \(10C+D\):
\[
101A+10B=101.
\]

The only digit solution is
\[
A=1,\quad B=0.
\]

Therefore
\[
A+B=1+0=1.
\]

ANSWER 5: A

Problem 6:  
There are 48 girls total, and 20 of the girls are from Jonas Middle School. So the number of girls from Clay Middle School is
\[
48-20=28.
\]

There are 60 students total from Clay Middle School, so the number of boys from Clay is
\[
60-28=32.
\]

ANSWER 6: B

Problem 7:  
Pair the terms:
\[
(1901+1902+\cdots+1993)-(101+102+\cdots+193)
\]
\[
=(1901-101)+(1902-102)+\cdots+(1993-193).
\]

Each pair is
\[
1800.
\]

Now count the number of terms:
\[
193-101+1=93.
\]

So the value is
\[
93\cdot 1800=167400.
\]

ANSWER 7: A

Problem 8:  
We need the tens digit of \(7^{2011}\), so we need the last two digits.

Powers of 7 modulo 100 cycle:
\[
7^1=07,
\]
\[
7^2=49,
\]
\[
7^3=343\equiv 43,
\]
\[
7^4=2401\equiv 01.
\]

Then the cycle repeats:
\[
07,\ 49,\ 43,\ 01.
\]

Since
\[
2011\equiv 3 \pmod 4,
\]
we have
\[
7^{2011}\equiv 7^3\equiv 43 \pmod{100}.
\]

The last two digits are 43, so the tens digit is 4.

ANSWER 8: D

Problem 9:  
The average of 10 different positive whole numbers is 10, so their sum is
\[
10\cdot 10=100.
\]

To make one number as large as possible, make the other 9 numbers as small as possible while still being different positive whole numbers. The smallest 9 distinct positive whole numbers are
\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is
\[
1+2+\cdots+9=45.
\]

So the largest number can be
\[
100-45=55.
\]

This is possible: \(1,2,3,4,5,6,7,8,9,55\) are all different positive whole numbers.

ANSWER 9: C

Problem 10:  
First count total student-days worked:

- Allen: \(7\cdot 3=21\) student-days
- Balboa: \(4\cdot 5=20\) student-days
- Carver: \(5\cdot 9=45\) student-days

Total:
\[
21+20+45=86
\]
student-days.

The total payment is \$774, so the daily rate is
\[
774\div 86=9
\]
dollars per student-day.

Balboa students worked 20 student-days, so they earned
\[
20\cdot 9=180
\]
dollars.

ANSWER 10: C

# Problem 1:
We need a statement always true about o² + no, where o is odd.

Factor: o² + no = o(o + n).

Since o is odd, the parity of the product is the parity of (o + n).
- If n is even, o + n is odd → product = odd·odd = **odd**.
- If n is odd, o + n is even → product = odd·even = **even**.

So the expression is odd exactly when n is even.

Ruling out others: A fails (n odd gives even), B fails (n even gives odd), C says "even only if n is even" — but it's even when n is *odd*, false. D says "odd only if n is odd" — false, it's odd when n is even. E: "odd only if n is even" — true.

ANSWER 1: E

# Problem 2:
The redemption days are 0, 10, 20, 30, 40, 50 days after the first.

Reduce mod 7 (days of week shift): 0, 3, 6, 2, 5, 1.

These residues are {0, 1, 2, 3, 5, 6} — every residue except **4**. So the only day of the week never hit is the day 4 days after the starting day. For no coupon to land on Sunday, Sunday must be that missed day:

start + 4 ≡ Sunday.

Counting back 4 days from Sunday: Saturday, Friday, Thursday, **Wednesday**.

Check: Wed → +10 Sat → +20 Tue → +30 Fri → +40 Mon → +50 Thu. No Sundays. ✓

ANSWER 2: C

# Problem 3:
The product is (3/2)·(4/3)·(5/4)·(6/5)·…·(a/b), where each numerator is one more than its denominator, so b = a − 1.

Telescoping: everything cancels except the last numerator and first denominator:
a/2 = 9 → a = 18, so b = 17.

a + b = 18 + 17 = 35.

(Choice E, 37, would come from mistakenly taking a = 19.)

ANSWER 3: D

# Problem 4:
Total value of all 60 stamps: 20(5 + 10 + 25) = 20·40 = 800 cents.

We need 710 cents, so we must *remove* stamps worth exactly 800 − 710 = 90 cents, using as **few** stamps as possible (to keep the most stamps).

Can 90 be made with 4 or fewer stamps? Max with 4 stamps is 100; possible 4-stamp totals near 90: 100, 85, 80, 75, 70… none equal 90. Three or fewer is at most 75. So 4 or fewer is impossible.

With 5 stamps: 25 + 25 + 25 + 10 + 5 = 90 ✓

So remove 5 stamps: 60 − 5 = 55 stamps.

Check: 17(25) + 19(10) + 19(5) = 425 + 190 + 95 = 710 cents, using 17 + 19 + 19 = 55 stamps.

ANSWER 4: E

# Problem 5:
CDCD as a number = CD·100 + CD = CD·101.

So ABA × CD = 101 × CD, and since CD ≠ 0, we get ABA = 101.

Thus A = 1, B = 0 (and indeed A, B, C, D distinct works, e.g. 101 × 23 = 2323).

A + B = 1 + 0 = 1.

ANSWER 5: A

# Problem 6:
Jonas total = 40, and 20 of those were girls, so boys from Jonas = 40 − 20 = 20.

Total boys = 52, so boys from Clay = 52 − 20 = 32.

(Check: girls from Clay = 48 − 20 = 28; 32 + 28 = 60 = Clay total ✓.)

Choice C (40) is the Jonas total, and E (52) is all boys — traps.

ANSWER 6: B

# Problem 7:
Pair each term: 1901 − 101 = 1800, 1902 − 102 = 1800, …, each pair differs by 1800.

Number of terms: 1993 − 1901 + 1 = 93.

Total = 93 × 1800 = 167,400.

ANSWER 7: A

# Problem 8:
Look at 7ⁿ mod 100:
- 7¹ = 07
- 7² = 49
- 7³ = 343 → 43
- 7⁴ = 2401 → 01
- then the pattern repeats with period 4.

2011 ÷ 4 = 502 remainder 3, so 7²⁰¹¹ ≡ 7³ ≡ 43 (mod 100).

Last two digits are 43, so the tens digit is 4.

ANSWER 8: D

# Problem 9:
Sum of the 10 numbers = 10 × 10 = 100.

To make one number as large as possible, make the other nine as small as possible while distinct and positive: 1 + 2 + … + 9 = 45.

Largest = 100 − 45 = 55.

(90 or 91 would require repeats or a zero, which isn't allowed.)

ANSWER 9: C

# Problem 10:
Total student-days:
- Allen: 7 × 3 = 21
- Balboa: 4 × 5 = 20
- Carver: 5 × 9 = 45

Total = 86 student-days.

Daily wage = 774 ÷ 86 = $9.00 per student-day.

Balboa earnings = 20 × 9 = $180.00.

(A is one day's pay for one student; E is Carver's total.)

ANSWER 10: C

Problem 1:

We need determine the parity of

\[
o^2+no.
\]

Factor out \(o\):

\[
o^2+no=o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on \(o+n\).

- If \(n\) is even, then odd \(+\) even \(=\) odd, so \(o+n\) is odd. Thus odd \(\cdot\) odd is odd.
- If \(n\) is odd, then odd \(+\) odd \(=\) even, so the product is even.

Therefore, the expression is odd only when \(n\) is even.

- A is wrong because the expression is not always odd.
- B is wrong because it is odd when \(n\) is even.
- C reverses the condition for being even.
- D reverses the condition for being odd.

ANSWER 1: E

Problem 2:

Isabella uses a coupon every 10 days. Since

\[
10\equiv 3\pmod 7,
\]

each coupon date moves forward by 3 days of the week.

If the first coupon is redeemed on day \(d\), the six redemption days are:

\[
d,\ d+3,\ d+6,\ d+2,\ d+5,\ d+1.
\]

These are six different days of the week. The only day missing is

\[
d+4.
\]

Since none of the six coupon dates is Sunday, the missing day must be Sunday. Thus,

\[
d+4=\text{Sunday}.
\]

Four days before Sunday is Wednesday.

Indeed, beginning on Wednesday gives:

\[
\text{Wednesday, Saturday, Tuesday, Friday, Monday, Thursday}.
\]

Sunday is the only missing day.

ANSWER 2: C

Problem 3:

The product is

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9.
\]

Almost every numerator cancels with a denominator:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab
=\frac a2.
\]

So

\[
\frac a2=9.
\]

Therefore,

\[
a=18.
\]

The final fraction must be

\[
\frac{18}{17},
\]

so \(b=17\). Therefore,

\[
a+b=18+17=35.
\]

Choices 11, 13, and 17 are too small, while 37 would correspond to a different ending fraction.

ANSWER 3: D

Problem 4:

Let

- \(x\) = number of 5-cent stamps,
- \(y\) = number of 10-cent stamps,
- \(z\) = number of 25-cent stamps.

In units of 5 cents, the postage total is

\[
\frac{710}{5}=142.
\]

Thus,

\[
x+2y+5z=142.
\]

We want to maximize

\[
x+y+z.
\]

Since there are at most 20 five-cent stamps, \(x\leq 20\). Therefore,

\[
2y+5z=142-x\geq 142-20=122.
\]

Since \(y\leq20\), if \(z\leq16\), then

\[
2y+5z\leq 2(20)+5(16)=40+80=120,
\]

which is not enough. So \(z\geq17\).

If \(z=17\), then

\[
2y+5(17)\geq122,
\]

so

\[
2y+85\geq122,
\]

\[
2y\geq37,
\]

and hence \(y\geq19\).

Now use

\[
x+y+z=142-y-4z.
\]

Since \(y\geq19\) and \(z\geq17\),

\[
x+y+z\leq142-19-4(17)
=142-19-68
=55.
\]

This maximum is attainable with

\[
x=19,\qquad y=19,\qquad z=17.
\]

Check:

\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So 55 stamps can be used. Choices 45, 46, 51, and 54 are possible totals to consider, but 55 is the greatest possible.

ANSWER 4: E

Problem 5:

We are told

\[
ABA\times CD=CDCD.
\]

The number \(ABA\) is

\[
100A+10B+A=101A+10B.
\]

The number \(CDCD\) is

\[
1000C+100D+10C+D.
\]

Group the \(C\)'s and \(D\)'s:

\[
CDCD=1010C+101D=101(10C+D)=101\cdot CD.
\]

Thus the equation becomes

\[
ABA\cdot CD=101\cdot CD.
\]

Since \(CD\) is a two-digit number and is not zero, divide both sides by \(CD\):

\[
ABA=101.
\]

Therefore,

\[
A=1,\qquad B=0.
\]

So

\[
A+B=1+0=1.
\]

For example, \(101\times23=2323\), and the digits \(1,0,2,3\) are all different.

ANSWER 5: A

Problem 6:

There are 40 students from Jonas Middle School, and 20 of those are girls.

Therefore the number of boys from Jonas is

\[
40-20=20.
\]

There are 52 boys altogether. Therefore the number of boys from Clay Middle School is

\[
52-20=32.
\]

The other choices do not correctly account for the 20 Jonas girls and the total of 52 boys.

ANSWER 6: B

Problem 7:

There are

\[
1993-1901+1=93
\]

terms in the first sum.

There are also

\[
193-101+1=93
\]

terms in the second sum.

Pair corresponding terms:

\[
1901-101=1800,
\]

\[
1902-102=1800,
\]

and so on, through

\[
1993-193=1800.
\]

Thus there are 93 differences of 1800:

\[
93\cdot1800=167400.
\]

Therefore the value is

\[
167,\!400.
\]

ANSWER 7: A

Problem 8:

We need the tens digit of

\[
7^{2011}.
\]

It is enough to examine powers of 7 modulo 100:

\[
7^1=7,
\]

\[
7^2=49,
\]

\[
7^3=343,
\]

whose last two digits are 43, and

\[
7^4=2401,
\]

whose last two digits are 01.

Thus the last two digits repeat every 4 powers because

\[
7^4\equiv1\pmod{100}.
\]

Now,

\[
2011\div4
\]

leaves remainder 3, since

\[
2011=4(502)+3.
\]

Therefore,

\[
7^{2011}
\]

has the same last two digits as

\[
7^3=343,
\]

which are 43. The tens digit is 4.

ANSWER 8: D

Problem 9:

The average of 10 different positive whole numbers is 10, so their total sum is

\[
10\cdot10=100.
\]

To make one number as large as possible, make the other nine numbers as small as possible.

The nine smallest different positive whole numbers are

\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is

\[
1+2+3+4+5+6+7+8+9=45.
\]

So the largest possible tenth number is

\[
100-45=55.
\]

This works because the ten numbers can be

\[
1,2,3,4,5,6,7,8,9,55,
\]

which are all different and have total 100.

A value such as 90 or 91 would leave too small a total for nine distinct positive whole numbers.

ANSWER 9: C

Problem 10:

Compute the total number of student-days worked.

Allen students worked:

\[
7\cdot3=21
\]

student-days.

Balboa students worked:

\[
4\cdot5=20
\]

student-days.

Carver students worked:

\[
5\cdot9=45
\]

student-days.

So the total number of student-days is

\[
21+20+45=86.
\]

The total pay was \$774, so the pay per student per day was

\[
\frac{774}{86}=9.
\]

Balboa students worked 20 student-days, so they earned

\[
20\cdot9=180.
\]

Thus the Balboa students earned \$180.00 altogether.

ANSWER 10: C

Problem 1:
We are asked about the parity of $o^2+no$ where $o$ is odd and $n$ is any whole number.

Factor:
$$o^2+no = o(o+n).$$

$o$ is odd, so $o^2 = o\cdot o$ is odd times odd = odd.

Hence parity is odd + $n\cdot o$.
$n\cdot o$ has the same parity as $n$, because $o$ is odd: if $n$ is even, $n o$ is even; if $n$ is odd, $n o$ is odd.

So:
- if $n$ even: odd + even = odd.
- if $n$ odd: odd + odd = even.

Thus the number is odd iff $n$ is even, and even iff $n$ is odd.

Check choices:
A. always odd - false, e.g. $o=1,n=1$ gives $1+1=2$ even.
B. always even - false, e.g. $o=1,n=2$ gives $1+2=3$ odd.
C. it is even only if $n$ is even - means even $\implies$ $n$ even. False, even happens when $n$ is odd.
D. it is odd only if $n$ is odd - means odd $\implies$ $n$ odd. False, odd happens when $n$ is even.
E. it is odd only if $n$ is even - means odd $\implies$ $n$ even. True by above.

ANSWER 1: E

Problem 2:
Isabella redeems on days $d, d+10, d+20, d+30, d+40, d+50$.
No date is a Sunday.

Modulo 7, $10\equiv 3$. So successive dates advance by 3 weekdays.

Starting from residue $r$, the residues are:
$$r,\ r+3,\ r+6,\ r+9=r+2,\ r+12=r+5,\ r+15=r+1 \pmod 7.$$
These 6 are all distinct, because adding 3 mod 7 cycles through all 7 residues. Continuing one more step would give $r+18=r+4$.

Thus the 6 dates occupy 6 distinct days of the week, missing exactly the 7th residue $r+4$.

Since none of the 6 is Sunday, the missing one must be Sunday. So:
$$r+4 \equiv \text{Sunday}.$$

Hence $r$ is 4 days before Sunday, i.e. Wednesday.

Check: Wed $\to$ Wed+3=Sat $\to$ +3=Tue $\to$ +3=Fri $\to$ +3=Mon $\to$ +3=Thu. That is Wed,Sat,Tue,Fri,Mon,Thu - no Sunday. Any other start would include Sunday.

ANSWER 2: C

Problem 3:
We interpret $3\ 2\cdot 4\ 3\cdots$ as fractions:
$$\frac32 \cdot \frac43 \cdot \frac54 \cdot \frac65 \cdots \frac{a}{b}=9.$$

The pattern is numerator $3,4,5,6,\dots$ and denominator $2,3,4,5,\dots$, i.e. term $k/(k-1)$. The last term is $a/b$, so we must have $b=a-1$ for the pattern to continue.

Then the product telescopes:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}
= \frac{a}{2},$$
because all intermediate $3,4,\dots,a-1$ cancel.

Set equal to 9:
$$\frac{a}{2}=9 \implies a=18.$$
Then $b=a-1=17$.

Sum $a+b=18+17=35$.

This rules out A,B,C,E.

ANSWER 3: D

Problem 4:
Let $x,y,z$ be numbers of 5-cent, 10-cent, 25-cent stamps used. $0\le x,y,z\le 20$ integers.
Value $710$ cents:
$$5x+10y+25z=710.$$
Divide by 5:
$$x+2y+5z=142. \tag{1}$$
Maximize $N=x+y+z$.

From (1), $x=142-2y-5z$, so $0\le x\le20$ gives:
$$122\le 2y+5z\le 142.$$
Also $N=x+y+z=142-y-4z$, so $y+4z=142-N$.

To use many stamps we want small denominations. Try $x=20,y=20$ fails to give integer $z$. Search systematically.

Since $y\le20$, to satisfy $2y+5z\ge122$ we need $z$ large. If $z\le16$, then $2y+5z\le40+80=120<122$, impossible. So $z\ge17$.

Try $z=17,y=19$: then $2y+5z=38+85=123$, so $x=142-123=19$, within $0-20$. This gives:
$$19\cdot5+19\cdot10+17\cdot25=95+190+425=710,$$
with $N=19+19+17=55$ stamps. So 55 is achievable.

Can we do 56 or more? Let $N\ge56$.
Then $y+4z=142-N\le86$ and $x=N-y-z\le20$ gives $y+z\ge N-20$.
Subtract: $3z=(y+4z)-(y+z)\le (142-N)-(N-20)=162-2N$.
So $z\le(162-2N)/3$.
Also $y=142-N-4z\le20$ gives $z\ge(122-N)/4$.

For $N=56$: $z\ge\lceil66/4\rceil=17$ and $z\le\lfloor50/3\rfloor=16$, contradiction. For $N>56$ the interval is even emptier; in general need $(122-N)/4\le(162-2N)/3$, i.e. $5N\le282$, so $N\le56.4$. Thus $N\ge57$ impossible, and $N=56$ impossible by the integer contradiction.

Hence 55 is greatest possible.

ANSWER 4: E

Problem 5:
We have $ABA\times CD = CDCD$, where $A,B,C,D$ are different digits, $A\ne0$, $C\ne0$.

Write $CD$ as the 2-digit number $10C+D$.
$CDCD$ as a 4-digit number is:
$$1000C+100D+10C+D =1010C+101D=101(10C+D)=101\cdot CD.$$

So:
$$ABA \times CD = 101\times CD.$$
Since $CD\ne0$, divide:
$$ABA =101.$$

Thus $A=1$, $B=0$, and $A+B=1$.

This is consistent with distinctness: we can choose e.g. $C=2,D=3$, all of $1,0,2,3$ different, and $101\times23=2323$. So valid.

Other sums are ruled out.

ANSWER 5: A

Problem 6:
Total 100: 52 boys, 48 girls. 40 from Jonas, 60 from Clay.

20 of the girls are from Jonas. So girls from Clay = total girls $-$ girls from Jonas:
$$48-20=28.$$

Clay total 60 = boys from Clay + girls from Clay, so:
$$\text{boys from Clay}=60-28=32.$$

Check: boys from Jonas $=52-32=20$, Jonas total $=20+20=40$. Works.

ANSWER 6: B

Problem 7:
Compute $(1901+\cdots+1993)-(101+\cdots+193)$.

Both sums have $1993-1901+1=93$ terms and $193-101+1=93$ terms.

Pair termwise: for $k=0,\dots,92$,
$$(1901+k)-(101+k)=1800.$$

So difference $=93\times1800$.
$93\times18=1674$, so $93\times1800=167{,}400$.

Other choices come from miscounting terms or arithmetic error.

ANSWER 7: A

Problem 8:
Find tens digit of $7^{2011}$, i.e. $7^{2011}\pmod{100}$.

Compute powers mod 100:
$7^1=7$,
$7^2=49$,
$7^3=343\equiv43$,
$7^4\equiv43\times7=301\equiv1$.

Since $7^4\equiv1\pmod{100}$, the last two digits repeat with period 4.

$2011 = 4\cdot502+3$, so:
$$7^{2011}=(7^4)^{502}\cdot7^3\equiv1^{502}\cdot43\equiv43\pmod{100}.$$

Thus the number ends $\dots43$, tens digit $4$.

ANSWER 8: D

Problem 9:
10 different positive whole numbers have mean 10, so sum $=10\times10=100$.

To make one number as large as possible, make the other nine as small as possible while distinct and positive.

Smallest 9 distinct positives are $1,2,3,4,5,6,7,8,9$, sum $=45$.

Then the tenth number $=100-45=55$, which is distinct from the others and valid. Any larger value would force the other nine to sum to $<45$, impossible with distinct positives.

So largest possible is 55.

ANSWER 9: C

Problem 10:
Count student-days (one student working one day).

Allen: $7$ students $\times$ $3$ days $=21$.
Balboa: $4\times5=20$.
Carver: $5\times9=45$.
Total $=21+20+45=86$ student-days.

Total pay $\$774$ for $86$ student-days at equal rate:
$$\text{rate}=774/86=9 \text{ dollars per student-day}.$$

Bal

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that $o$ is an odd whole number and $n$ is any whole number. We must determine which statement about $o^2 + no$ is always true.

**Step-by-Step Solution:**
1. Factor the expression:
   $$o^2 + no = o(o + n)$$
2. Since $o$ is an odd whole number, the parity of the product $o(o + n)$ is entirely determined by the parity of $(o + n)$:
   - If $n$ is **odd**, then $\text{odd} + \text{odd} = \text{even}$, so $o(o+n)$ is $\text{odd} \times \text{even} = \textbf{even}$.
   - If $n$ is **even**, then $\text{odd} + \text{even} = \text{odd}$, so $o(o+n)$ is $\text{odd} \times \text{odd} = \textbf{odd}$.
3. Therefore, the expression $o^2 + no$ is odd if and only if $n$ is even. 
4. This matches the statement: "it is odd only if $n$ is even."

ANSWER 1: E

---

### Problem 2:
**Restatement:** Isabella redeems 6 coupons, one every 10 days. Pete's is closed on Sundays, and none of the 6 dates falls on a Sunday. We need to find the day of the week she redeems her first coupon.

**Step-by-Step Solution:**
1. There are 7 days in a week. Moving forward by 10 days advances the day of the week by:
   $$10 \equiv 3 \pmod 7$$
2. Let $d$ represent the day of the week of the first coupon ($0 = \text{Sunday}, 1 = \text{Monday}, \dots, 6 = \text{Saturday}$).
3. The days of the week for the 6 coupons will be:
   - Coupon 1: $d$
   - Coupon 2: $d + 3$
   - Coupon 3: $d + 6$
   - Coupon 4: $d + 9 \equiv d + 2 \pmod 7$
   - Coupon 5: $d + 12 \equiv d + 5 \pmod 7$
   - Coupon 6: $d + 15 \equiv d + 1 \pmod 7$
4. The six days of the week visited are $d, d+1, d+2, d+3, d+5, d+6$. Notice that the only day of the week omitted from this list is $d+4$.
5. Since no circled date falls on a Sunday, Sunday must be the omitted day:
   $$d + 4 \equiv 0 \pmod 7 \implies d \equiv 3 \pmod 7$$
6. Day 3 corresponds to **Wednesday**.
   - Checking: Wed $\to$ Sat $\to$ Tue $\to$ Fri $\to$ Mon $\to$ Thu. None of these is Sunday.

ANSWER 2: C

---

### Problem 3:
**Restatement:** We are given the telescoping product $\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdots \frac{a}{b} = 9$. We must find the sum $a + b$.

**Step-by-Step Solution:**
1. Each fraction is of the form $\frac{k+1}{k}$, where the numerator of each term cancels with the denominator of the subsequent term:
   $$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{a}{b} = \frac{a}{2}$$
   where $b = a - 1$.
2. We are given that this product equals 9:
   $$\frac{a}{2} = 9 \implies a = 18$$
3. Since $b = a - 1$, we have $b = 17$.
4. The sum of $a$ and $b$ is:
   $$a + b = 18 + 17 = 35$$

ANSWER 3: D

---

### Problem 4:
**Restatement:** Nicolas has 20 stamps each of 5¢, 10¢, and 25¢. We want to find the greatest number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-Step Solution:**
1. Calculate the total value and count of all stamps:
   - $20 \times 5¢ = 100¢$
   - $20 \times 10¢ = 200¢$
   - $20 \times 25¢ = 500¢$
   - Total value $= 100 + 200 + 500 = 800¢$ using 60 stamps.
2. To use the **maximum** number of stamps to make 710¢, Nicolas must **minimize** the number of stamps removed to subtract:
   $$800¢ - 710¢ = 90¢$$
3. To remove the fewest stamps totaling 90¢, we try to use the largest denomination stamps (25¢):
   - With 3 stamps of 25¢ ($75¢$), the remaining amount needed is $15¢$.
   - Since Nicolas has only 5¢, 10¢, and 25¢ stamps, 15¢ cannot be made with a single stamp. It requires at least two stamps ($10¢ + 5¢$).
   - Thus, 90¢ can be formed with $3 \times 25¢ + 1 \times 10¢ + 1 \times 5¢$, which uses $3 + 1 + 1 = 5$ stamps.
   - Can 90¢ be formed using 4 stamps? The maximum value of 4 stamps less than or equal to 90¢ would be $3 \times 25¢ + 10¢ = 85¢ \ne 90¢$. Thus, 4 stamps is impossible.
4. The minimum number of stamps to remove is 5.
5. Therefore, the maximum number of stamps used is:
   $$60 - 5 = 55$$

ANSWER 4: E

---

### Problem 5:
**Restatement:** In the multiplication $ABA \times CD = CDCD$, $A, B, C, D$ are distinct digits. We need to find $A + B$.

**Step-by-Step Solution:**
1. Express $CDCD$ in terms of the two-digit number $CD$:
   $$CDCD = CD \times 100 + CD = CD \times 101$$
2. Substitute this into the equation:
   $$ABA \times CD = 101 \times CD$$
3. Since $CD$ is a non-zero two-digit number, divide both sides by $CD$:
   $$ABA = 101$$
4. Comparing digits, we have $A = 1$ and $B = 0$.
5. Thus:
   $$A + B = 1 + 0 = 1$$

ANSWER 5: A

---

### Problem 6:
**Restatement:** Out of 100 students at a camp, 52 are boys and 48 are girls. Also, 40 are from Jonas Middle School and 60 from Clay Middle School. Given that 20 of the girls are from Jonas, find how many boys are from Clay.

**Step-by-Step Solution:**
1. Find the number of girls from Clay Middle School:
   $$\text{Total girls} - \text{Girls from Jonas} = 48 - 20 = 28\text{ girls from Clay}$$
2. Find the number of boys from Clay Middle School:
   $$\text{Total students from Clay} - \text{Girls from Clay} = 60 - 28 = 32\text{ boys from Clay}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** Compute the value of:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-Step Solution:**
1. Pair each term in the first sum with the corresponding term in the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
2. Each paired difference is equal to:
   $$1900 + k - (100 + k) = 1800$$
3. Find the number of terms: from $k = 1$ to $k = 93$, there are:
   $$93 - 1 + 1 = 93\text{ terms}$$
4. Multiply the difference by the number of terms:
   $$93 \times 1800 = 93 \times 18 \times 100 = 1674 \times 100 = 167,400$$

ANSWER 7: A

---

### Problem 8:
**Restatement:** Find the tens digit of $7^{2011}$.

**Step-by-Step Solution:**
1. Find the pattern of the last two digits of powers of 7 by computing $7^n \pmod{100}$:
   - $7^1 = 07$
   - $7^2 = 49$
   - $7^3 = 343 \equiv 43 \pmod{100}$
   - $7^4 = 43

# Problem 1:
We must decide when $o^2 + no$ is odd or even, given $o$ is odd and $n$ is any whole number.

**Key facts:** odd × odd = odd, so $o^2$ is **odd**. Since $o$ is odd, multiplying by $o$ does not change $n$'s parity, so $no$ has the **same parity as $n$**.

**Case 1: $n$ even.** Then $no$ is even, so $o^2 + no = \text{odd} + \text{even} = \textbf{odd}$.
**Case 2: $n$ odd.** Then $no$ is odd, so $o^2 + no = \text{odd} + \text{odd} = \textbf{even}$.

So the number is odd exactly when $n$ is even. In particular, **if the number is odd, $n$ must be even**.

Ruling out traps: (A) fails since $o=1, n=1$ gives $1+1=2$, even; (B) fails since $n=2$ gives $1+2=3$, odd; (C) and (D) state the exact opposite of the truth. Only (E) matches.

ANSWER 1: E

# Problem 2:
The 6 redemption dates are 10 days apart, so their days of the week shift by $10 \bmod 7 = 3$ each time.

If the first date falls on weekday $W$, the six dates fall on (mod 7):
$$W,\; W+3,\; W+6,\; W+2,\; W+5,\; W+1$$
That is **six of the seven weekdays**, and the only weekday missed is $W+4$.

Since no circled date is a Sunday, the one missed weekday must be Sunday:
$$W + 4 = \text{Sunday} \implies W = \text{Wednesday}.$$

**Check:** Starting Wednesday, the dates fall on Wed, Sat, Tue, Fri, Mon, Thu — no Sunday. (Trap: starting Thursday gives Thu, Sun, ... — fails immediately.)

ANSWER 2: C

# Problem 3:
The product is a chain of fractions (the bars were lost in formatting):
$$\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdots\frac{a}{b} = 9.$$

Each numerator cancels the denominator before it (e.g., $\frac{3}{2}\cdot\frac{4}{3} = \frac{4}{2}$), and each fraction has the form $\frac{k+1}{k}$, so $b = a-1$. Everything telescopes:
$$\text{Product} = \frac{a}{2} = 9 \implies a = 18, \quad b = 17.$$
Then $a + b = 18 + 17 = 35$.

Trap check: $a+b=17$ would mean $a=9$, giving product $\tfrac92$, not $9$; only $35$ fits the pattern.

ANSWER 3: D

# Problem 4:
Work in units of 5¢: we need total value $710¢ = 142$ units, using stamps worth $1, 2, 5$ units (i.e., 5¢, 10¢, 25¢), at most 20 of each. To **maximize the count**, use as many cheap stamps as possible.

**Building a 55-stamp solution:** Take nineteen 1's and nineteen 2's: value $19 + 38 = 57$ units with 38 stamps. Remaining value $142 - 57 = 85 = 17\times 5$: seventeen 5's.
$$19\times 5¢ + 19\times 10¢ + 17\times 25¢ = 95 + 190 + 425 = 710¢ = \$7.10.\; ✓$$
That's $19+19+17 = 55$ stamps, all counts $\le 20$. ✓

**Why 56 is impossible:** with 56 stamps, at least $56 - 40 = 16$ must be 25¢ stamps (only 40 cheap ones exist). Let $z$ = number of 25¢ stamps:
- $z=16$: max value $= 100 + 200 + 400 = 700 < 710$. ✗
- $z=17$: need $5x+10y=285$, $x+y=39 \Rightarrow x=21>20$. ✗
- $z=18$: $x=24>20$. ✗  $z=19$: $x=27>20$. ✗  $z=20$: $x=30>20$. ✗

Traps: 54 comes from $20×5+16×10+18×25 = 710$; 51 from $20×5+11×10+20×25 = 710$ — both valid but not maximal.

ANSWER 4: E

# Problem 5:
Write $CDCD$ as a number:
$$CDCD = 1000C + 100D + 10C + D = 101(10C + D) = 101 \times CD.$$
The given equation is
$$ABA \times CD = CDCD = 101 \times CD.$$
Since $CD \neq 0$ (it's a two-digit number), we can divide both sides by $CD$:
$$ABA = 101.$$
So $A = 1$ and $B = 0$, giving $A + B = 1$.

**Consistency check:** $C, D$ just need to be distinct from $0, 1$ and each other, e.g., $101 \times 23 = 2323$. ✓ (Trap: guessing digits without using the factorization 101 wastes time or gives 9.)

ANSWER 5: A

# Problem 6:
Set up a two-way table with 100 total students.

- Girls: 48 total, and 20 girls are from Jonas $\Rightarrow$ girls from Clay $= 48 - 20 = 28$.
- Jonas has 40 students, of whom 20 are girls $\Rightarrow$ boys from Jonas $= 40 - 20 = 20$.
- Boys: 52 total $\Rightarrow$ boys from Clay $= 52 - 20 = \textbf{32}$.

**Check:** Jonas $20+20=40$ ✓; Clay $28+32=60$ ✓; boys $20+32=52$ ✓; girls $20+28=48$ ✓; total $100$ ✓. (Trap: 20 counts the boys from Jonas, not Clay.)

ANSWER 6: B

# Problem 7:
Both sums have the same number of terms: $1901$ to $1993$ has $1993-1901+1 = 93$ terms, and $101$ to $193$ has $193-101+1 = 93$ terms.

Pair them up: each term in the first sum is exactly $1800$ more than its partner ($1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$). So the difference is
$$93 \times 1800 = 167{,}400.$$

**Check via formulas:** $\frac{93(1901+1993)}{2} - \frac{93(101+193)}{2} = 181{,}071 - 13{,}671 = 167{,}400$. ✓ (Trap: 181,071 is the first sum alone — forgetting to subtract.)

ANSWER 7: A

# Problem 8:
Only the tens digit matters, so track the last two digits of $7^{k}$, which cycle with period 4:
$$7^1 = 07,\quad 7^2 = 49,\quad 7^3 = 343 \to 43,\quad 7^4 = 2401 \to 01,\quad 7^5 \to 07, \ldots$$

Find where $2011$ falls in the cycle:
$$2011 = 4 \times 502 + 3,$$
so $7^{2011}$ ends in the same two digits as $7^3$, namely $\ldots 43$.

The tens digit is $\textbf{4}$. (Trap: $7^{2012}$, a multiple of 4, ends in $01$, tens digit 0 — choice A; that would require the exponent to be divisible by 4.)

ANSWER 8: D

# Problem 9:
The ten **different** positive whole numbers have average 10, so their sum is $10 \times 10 = 100$.

To make one number as large as possible, make the other nine as small as possible. Since they must be **distinct positive** whole numbers, the smallest possible set is $1, 2, 3, \ldots, 9$, with sum
$$1+2+\cdots+9 = 45.$$
So the largest possible value is
$$100 - 45 = 55.$$

**Check:** $1,2,\ldots,9,55$ are ten different positive whole numbers with sum $100$ and average $10$. ✓ (Trap: 91 arises if you ignore "different" and use nine 1's; but the numbers must be distinct.)

ANSWER 9: C

# Problem 10:
Count student-days (the unit being paid):
- Allen: $7 \times 3 = 21$ student-days
- Balboa: $4 \times 5 = 20$ student-days
- Carver: $5 \times 9 = 45$ student-days

Total: $21 + 20 + 45 = 86$ student-days, paid $\$774$ altogether. So the daily rate is
$$774 \div 86 = 9 \text{ dollars per student-day}.$$
Balboa's students worked 20 student-days, so they earned
$$20 \times 9 = 18

Problem 1:  
We need the parity of \(o^2+no\), where \(o\) is odd and \(n\) is any whole number.

Factor:
\[
o^2+no=o(o+n).
\]

Since \(o\) is odd:

- If \(n\) is even, then \(o+n\) is odd, so \(o(o+n)\) is odd.
- If \(n\) is odd, then \(o+n\) is even, so \(o(o+n)\) is even.

Thus the expression is odd exactly when \(n\) is even.

Check the choices:
- A and B are false because the parity depends on \(n\).
- C says “even only if \(n\) is even,” but when the expression is even, \(n\) is actually odd.
- D says “odd only if \(n\) is odd,” but when the expression is odd, \(n\) is even.
- E says “odd only if \(n\) is even,” which is true.

ANSWER 1: E

Problem 2:  
Isabella redeems coupons 10 days apart. Since
\[
10\equiv 3 \pmod 7,
\]
the six circled days are the first day plus
\[
0,3,6,9,12,15
\]
days, which modulo 7 gives offsets
\[
0,3,6,2,5,1.
\]

So the six circled days are all weekdays except the one 4 days after the first day. Since none of the circled days is a Sunday, Sunday must be the un-circled day: 4 days after the first day.

Therefore the first coupon is 4 days before Sunday, which is Wednesday.

Check: Wednesday, Saturday, Tuesday, Friday, Monday, Thursday — no Sunday.

ANSWER 2: C

Problem 3:  
The product is the telescoping product of fractions:
\[
\frac32\cdot \frac43\cdot \frac54\cdot \frac65\cdots \frac{a}{b}=9.
\]

The numerators cancel the next denominators:
\[
\frac32\cdot \frac43=\frac42,
\]
\[
\frac42\cdot \frac54=\frac52,
\]
and continuing, the whole product simplifies to
\[
\frac{a}{2}.
\]

So
\[
\frac{a}{2}=9 \implies a=18.
\]

By the pattern, the denominator is one less than the numerator, so
\[
b=17.
\]

Thus
\[
a+b=18+17=35.
\]

ANSWER 3: D

Problem 4:  
Nicolas has 20 of each type of stamp, so he has
\[
20+20+20=60
\]
stamps total. Their total value is
\[
20(5+10+25)=20(40)=800
\]
cents.

He wants postage of
\[
\$7.10=710
\]
cents, so the stamps he does **not** use must total
\[
800-710=90
\]
cents.

To maximize the number of stamps he uses, we minimize the number of stamps he leaves unused.

Can 4 unused stamps total 90 cents? Let \(x,y,z\) be the numbers of 25-, 10-, and 5-cent unused stamps. Then
\[
x+y+z=4
\]
and
\[
25x+10y+5z=90.
\]

Divide by 5:
\[
5x+2y+z=18.
\]

Subtract \(x+y+z=4\):
\[
4x+y=14.
\]

Testing possible \(x=0,1,2,3,4\):

- \(x=0\Rightarrow y=14\), impossible.
- \(x=1\Rightarrow y=10\), impossible.
- \(x=2\Rightarrow y=6\), impossible.
- \(x=3\Rightarrow y=2\), then \(z=4-3-2=-1\), impossible.
- \(x=4\Rightarrow y=-2\), impossible.

So 4 unused stamps cannot total 90 cents.

Can 5 unused stamps total 90 cents? Yes:
\[
25+25+25+10+5=90.
\]

Then he uses
\[
60-5=55
\]
stamps. For example, he could use 19 five-cent stamps, 19 ten-cent stamps, and 17 twenty-five-cent stamps:
\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So the greatest possible number of stamps is 55. Choice D, 54, is possible but not greatest.

ANSWER 4: E

Problem 5:  
We are told
\[
ABA\times CD=CDCD,
\]
where \(A,B,C,D\) are different digits.

Write the numbers in place-value form:
\[
ABA=101A+10B,
\]
\[
CD=10C+D,
\]
\[
CDCD=1000C+100D+10C+D=101(10C+D).
\]

So
\[
(101A+10B)(10C+D)=101(10C+D).
\]

Since \(CD\) is a two-digit number, \(10C+D\neq 0\), so we can divide both sides by \(10C+D\):
\[
101A+10B=101.
\]

The only digit solution is
\[
A=1,\quad B=0.
\]

Therefore
\[
A+B=1+0=1.
\]

ANSWER 5: A

Problem 6:  
There are 48 girls total, and 20 of the girls are from Jonas Middle School. So the number of girls from Clay Middle School is
\[
48-20=28.
\]

There are 60 students total from Clay Middle School, so the number of boys from Clay is
\[
60-28=32.
\]

ANSWER 6: B

Problem 7:  
Pair the terms:
\[
(1901+1902+\cdots+1993)-(101+102+\cdots+193)
\]
\[
=(1901-101)+(1902-102)+\cdots+(1993-193).
\]

Each pair is
\[
1800.
\]

Now count the number of terms:
\[
193-101+1=93.
\]

So the value is
\[
93\cdot 1800=167400.
\]

ANSWER 7: A

Problem 8:  
We need the tens digit of \(7^{2011}\), so we need the last two digits.

Powers of 7 modulo 100 cycle:
\[
7^1=07,
\]
\[
7^2=49,
\]
\[
7^3=343\equiv 43,
\]
\[
7^4=2401\equiv 01.
\]

Then the cycle repeats:
\[
07,\ 49,\ 43,\ 01.
\]

Since
\[
2011\equiv 3 \pmod 4,
\]
we have
\[
7^{2011}\equiv 7^3\equiv 43 \pmod{100}.
\]

The last two digits are 43, so the tens digit is 4.

ANSWER 8: D

Problem 9:  
The average of 10 different positive whole numbers is 10, so their sum is
\[
10\cdot 10=100.
\]

To make one number as large as possible, make the other 9 numbers as small as possible while still being different positive whole numbers. The smallest 9 distinct positive whole numbers are
\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is
\[
1+2+\cdots+9=45.
\]

So the largest number can be
\[
100-45=55.
\]

This is possible: \(1,2,3,4,5,6,7,8,9,55\) are all different positive whole numbers.

ANSWER 9: C

Problem 10:  
First count total student-days worked:

- Allen: \(7\cdot 3=21\) student-days
- Balboa: \(4\cdot 5=20\) student-days
- Carver: \(5\cdot 9=45\) student-days

Total:
\[
21+20+45=86
\]
student-days.

The total payment is \$774, so the daily rate is
\[
774\div 86=9
\]
dollars per student-day.

Balboa students worked 20 student-days, so they earned
\[
20\cdot 9=180
\]
dollars.

ANSWER 10: C

# Problem 1:
We need a statement always true about o² + no, where o is odd.

Factor: o² + no = o(o + n).

Since o is odd, the parity of the product is the parity of (o + n).
- If n is even, o + n is odd → product = odd·odd = **odd**.
- If n is odd, o + n is even → product = odd·even = **even**.

So the expression is odd exactly when n is even.

Ruling out others: A fails (n odd gives even), B fails (n even gives odd), C says "even only if n is even" — but it's even when n is *odd*, false. D says "odd only if n is odd" — false, it's odd when n is even. E: "odd only if n is even" — true.

ANSWER 1: E

# Problem 2:
The redemption days are 0, 10, 20, 30, 40, 50 days after the first.

Reduce mod 7 (days of week shift): 0, 3, 6, 2, 5, 1.

These residues are {0, 1, 2, 3, 5, 6} — every residue except **4**. So the only day of the week never hit is the day 4 days after the starting day. For no coupon to land on Sunday, Sunday must be that missed day:

start + 4 ≡ Sunday.

Counting back 4 days from Sunday: Saturday, Friday, Thursday, **Wednesday**.

Check: Wed → +10 Sat → +20 Tue → +30 Fri → +40 Mon → +50 Thu. No Sundays. ✓

ANSWER 2: C

# Problem 3:
The product is (3/2)·(4/3)·(5/4)·(6/5)·…·(a/b), where each numerator is one more than its denominator, so b = a − 1.

Telescoping: everything cancels except the last numerator and first denominator:
a/2 = 9 → a = 18, so b = 17.

a + b = 18 + 17 = 35.

(Choice E, 37, would come from mistakenly taking a = 19.)

ANSWER 3: D

# Problem 4:
Total value of all 60 stamps: 20(5 + 10 + 25) = 20·40 = 800 cents.

We need 710 cents, so we must *remove* stamps worth exactly 800 − 710 = 90 cents, using as **few** stamps as possible (to keep the most stamps).

Can 90 be made with 4 or fewer stamps? Max with 4 stamps is 100; possible 4-stamp totals near 90: 100, 85, 80, 75, 70… none equal 90. Three or fewer is at most 75. So 4 or fewer is impossible.

With 5 stamps: 25 + 25 + 25 + 10 + 5 = 90 ✓

So remove 5 stamps: 60 − 5 = 55 stamps.

Check: 17(25) + 19(10) + 19(5) = 425 + 190 + 95 = 710 cents, using 17 + 19 + 19 = 55 stamps.

ANSWER 4: E

# Problem 5:
CDCD as a number = CD·100 + CD = CD·101.

So ABA × CD = 101 × CD, and since CD ≠ 0, we get ABA = 101.

Thus A = 1, B = 0 (and indeed A, B, C, D distinct works, e.g. 101 × 23 = 2323).

A + B = 1 + 0 = 1.

ANSWER 5: A

# Problem 6:
Jonas total = 40, and 20 of those were girls, so boys from Jonas = 40 − 20 = 20.

Total boys = 52, so boys from Clay = 52 − 20 = 32.

(Check: girls from Clay = 48 − 20 = 28; 32 + 28 = 60 = Clay total ✓.)

Choice C (40) is the Jonas total, and E (52) is all boys — traps.

ANSWER 6: B

# Problem 7:
Pair each term: 1901 − 101 = 1800, 1902 − 102 = 1800, …, each pair differs by 1800.

Number of terms: 1993 − 1901 + 1 = 93.

Total = 93 × 1800 = 167,400.

ANSWER 7: A

# Problem 8:
Look at 7ⁿ mod 100:
- 7¹ = 07
- 7² = 49
- 7³ = 343 → 43
- 7⁴ = 2401 → 01
- then the pattern repeats with period 4.

2011 ÷ 4 = 502 remainder 3, so 7²⁰¹¹ ≡ 7³ ≡ 43 (mod 100).

Last two digits are 43, so the tens digit is 4.

ANSWER 8: D

# Problem 9:
Sum of the 10 numbers = 10 × 10 = 100.

To make one number as large as possible, make the other nine as small as possible while distinct and positive: 1 + 2 + … + 9 = 45.

Largest = 100 − 45 = 55.

(90 or 91 would require repeats or a zero, which isn't allowed.)

ANSWER 9: C

# Problem 10:
Total student-days:
- Allen: 7 × 3 = 21
- Balboa: 4 × 5 = 20
- Carver: 5 × 9 = 45

Total = 86 student-days.

Daily wage = 774 ÷ 86 = $9.00 per student-day.

Balboa earnings = 20 × 9 = $180.00.

(A is one day's pay for one student; E is Carver's total.)

ANSWER 10: C

Problem 1:

We need determine the parity of

\[
o^2+no.
\]

Factor out \(o\):

\[
o^2+no=o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on \(o+n\).

- If \(n\) is even, then odd \(+\) even \(=\) odd, so \(o+n\) is odd. Thus odd \(\cdot\) odd is odd.
- If \(n\) is odd, then odd \(+\) odd \(=\) even, so the product is even.

Therefore, the expression is odd only when \(n\) is even.

- A is wrong because the expression is not always odd.
- B is wrong because it is odd when \(n\) is even.
- C reverses the condition for being even.
- D reverses the condition for being odd.

ANSWER 1: E

Problem 2:

Isabella uses a coupon every 10 days. Since

\[
10\equiv 3\pmod 7,
\]

each coupon date moves forward by 3 days of the week.

If the first coupon is redeemed on day \(d\), the six redemption days are:

\[
d,\ d+3,\ d+6,\ d+2,\ d+5,\ d+1.
\]

These are six different days of the week. The only day missing is

\[
d+4.
\]

Since none of the six coupon dates is Sunday, the missing day must be Sunday. Thus,

\[
d+4=\text{Sunday}.
\]

Four days before Sunday is Wednesday.

Indeed, beginning on Wednesday gives:

\[
\text{Wednesday, Saturday, Tuesday, Friday, Monday, Thursday}.
\]

Sunday is the only missing day.

ANSWER 2: C

Problem 3:

The product is

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9.
\]

Almost every numerator cancels with a denominator:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab
=\frac a2.
\]

So

\[
\frac a2=9.
\]

Therefore,

\[
a=18.
\]

The final fraction must be

\[
\frac{18}{17},
\]

so \(b=17\). Therefore,

\[
a+b=18+17=35.
\]

Choices 11, 13, and 17 are too small, while 37 would correspond to a different ending fraction.

ANSWER 3: D

Problem 4:

Let

- \(x\) = number of 5-cent stamps,
- \(y\) = number of 10-cent stamps,
- \(z\) = number of 25-cent stamps.

In units of 5 cents, the postage total is

\[
\frac{710}{5}=142.
\]

Thus,

\[
x+2y+5z=142.
\]

We want to maximize

\[
x+y+z.
\]

Since there are at most 20 five-cent stamps, \(x\leq 20\). Therefore,

\[
2y+5z=142-x\geq 142-20=122.
\]

Since \(y\leq20\), if \(z\leq16\), then

\[
2y+5z\leq 2(20)+5(16)=40+80=120,
\]

which is not enough. So \(z\geq17\).

If \(z=17\), then

\[
2y+5(17)\geq122,
\]

so

\[
2y+85\geq122,
\]

\[
2y\geq37,
\]

and hence \(y\geq19\).

Now use

\[
x+y+z=142-y-4z.
\]

Since \(y\geq19\) and \(z\geq17\),

\[
x+y+z\leq142-19-4(17)
=142-19-68
=55.
\]

This maximum is attainable with

\[
x=19,\qquad y=19,\qquad z=17.
\]

Check:

\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So 55 stamps can be used. Choices 45, 46, 51, and 54 are possible totals to consider, but 55 is the greatest possible.

ANSWER 4: E

Problem 5:

We are told

\[
ABA\times CD=CDCD.
\]

The number \(ABA\) is

\[
100A+10B+A=101A+10B.
\]

The number \(CDCD\) is

\[
1000C+100D+10C+D.
\]

Group the \(C\)'s and \(D\)'s:

\[
CDCD=1010C+101D=101(10C+D)=101\cdot CD.
\]

Thus the equation becomes

\[
ABA\cdot CD=101\cdot CD.
\]

Since \(CD\) is a two-digit number and is not zero, divide both sides by \(CD\):

\[
ABA=101.
\]

Therefore,

\[
A=1,\qquad B=0.
\]

So

\[
A+B=1+0=1.
\]

For example, \(101\times23=2323\), and the digits \(1,0,2,3\) are all different.

ANSWER 5: A

Problem 6:

There are 40 students from Jonas Middle School, and 20 of those are girls.

Therefore the number of boys from Jonas is

\[
40-20=20.
\]

There are 52 boys altogether. Therefore the number of boys from Clay Middle School is

\[
52-20=32.
\]

The other choices do not correctly account for the 20 Jonas girls and the total of 52 boys.

ANSWER 6: B

Problem 7:

There are

\[
1993-1901+1=93
\]

terms in the first sum.

There are also

\[
193-101+1=93
\]

terms in the second sum.

Pair corresponding terms:

\[
1901-101=1800,
\]

\[
1902-102=1800,
\]

and so on, through

\[
1993-193=1800.
\]

Thus there are 93 differences of 1800:

\[
93\cdot1800=167400.
\]

Therefore the value is

\[
167,\!400.
\]

ANSWER 7: A

Problem 8:

We need the tens digit of

\[
7^{2011}.
\]

It is enough to examine powers of 7 modulo 100:

\[
7^1=7,
\]

\[
7^2=49,
\]

\[
7^3=343,
\]

whose last two digits are 43, and

\[
7^4=2401,
\]

whose last two digits are 01.

Thus the last two digits repeat every 4 powers because

\[
7^4\equiv1\pmod{100}.
\]

Now,

\[
2011\div4
\]

leaves remainder 3, since

\[
2011=4(502)+3.
\]

Therefore,

\[
7^{2011}
\]

has the same last two digits as

\[
7^3=343,
\]

which are 43. The tens digit is 4.

ANSWER 8: D

Problem 9:

The average of 10 different positive whole numbers is 10, so their total sum is

\[
10\cdot10=100.
\]

To make one number as large as possible, make the other nine numbers as small as possible.

The nine smallest different positive whole numbers are

\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is

\[
1+2+3+4+5+6+7+8+9=45.
\]

So the largest possible tenth number is

\[
100-45=55.
\]

This works because the ten numbers can be

\[
1,2,3,4,5,6,7,8,9,55,
\]

which are all different and have total 100.

A value such as 90 or 91 would leave too small a total for nine distinct positive whole numbers.

ANSWER 9: C

Problem 10:

Compute the total number of student-days worked.

Allen students worked:

\[
7\cdot3=21
\]

student-days.

Balboa students worked:

\[
4\cdot5=20
\]

student-days.

Carver students worked:

\[
5\cdot9=45
\]

student-days.

So the total number of student-days is

\[
21+20+45=86.
\]

The total pay was \$774, so the pay per student per day was

\[
\frac{774}{86}=9.
\]

Balboa students worked 20 student-days, so they earned

\[
20\cdot9=180.
\]

Thus the Balboa students earned \$180.00 altogether.

ANSWER 10: C

Problem 1:
We are asked about the parity of $o^2+no$ where $o$ is odd and $n$ is any whole number.

Factor:
$$o^2+no = o(o+n).$$

$o$ is odd, so $o^2 = o\cdot o$ is odd times odd = odd.

Hence parity is odd + $n\cdot o$.
$n\cdot o$ has the same parity as $n$, because $o$ is odd: if $n$ is even, $n o$ is even; if $n$ is odd, $n o$ is odd.

So:
- if $n$ even: odd + even = odd.
- if $n$ odd: odd + odd = even.

Thus the number is odd iff $n$ is even, and even iff $n$ is odd.

Check choices:
A. always odd - false, e.g. $o=1,n=1$ gives $1+1=2$ even.
B. always even - false, e.g. $o=1,n=2$ gives $1+2=3$ odd.
C. it is even only if $n$ is even - means even $\implies$ $n$ even. False, even happens when $n$ is odd.
D. it is odd only if $n$ is odd - means odd $\implies$ $n$ odd. False, odd happens when $n$ is even.
E. it is odd only if $n$ is even - means odd $\implies$ $n$ even. True by above.

ANSWER 1: E

Problem 2:
Isabella redeems on days $d, d+10, d+20, d+30, d+40, d+50$.
No date is a Sunday.

Modulo 7, $10\equiv 3$. So successive dates advance by 3 weekdays.

Starting from residue $r$, the residues are:
$$r,\ r+3,\ r+6,\ r+9=r+2,\ r+12=r+5,\ r+15=r+1 \pmod 7.$$
These 6 are all distinct, because adding 3 mod 7 cycles through all 7 residues. Continuing one more step would give $r+18=r+4$.

Thus the 6 dates occupy 6 distinct days of the week, missing exactly the 7th residue $r+4$.

Since none of the 6 is Sunday, the missing one must be Sunday. So:
$$r+4 \equiv \text{Sunday}.$$

Hence $r$ is 4 days before Sunday, i.e. Wednesday.

Check: Wed $\to$ Wed+3=Sat $\to$ +3=Tue $\to$ +3=Fri $\to$ +3=Mon $\to$ +3=Thu. That is Wed,Sat,Tue,Fri,Mon,Thu - no Sunday. Any other start would include Sunday.

ANSWER 2: C

Problem 3:
We interpret $3\ 2\cdot 4\ 3\cdots$ as fractions:
$$\frac32 \cdot \frac43 \cdot \frac54 \cdot \frac65 \cdots \frac{a}{b}=9.$$

The pattern is numerator $3,4,5,6,\dots$ and denominator $2,3,4,5,\dots$, i.e. term $k/(k-1)$. The last term is $a/b$, so we must have $b=a-1$ for the pattern to continue.

Then the product telescopes:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}
= \frac{a}{2},$$
because all intermediate $3,4,\dots,a-1$ cancel.

Set equal to 9:
$$\frac{a}{2}=9 \implies a=18.$$
Then $b=a-1=17$.

Sum $a+b=18+17=35$.

This rules out A,B,C,E.

ANSWER 3: D

Problem 4:
Let $x,y,z$ be numbers of 5-cent, 10-cent, 25-cent stamps used. $0\le x,y,z\le 20$ integers.
Value $710$ cents:
$$5x+10y+25z=710.$$
Divide by 5:
$$x+2y+5z=142. \tag{1}$$
Maximize $N=x+y+z$.

From (1), $x=142-2y-5z$, so $0\le x\le20$ gives:
$$122\le 2y+5z\le 142.$$
Also $N=x+y+z=142-y-4z$, so $y+4z=142-N$.

To use many stamps we want small denominations. Try $x=20,y=20$ fails to give integer $z$. Search systematically.

Since $y\le20$, to satisfy $2y+5z\ge122$ we need $z$ large. If $z\le16$, then $2y+5z\le40+80=120<122$, impossible. So $z\ge17$.

Try $z=17,y=19$: then $2y+5z=38+85=123$, so $x=142-123=19$, within $0-20$. This gives:
$$19\cdot5+19\cdot10+17\cdot25=95+190+425=710,$$
with $N=19+19+17=55$ stamps. So 55 is achievable.

Can we do 56 or more? Let $N\ge56$.
Then $y+4z=142-N\le86$ and $x=N-y-z\le20$ gives $y+z\ge N-20$.
Subtract: $3z=(y+4z)-(y+z)\le (142-N)-(N-20)=162-2N$.
So $z\le(162-2N)/3$.
Also $y=142-N-4z\le20$ gives $z\ge(122-N)/4$.

For $N=56$: $z\ge\lceil66/4\rceil=17$ and $z\le\lfloor50/3\rfloor=16$, contradiction. For $N>56$ the interval is even emptier; in general need $(122-N)/4\le(162-2N)/3$, i.e. $5N\le282$, so $N\le56.4$. Thus $N\ge57$ impossible, and $N=56$ impossible by the integer contradiction.

Hence 55 is greatest possible.

ANSWER 4: E

Problem 5:
We have $ABA\times CD = CDCD$, where $A,B,C,D$ are different digits, $A\ne0$, $C\ne0$.

Write $CD$ as the 2-digit number $10C+D$.
$CDCD$ as a 4-digit number is:
$$1000C+100D+10C+D =1010C+101D=101(10C+D)=101\cdot CD.$$

So:
$$ABA \times CD = 101\times CD.$$
Since $CD\ne0$, divide:
$$ABA =101.$$

Thus $A=1$, $B=0$, and $A+B=1$.

This is consistent with distinctness: we can choose e.g. $C=2,D=3$, all of $1,0,2,3$ different, and $101\times23=2323$. So valid.

Other sums are ruled out.

ANSWER 5: A

Problem 6:
Total 100: 52 boys, 48 girls. 40 from Jonas, 60 from Clay.

20 of the girls are from Jonas. So girls from Clay = total girls $-$ girls from Jonas:
$$48-20=28.$$

Clay total 60 = boys from Clay + girls from Clay, so:
$$\text{boys from Clay}=60-28=32.$$

Check: boys from Jonas $=52-32=20$, Jonas total $=20+20=40$. Works.

ANSWER 6: B

Problem 7:
Compute $(1901+\cdots+1993)-(101+\cdots+193)$.

Both sums have $1993-1901+1=93$ terms and $193-101+1=93$ terms.

Pair termwise: for $k=0,\dots,92$,
$$(1901+k)-(101+k)=1800.$$

So difference $=93\times1800$.
$93\times18=1674$, so $93\times1800=167{,}400$.

Other choices come from miscounting terms or arithmetic error.

ANSWER 7: A

Problem 8:
Find tens digit of $7^{2011}$, i.e. $7^{2011}\pmod{100}$.

Compute powers mod 100:
$7^1=7$,
$7^2=49$,
$7^3=343\equiv43$,
$7^4\equiv43\times7=301\equiv1$.

Since $7^4\equiv1\pmod{100}$, the last two digits repeat with period 4.

$2011 = 4\cdot502+3$, so:
$$7^{2011}=(7^4)^{502}\cdot7^3\equiv1^{502}\cdot43\equiv43\pmod{100}.$$

Thus the number ends $\dots43$, tens digit $4$.

ANSWER 8: D

Problem 9:
10 different positive whole numbers have mean 10, so sum $=10\times10=100$.

To make one number as large as possible, make the other nine as small as possible while distinct and positive.

Smallest 9 distinct positives are $1,2,3,4,5,6,7,8,9$, sum $=45$.

Then the tenth number $=100-45=55$, which is distinct from the others and valid. Any larger value would force the other nine to sum to $<45$, impossible with distinct positives.

So largest possible is 55.

ANSWER 9: C

Problem 10:
Count student-days (one student working one day).

Allen: $7$ students $\times$ $3$ days $=21$.
Balboa: $4\times5=20$.
Carver: $5\times9=45$.
Total $=21+20+45=86$ student-days.

Total pay $\$774$ for $86$ student-days at equal rate:
$$\text{rate}=774/86=9 \text{ dollars per student-day}.$$

Bal

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that $o$ is an odd whole number and $n$ is any whole number. We must determine which statement about $o^2 + no$ is always true.

**Step-by-Step Solution:**
1. Factor the expression:
   $$o^2 + no = o(o + n)$$
2. Since $o$ is an odd whole number, the parity of the product $o(o + n)$ is entirely determined by the parity of $(o + n)$:
   - If $n$ is **odd**, then $\text{odd} + \text{odd} = \text{even}$, so $o(o+n)$ is $\text{odd} \times \text{even} = \textbf{even}$.
   - If $n$ is **even**, then $\text{odd} + \text{even} = \text{odd}$, so $o(o+n)$ is $\text{odd} \times \text{odd} = \textbf{odd}$.
3. Therefore, the expression $o^2 + no$ is odd if and only if $n$ is even. 
4. This matches the statement: "it is odd only if $n$ is even."

ANSWER 1: E

---

### Problem 2:
**Restatement:** Isabella redeems 6 coupons, one every 10 days. Pete's is closed on Sundays, and none of the 6 dates falls on a Sunday. We need to find the day of the week she redeems her first coupon.

**Step-by-Step Solution:**
1. There are 7 days in a week. Moving forward by 10 days advances the day of the week by:
   $$10 \equiv 3 \pmod 7$$
2. Let $d$ represent the day of the week of the first coupon ($0 = \text{Sunday}, 1 = \text{Monday}, \dots, 6 = \text{Saturday}$).
3. The days of the week for the 6 coupons will be:
   - Coupon 1: $d$
   - Coupon 2: $d + 3$
   - Coupon 3: $d + 6$
   - Coupon 4: $d + 9 \equiv d + 2 \pmod 7$
   - Coupon 5: $d + 12 \equiv d + 5 \pmod 7$
   - Coupon 6: $d + 15 \equiv d + 1 \pmod 7$
4. The six days of the week visited are $d, d+1, d+2, d+3, d+5, d+6$. Notice that the only day of the week omitted from this list is $d+4$.
5. Since no circled date falls on a Sunday, Sunday must be the omitted day:
   $$d + 4 \equiv 0 \pmod 7 \implies d \equiv 3 \pmod 7$$
6. Day 3 corresponds to **Wednesday**.
   - Checking: Wed $\to$ Sat $\to$ Tue $\to$ Fri $\to$ Mon $\to$ Thu. None of these is Sunday.

ANSWER 2: C

---

### Problem 3:
**Restatement:** We are given the telescoping product $\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdots \frac{a}{b} = 9$. We must find the sum $a + b$.

**Step-by-Step Solution:**
1. Each fraction is of the form $\frac{k+1}{k}$, where the numerator of each term cancels with the denominator of the subsequent term:
   $$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{a}{b} = \frac{a}{2}$$
   where $b = a - 1$.
2. We are given that this product equals 9:
   $$\frac{a}{2} = 9 \implies a = 18$$
3. Since $b = a - 1$, we have $b = 17$.
4. The sum of $a$ and $b$ is:
   $$a + b = 18 + 17 = 35$$

ANSWER 3: D

---

### Problem 4:
**Restatement:** Nicolas has 20 stamps each of 5¢, 10¢, and 25¢. We want to find the greatest number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-Step Solution:**
1. Calculate the total value and count of all stamps:
   - $20 \times 5¢ = 100¢$
   - $20 \times 10¢ = 200¢$
   - $20 \times 25¢ = 500¢$
   - Total value $= 100 + 200 + 500 = 800¢$ using 60 stamps.
2. To use the **maximum** number of stamps to make 710¢, Nicolas must **minimize** the number of stamps removed to subtract:
   $$800¢ - 710¢ = 90¢$$
3. To remove the fewest stamps totaling 90¢, we try to use the largest denomination stamps (25¢):
   - With 3 stamps of 25¢ ($75¢$), the remaining amount needed is $15¢$.
   - Since Nicolas has only 5¢, 10¢, and 25¢ stamps, 15¢ cannot be made with a single stamp. It requires at least two stamps ($10¢ + 5¢$).
   - Thus, 90¢ can be formed with $3 \times 25¢ + 1 \times 10¢ + 1 \times 5¢$, which uses $3 + 1 + 1 = 5$ stamps.
   - Can 90¢ be formed using 4 stamps? The maximum value of 4 stamps less than or equal to 90¢ would be $3 \times 25¢ + 10¢ = 85¢ \ne 90¢$. Thus, 4 stamps is impossible.
4. The minimum number of stamps to remove is 5.
5. Therefore, the maximum number of stamps used is:
   $$60 - 5 = 55$$

ANSWER 4: E

---

### Problem 5:
**Restatement:** In the multiplication $ABA \times CD = CDCD$, $A, B, C, D$ are distinct digits. We need to find $A + B$.

**Step-by-Step Solution:**
1. Express $CDCD$ in terms of the two-digit number $CD$:
   $$CDCD = CD \times 100 + CD = CD \times 101$$
2. Substitute this into the equation:
   $$ABA \times CD = 101 \times CD$$
3. Since $CD$ is a non-zero two-digit number, divide both sides by $CD$:
   $$ABA = 101$$
4. Comparing digits, we have $A = 1$ and $B = 0$.
5. Thus:
   $$A + B = 1 + 0 = 1$$

ANSWER 5: A

---

### Problem 6:
**Restatement:** Out of 100 students at a camp, 52 are boys and 48 are girls. Also, 40 are from Jonas Middle School and 60 from Clay Middle School. Given that 20 of the girls are from Jonas, find how many boys are from Clay.

**Step-by-Step Solution:**
1. Find the number of girls from Clay Middle School:
   $$\text{Total girls} - \text{Girls from Jonas} = 48 - 20 = 28\text{ girls from Clay}$$
2. Find the number of boys from Clay Middle School:
   $$\text{Total students from Clay} - \text{Girls from Clay} = 60 - 28 = 32\text{ boys from Clay}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** Compute the value of:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-Step Solution:**
1. Pair each term in the first sum with the corresponding term in the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
2. Each paired difference is equal to:
   $$1900 + k - (100 + k) = 1800$$
3. Find the number of terms: from $k = 1$ to $k = 93$, there are:
   $$93 - 1 + 1 = 93\text{ terms}$$
4. Multiply the difference by the number of terms:
   $$93 \times 1800 = 93 \times 18 \times 100 = 1674 \times 100 = 167,400$$

ANSWER 7: A

---

### Problem 8:
**Restatement:** Find the tens digit of $7^{2011}$.

**Step-by-Step Solution:**
1. Find the pattern of the last two digits of powers of 7 by computing $7^n \pmod{100}$:
   - $7^1 = 07$
   - $7^2 = 49$
   - $7^3 = 343 \equiv 43 \pmod{100}$
   - $7^4 = 43

# Problem 1:
We must decide when $o^2 + no$ is odd or even, given $o$ is odd and $n$ is any whole number.

**Key facts:** odd × odd = odd, so $o^2$ is **odd**. Since $o$ is odd, multiplying by $o$ does not change $n$'s parity, so $no$ has the **same parity as $n$**.

**Case 1: $n$ even.** Then $no$ is even, so $o^2 + no = \text{odd} + \text{even} = \textbf{odd}$.
**Case 2: $n$ odd.** Then $no$ is odd, so $o^2 + no = \text{odd} + \text{odd} = \textbf{even}$.

So the number is odd exactly when $n$ is even. In particular, **if the number is odd, $n$ must be even**.

Ruling out traps: (A) fails since $o=1, n=1$ gives $1+1=2$, even; (B) fails since $n=2$ gives $1+2=3$, odd; (C) and (D) state the exact opposite of the truth. Only (E) matches.

ANSWER 1: E

# Problem 2:
The 6 redemption dates are 10 days apart, so their days of the week shift by $10 \bmod 7 = 3$ each time.

If the first date falls on weekday $W$, the six dates fall on (mod 7):
$$W,\; W+3,\; W+6,\; W+2,\; W+5,\; W+1$$
That is **six of the seven weekdays**, and the only weekday missed is $W+4$.

Since no circled date is a Sunday, the one missed weekday must be Sunday:
$$W + 4 = \text{Sunday} \implies W = \text{Wednesday}.$$

**Check:** Starting Wednesday, the dates fall on Wed, Sat, Tue, Fri, Mon, Thu — no Sunday. (Trap: starting Thursday gives Thu, Sun, ... — fails immediately.)

ANSWER 2: C

# Problem 3:
The product is a chain of fractions (the bars were lost in formatting):
$$\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdots\frac{a}{b} = 9.$$

Each numerator cancels the denominator before it (e.g., $\frac{3}{2}\cdot\frac{4}{3} = \frac{4}{2}$), and each fraction has the form $\frac{k+1}{k}$, so $b = a-1$. Everything telescopes:
$$\text{Product} = \frac{a}{2} = 9 \implies a = 18, \quad b = 17.$$
Then $a + b = 18 + 17 = 35$.

Trap check: $a+b=17$ would mean $a=9$, giving product $\tfrac92$, not $9$; only $35$ fits the pattern.

ANSWER 3: D

# Problem 4:
Work in units of 5¢: we need total value $710¢ = 142$ units, using stamps worth $1, 2, 5$ units (i.e., 5¢, 10¢, 25¢), at most 20 of each. To **maximize the count**, use as many cheap stamps as possible.

**Building a 55-stamp solution:** Take nineteen 1's and nineteen 2's: value $19 + 38 = 57$ units with 38 stamps. Remaining value $142 - 57 = 85 = 17\times 5$: seventeen 5's.
$$19\times 5¢ + 19\times 10¢ + 17\times 25¢ = 95 + 190 + 425 = 710¢ = \$7.10.\; ✓$$
That's $19+19+17 = 55$ stamps, all counts $\le 20$. ✓

**Why 56 is impossible:** with 56 stamps, at least $56 - 40 = 16$ must be 25¢ stamps (only 40 cheap ones exist). Let $z$ = number of 25¢ stamps:
- $z=16$: max value $= 100 + 200 + 400 = 700 < 710$. ✗
- $z=17$: need $5x+10y=285$, $x+y=39 \Rightarrow x=21>20$. ✗
- $z=18$: $x=24>20$. ✗  $z=19$: $x=27>20$. ✗  $z=20$: $x=30>20$. ✗

Traps: 54 comes from $20×5+16×10+18×25 = 710$; 51 from $20×5+11×10+20×25 = 710$ — both valid but not maximal.

ANSWER 4: E

# Problem 5:
Write $CDCD$ as a number:
$$CDCD = 1000C + 100D + 10C + D = 101(10C + D) = 101 \times CD.$$
The given equation is
$$ABA \times CD = CDCD = 101 \times CD.$$
Since $CD \neq 0$ (it's a two-digit number), we can divide both sides by $CD$:
$$ABA = 101.$$
So $A = 1$ and $B = 0$, giving $A + B = 1$.

**Consistency check:** $C, D$ just need to be distinct from $0, 1$ and each other, e.g., $101 \times 23 = 2323$. ✓ (Trap: guessing digits without using the factorization 101 wastes time or gives 9.)

ANSWER 5: A

# Problem 6:
Set up a two-way table with 100 total students.

- Girls: 48 total, and 20 girls are from Jonas $\Rightarrow$ girls from Clay $= 48 - 20 = 28$.
- Jonas has 40 students, of whom 20 are girls $\Rightarrow$ boys from Jonas $= 40 - 20 = 20$.
- Boys: 52 total $\Rightarrow$ boys from Clay $= 52 - 20 = \textbf{32}$.

**Check:** Jonas $20+20=40$ ✓; Clay $28+32=60$ ✓; boys $20+32=52$ ✓; girls $20+28=48$ ✓; total $100$ ✓. (Trap: 20 counts the boys from Jonas, not Clay.)

ANSWER 6: B

# Problem 7:
Both sums have the same number of terms: $1901$ to $1993$ has $1993-1901+1 = 93$ terms, and $101$ to $193$ has $193-101+1 = 93$ terms.

Pair them up: each term in the first sum is exactly $1800$ more than its partner ($1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$). So the difference is
$$93 \times 1800 = 167{,}400.$$

**Check via formulas:** $\frac{93(1901+1993)}{2} - \frac{93(101+193)}{2} = 181{,}071 - 13{,}671 = 167{,}400$. ✓ (Trap: 181,071 is the first sum alone — forgetting to subtract.)

ANSWER 7: A

# Problem 8:
Only the tens digit matters, so track the last two digits of $7^{k}$, which cycle with period 4:
$$7^1 = 07,\quad 7^2 = 49,\quad 7^3 = 343 \to 43,\quad 7^4 = 2401 \to 01,\quad 7^5 \to 07, \ldots$$

Find where $2011$ falls in the cycle:
$$2011 = 4 \times 502 + 3,$$
so $7^{2011}$ ends in the same two digits as $7^3$, namely $\ldots 43$.

The tens digit is $\textbf{4}$. (Trap: $7^{2012}$, a multiple of 4, ends in $01$, tens digit 0 — choice A; that would require the exponent to be divisible by 4.)

ANSWER 8: D

# Problem 9:
The ten **different** positive whole numbers have average 10, so their sum is $10 \times 10 = 100$.

To make one number as large as possible, make the other nine as small as possible. Since they must be **distinct positive** whole numbers, the smallest possible set is $1, 2, 3, \ldots, 9$, with sum
$$1+2+\cdots+9 = 45.$$
So the largest possible value is
$$100 - 45 = 55.$$

**Check:** $1,2,\ldots,9,55$ are ten different positive whole numbers with sum $100$ and average $10$. ✓ (Trap: 91 arises if you ignore "different" and use nine 1's; but the numbers must be distinct.)

ANSWER 9: C

# Problem 10:
Count student-days (the unit being paid):
- Allen: $7 \times 3 = 21$ student-days
- Balboa: $4 \times 5 = 20$ student-days
- Carver: $5 \times 9 = 45$ student-days

Total: $21 + 20 + 45 = 86$ student-days, paid $\$774$ altogether. So the daily rate is
$$774 \div 86 = 9 \text{ dollars per student-day}.$$
Balboa's students worked 20 student-days, so they earned
$$20 \times 9 = 18

Problem 1:  
We need the parity of \(o^2+no\), where \(o\) is odd and \(n\) is any whole number.

Factor:
\[
o^2+no=o(o+n).
\]

Since \(o\) is odd:

- If \(n\) is even, then \(o+n\) is odd, so \(o(o+n)\) is odd.
- If \(n\) is odd, then \(o+n\) is even, so \(o(o+n)\) is even.

Thus the expression is odd exactly when \(n\) is even.

Check the choices:
- A and B are false because the parity depends on \(n\).
- C says “even only if \(n\) is even,” but when the expression is even, \(n\) is actually odd.
- D says “odd only if \(n\) is odd,” but when the expression is odd, \(n\) is even.
- E says “odd only if \(n\) is even,” which is true.

ANSWER 1: E

Problem 2:  
Isabella redeems coupons 10 days apart. Since
\[
10\equiv 3 \pmod 7,
\]
the six circled days are the first day plus
\[
0,3,6,9,12,15
\]
days, which modulo 7 gives offsets
\[
0,3,6,2,5,1.
\]

So the six circled days are all weekdays except the one 4 days after the first day. Since none of the circled days is a Sunday, Sunday must be the un-circled day: 4 days after the first day.

Therefore the first coupon is 4 days before Sunday, which is Wednesday.

Check: Wednesday, Saturday, Tuesday, Friday, Monday, Thursday — no Sunday.

ANSWER 2: C

Problem 3:  
The product is the telescoping product of fractions:
\[
\frac32\cdot \frac43\cdot \frac54\cdot \frac65\cdots \frac{a}{b}=9.
\]

The numerators cancel the next denominators:
\[
\frac32\cdot \frac43=\frac42,
\]
\[
\frac42\cdot \frac54=\frac52,
\]
and continuing, the whole product simplifies to
\[
\frac{a}{2}.
\]

So
\[
\frac{a}{2}=9 \implies a=18.
\]

By the pattern, the denominator is one less than the numerator, so
\[
b=17.
\]

Thus
\[
a+b=18+17=35.
\]

ANSWER 3: D

Problem 4:  
Nicolas has 20 of each type of stamp, so he has
\[
20+20+20=60
\]
stamps total. Their total value is
\[
20(5+10+25)=20(40)=800
\]
cents.

He wants postage of
\[
\$7.10=710
\]
cents, so the stamps he does **not** use must total
\[
800-710=90
\]
cents.

To maximize the number of stamps he uses, we minimize the number of stamps he leaves unused.

Can 4 unused stamps total 90 cents? Let \(x,y,z\) be the numbers of 25-, 10-, and 5-cent unused stamps. Then
\[
x+y+z=4
\]
and
\[
25x+10y+5z=90.
\]

Divide by 5:
\[
5x+2y+z=18.
\]

Subtract \(x+y+z=4\):
\[
4x+y=14.
\]

Testing possible \(x=0,1,2,3,4\):

- \(x=0\Rightarrow y=14\), impossible.
- \(x=1\Rightarrow y=10\), impossible.
- \(x=2\Rightarrow y=6\), impossible.
- \(x=3\Rightarrow y=2\), then \(z=4-3-2=-1\), impossible.
- \(x=4\Rightarrow y=-2\), impossible.

So 4 unused stamps cannot total 90 cents.

Can 5 unused stamps total 90 cents? Yes:
\[
25+25+25+10+5=90.
\]

Then he uses
\[
60-5=55
\]
stamps. For example, he could use 19 five-cent stamps, 19 ten-cent stamps, and 17 twenty-five-cent stamps:
\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So the greatest possible number of stamps is 55. Choice D, 54, is possible but not greatest.

ANSWER 4: E

Problem 5:  
We are told
\[
ABA\times CD=CDCD,
\]
where \(A,B,C,D\) are different digits.

Write the numbers in place-value form:
\[
ABA=101A+10B,
\]
\[
CD=10C+D,
\]
\[
CDCD=1000C+100D+10C+D=101(10C+D).
\]

So
\[
(101A+10B)(10C+D)=101(10C+D).
\]

Since \(CD\) is a two-digit number, \(10C+D\neq 0\), so we can divide both sides by \(10C+D\):
\[
101A+10B=101.
\]

The only digit solution is
\[
A=1,\quad B=0.
\]

Therefore
\[
A+B=1+0=1.
\]

ANSWER 5: A

Problem 6:  
There are 48 girls total, and 20 of the girls are from Jonas Middle School. So the number of girls from Clay Middle School is
\[
48-20=28.
\]

There are 60 students total from Clay Middle School, so the number of boys from Clay is
\[
60-28=32.
\]

ANSWER 6: B

Problem 7:  
Pair the terms:
\[
(1901+1902+\cdots+1993)-(101+102+\cdots+193)
\]
\[
=(1901-101)+(1902-102)+\cdots+(1993-193).
\]

Each pair is
\[
1800.
\]

Now count the number of terms:
\[
193-101+1=93.
\]

So the value is
\[
93\cdot 1800=167400.
\]

ANSWER 7: A

Problem 8:  
We need the tens digit of \(7^{2011}\), so we need the last two digits.

Powers of 7 modulo 100 cycle:
\[
7^1=07,
\]
\[
7^2=49,
\]
\[
7^3=343\equiv 43,
\]
\[
7^4=2401\equiv 01.
\]

Then the cycle repeats:
\[
07,\ 49,\ 43,\ 01.
\]

Since
\[
2011\equiv 3 \pmod 4,
\]
we have
\[
7^{2011}\equiv 7^3\equiv 43 \pmod{100}.
\]

The last two digits are 43, so the tens digit is 4.

ANSWER 8: D

Problem 9:  
The average of 10 different positive whole numbers is 10, so their sum is
\[
10\cdot 10=100.
\]

To make one number as large as possible, make the other 9 numbers as small as possible while still being different positive whole numbers. The smallest 9 distinct positive whole numbers are
\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is
\[
1+2+\cdots+9=45.
\]

So the largest number can be
\[
100-45=55.
\]

This is possible: \(1,2,3,4,5,6,7,8,9,55\) are all different positive whole numbers.

ANSWER 9: C

Problem 10:  
First count total student-days worked:

- Allen: \(7\cdot 3=21\) student-days
- Balboa: \(4\cdot 5=20\) student-days
- Carver: \(5\cdot 9=45\) student-days

Total:
\[
21+20+45=86
\]
student-days.

The total payment is \$774, so the daily rate is
\[
774\div 86=9
\]
dollars per student-day.

Balboa students worked 20 student-days, so they earned
\[
20\cdot 9=180
\]
dollars.

ANSWER 10: C

# Problem 1:
We need a statement always true about o² + no, where o is odd.

Factor: o² + no = o(o + n).

Since o is odd, the parity of the product is the parity of (o + n).
- If n is even, o + n is odd → product = odd·odd = **odd**.
- If n is odd, o + n is even → product = odd·even = **even**.

So the expression is odd exactly when n is even.

Ruling out others: A fails (n odd gives even), B fails (n even gives odd), C says "even only if n is even" — but it's even when n is *odd*, false. D says "odd only if n is odd" — false, it's odd when n is even. E: "odd only if n is even" — true.

ANSWER 1: E

# Problem 2:
The redemption days are 0, 10, 20, 30, 40, 50 days after the first.

Reduce mod 7 (days of week shift): 0, 3, 6, 2, 5, 1.

These residues are {0, 1, 2, 3, 5, 6} — every residue except **4**. So the only day of the week never hit is the day 4 days after the starting day. For no coupon to land on Sunday, Sunday must be that missed day:

start + 4 ≡ Sunday.

Counting back 4 days from Sunday: Saturday, Friday, Thursday, **Wednesday**.

Check: Wed → +10 Sat → +20 Tue → +30 Fri → +40 Mon → +50 Thu. No Sundays. ✓

ANSWER 2: C

# Problem 3:
The product is (3/2)·(4/3)·(5/4)·(6/5)·…·(a/b), where each numerator is one more than its denominator, so b = a − 1.

Telescoping: everything cancels except the last numerator and first denominator:
a/2 = 9 → a = 18, so b = 17.

a + b = 18 + 17 = 35.

(Choice E, 37, would come from mistakenly taking a = 19.)

ANSWER 3: D

# Problem 4:
Total value of all 60 stamps: 20(5 + 10 + 25) = 20·40 = 800 cents.

We need 710 cents, so we must *remove* stamps worth exactly 800 − 710 = 90 cents, using as **few** stamps as possible (to keep the most stamps).

Can 90 be made with 4 or fewer stamps? Max with 4 stamps is 100; possible 4-stamp totals near 90: 100, 85, 80, 75, 70… none equal 90. Three or fewer is at most 75. So 4 or fewer is impossible.

With 5 stamps: 25 + 25 + 25 + 10 + 5 = 90 ✓

So remove 5 stamps: 60 − 5 = 55 stamps.

Check: 17(25) + 19(10) + 19(5) = 425 + 190 + 95 = 710 cents, using 17 + 19 + 19 = 55 stamps.

ANSWER 4: E

# Problem 5:
CDCD as a number = CD·100 + CD = CD·101.

So ABA × CD = 101 × CD, and since CD ≠ 0, we get ABA = 101.

Thus A = 1, B = 0 (and indeed A, B, C, D distinct works, e.g. 101 × 23 = 2323).

A + B = 1 + 0 = 1.

ANSWER 5: A

# Problem 6:
Jonas total = 40, and 20 of those were girls, so boys from Jonas = 40 − 20 = 20.

Total boys = 52, so boys from Clay = 52 − 20 = 32.

(Check: girls from Clay = 48 − 20 = 28; 32 + 28 = 60 = Clay total ✓.)

Choice C (40) is the Jonas total, and E (52) is all boys — traps.

ANSWER 6: B

# Problem 7:
Pair each term: 1901 − 101 = 1800, 1902 − 102 = 1800, …, each pair differs by 1800.

Number of terms: 1993 − 1901 + 1 = 93.

Total = 93 × 1800 = 167,400.

ANSWER 7: A

# Problem 8:
Look at 7ⁿ mod 100:
- 7¹ = 07
- 7² = 49
- 7³ = 343 → 43
- 7⁴ = 2401 → 01
- then the pattern repeats with period 4.

2011 ÷ 4 = 502 remainder 3, so 7²⁰¹¹ ≡ 7³ ≡ 43 (mod 100).

Last two digits are 43, so the tens digit is 4.

ANSWER 8: D

# Problem 9:
Sum of the 10 numbers = 10 × 10 = 100.

To make one number as large as possible, make the other nine as small as possible while distinct and positive: 1 + 2 + … + 9 = 45.

Largest = 100 − 45 = 55.

(90 or 91 would require repeats or a zero, which isn't allowed.)

ANSWER 9: C

# Problem 10:
Total student-days:
- Allen: 7 × 3 = 21
- Balboa: 4 × 5 = 20
- Carver: 5 × 9 = 45

Total = 86 student-days.

Daily wage = 774 ÷ 86 = $9.00 per student-day.

Balboa earnings = 20 × 9 = $180.00.

(A is one day's pay for one student; E is Carver's total.)

ANSWER 10: C

Problem 1:

We need determine the parity of

\[
o^2+no.
\]

Factor out \(o\):

\[
o^2+no=o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on \(o+n\).

- If \(n\) is even, then odd \(+\) even \(=\) odd, so \(o+n\) is odd. Thus odd \(\cdot\) odd is odd.
- If \(n\) is odd, then odd \(+\) odd \(=\) even, so the product is even.

Therefore, the expression is odd only when \(n\) is even.

- A is wrong because the expression is not always odd.
- B is wrong because it is odd when \(n\) is even.
- C reverses the condition for being even.
- D reverses the condition for being odd.

ANSWER 1: E

Problem 2:

Isabella uses a coupon every 10 days. Since

\[
10\equiv 3\pmod 7,
\]

each coupon date moves forward by 3 days of the week.

If the first coupon is redeemed on day \(d\), the six redemption days are:

\[
d,\ d+3,\ d+6,\ d+2,\ d+5,\ d+1.
\]

These are six different days of the week. The only day missing is

\[
d+4.
\]

Since none of the six coupon dates is Sunday, the missing day must be Sunday. Thus,

\[
d+4=\text{Sunday}.
\]

Four days before Sunday is Wednesday.

Indeed, beginning on Wednesday gives:

\[
\text{Wednesday, Saturday, Tuesday, Friday, Monday, Thursday}.
\]

Sunday is the only missing day.

ANSWER 2: C

Problem 3:

The product is

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9.
\]

Almost every numerator cancels with a denominator:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab
=\frac a2.
\]

So

\[
\frac a2=9.
\]

Therefore,

\[
a=18.
\]

The final fraction must be

\[
\frac{18}{17},
\]

so \(b=17\). Therefore,

\[
a+b=18+17=35.
\]

Choices 11, 13, and 17 are too small, while 37 would correspond to a different ending fraction.

ANSWER 3: D

Problem 4:

Let

- \(x\) = number of 5-cent stamps,
- \(y\) = number of 10-cent stamps,
- \(z\) = number of 25-cent stamps.

In units of 5 cents, the postage total is

\[
\frac{710}{5}=142.
\]

Thus,

\[
x+2y+5z=142.
\]

We want to maximize

\[
x+y+z.
\]

Since there are at most 20 five-cent stamps, \(x\leq 20\). Therefore,

\[
2y+5z=142-x\geq 142-20=122.
\]

Since \(y\leq20\), if \(z\leq16\), then

\[
2y+5z\leq 2(20)+5(16)=40+80=120,
\]

which is not enough. So \(z\geq17\).

If \(z=17\), then

\[
2y+5(17)\geq122,
\]

so

\[
2y+85\geq122,
\]

\[
2y\geq37,
\]

and hence \(y\geq19\).

Now use

\[
x+y+z=142-y-4z.
\]

Since \(y\geq19\) and \(z\geq17\),

\[
x+y+z\leq142-19-4(17)
=142-19-68
=55.
\]

This maximum is attainable with

\[
x=19,\qquad y=19,\qquad z=17.
\]

Check:

\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So 55 stamps can be used. Choices 45, 46, 51, and 54 are possible totals to consider, but 55 is the greatest possible.

ANSWER 4: E

Problem 5:

We are told

\[
ABA\times CD=CDCD.
\]

The number \(ABA\) is

\[
100A+10B+A=101A+10B.
\]

The number \(CDCD\) is

\[
1000C+100D+10C+D.
\]

Group the \(C\)'s and \(D\)'s:

\[
CDCD=1010C+101D=101(10C+D)=101\cdot CD.
\]

Thus the equation becomes

\[
ABA\cdot CD=101\cdot CD.
\]

Since \(CD\) is a two-digit number and is not zero, divide both sides by \(CD\):

\[
ABA=101.
\]

Therefore,

\[
A=1,\qquad B=0.
\]

So

\[
A+B=1+0=1.
\]

For example, \(101\times23=2323\), and the digits \(1,0,2,3\) are all different.

ANSWER 5: A

Problem 6:

There are 40 students from Jonas Middle School, and 20 of those are girls.

Therefore the number of boys from Jonas is

\[
40-20=20.
\]

There are 52 boys altogether. Therefore the number of boys from Clay Middle School is

\[
52-20=32.
\]

The other choices do not correctly account for the 20 Jonas girls and the total of 52 boys.

ANSWER 6: B

Problem 7:

There are

\[
1993-1901+1=93
\]

terms in the first sum.

There are also

\[
193-101+1=93
\]

terms in the second sum.

Pair corresponding terms:

\[
1901-101=1800,
\]

\[
1902-102=1800,
\]

and so on, through

\[
1993-193=1800.
\]

Thus there are 93 differences of 1800:

\[
93\cdot1800=167400.
\]

Therefore the value is

\[
167,\!400.
\]

ANSWER 7: A

Problem 8:

We need the tens digit of

\[
7^{2011}.
\]

It is enough to examine powers of 7 modulo 100:

\[
7^1=7,
\]

\[
7^2=49,
\]

\[
7^3=343,
\]

whose last two digits are 43, and

\[
7^4=2401,
\]

whose last two digits are 01.

Thus the last two digits repeat every 4 powers because

\[
7^4\equiv1\pmod{100}.
\]

Now,

\[
2011\div4
\]

leaves remainder 3, since

\[
2011=4(502)+3.
\]

Therefore,

\[
7^{2011}
\]

has the same last two digits as

\[
7^3=343,
\]

which are 43. The tens digit is 4.

ANSWER 8: D

Problem 9:

The average of 10 different positive whole numbers is 10, so their total sum is

\[
10\cdot10=100.
\]

To make one number as large as possible, make the other nine numbers as small as possible.

The nine smallest different positive whole numbers are

\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is

\[
1+2+3+4+5+6+7+8+9=45.
\]

So the largest possible tenth number is

\[
100-45=55.
\]

This works because the ten numbers can be

\[
1,2,3,4,5,6,7,8,9,55,
\]

which are all different and have total 100.

A value such as 90 or 91 would leave too small a total for nine distinct positive whole numbers.

ANSWER 9: C

Problem 10:

Compute the total number of student-days worked.

Allen students worked:

\[
7\cdot3=21
\]

student-days.

Balboa students worked:

\[
4\cdot5=20
\]

student-days.

Carver students worked:

\[
5\cdot9=45
\]

student-days.

So the total number of student-days is

\[
21+20+45=86.
\]

The total pay was \$774, so the pay per student per day was

\[
\frac{774}{86}=9.
\]

Balboa students worked 20 student-days, so they earned

\[
20\cdot9=180.
\]

Thus the Balboa students earned \$180.00 altogether.

ANSWER 10: C

Problem 1:
We are asked about the parity of $o^2+no$ where $o$ is odd and $n$ is any whole number.

Factor:
$$o^2+no = o(o+n).$$

$o$ is odd, so $o^2 = o\cdot o$ is odd times odd = odd.

Hence parity is odd + $n\cdot o$.
$n\cdot o$ has the same parity as $n$, because $o$ is odd: if $n$ is even, $n o$ is even; if $n$ is odd, $n o$ is odd.

So:
- if $n$ even: odd + even = odd.
- if $n$ odd: odd + odd = even.

Thus the number is odd iff $n$ is even, and even iff $n$ is odd.

Check choices:
A. always odd - false, e.g. $o=1,n=1$ gives $1+1=2$ even.
B. always even - false, e.g. $o=1,n=2$ gives $1+2=3$ odd.
C. it is even only if $n$ is even - means even $\implies$ $n$ even. False, even happens when $n$ is odd.
D. it is odd only if $n$ is odd - means odd $\implies$ $n$ odd. False, odd happens when $n$ is even.
E. it is odd only if $n$ is even - means odd $\implies$ $n$ even. True by above.

ANSWER 1: E

Problem 2:
Isabella redeems on days $d, d+10, d+20, d+30, d+40, d+50$.
No date is a Sunday.

Modulo 7, $10\equiv 3$. So successive dates advance by 3 weekdays.

Starting from residue $r$, the residues are:
$$r,\ r+3,\ r+6,\ r+9=r+2,\ r+12=r+5,\ r+15=r+1 \pmod 7.$$
These 6 are all distinct, because adding 3 mod 7 cycles through all 7 residues. Continuing one more step would give $r+18=r+4$.

Thus the 6 dates occupy 6 distinct days of the week, missing exactly the 7th residue $r+4$.

Since none of the 6 is Sunday, the missing one must be Sunday. So:
$$r+4 \equiv \text{Sunday}.$$

Hence $r$ is 4 days before Sunday, i.e. Wednesday.

Check: Wed $\to$ Wed+3=Sat $\to$ +3=Tue $\to$ +3=Fri $\to$ +3=Mon $\to$ +3=Thu. That is Wed,Sat,Tue,Fri,Mon,Thu - no Sunday. Any other start would include Sunday.

ANSWER 2: C

Problem 3:
We interpret $3\ 2\cdot 4\ 3\cdots$ as fractions:
$$\frac32 \cdot \frac43 \cdot \frac54 \cdot \frac65 \cdots \frac{a}{b}=9.$$

The pattern is numerator $3,4,5,6,\dots$ and denominator $2,3,4,5,\dots$, i.e. term $k/(k-1)$. The last term is $a/b$, so we must have $b=a-1$ for the pattern to continue.

Then the product telescopes:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}
= \frac{a}{2},$$
because all intermediate $3,4,\dots,a-1$ cancel.

Set equal to 9:
$$\frac{a}{2}=9 \implies a=18.$$
Then $b=a-1=17$.

Sum $a+b=18+17=35$.

This rules out A,B,C,E.

ANSWER 3: D

Problem 4:
Let $x,y,z$ be numbers of 5-cent, 10-cent, 25-cent stamps used. $0\le x,y,z\le 20$ integers.
Value $710$ cents:
$$5x+10y+25z=710.$$
Divide by 5:
$$x+2y+5z=142. \tag{1}$$
Maximize $N=x+y+z$.

From (1), $x=142-2y-5z$, so $0\le x\le20$ gives:
$$122\le 2y+5z\le 142.$$
Also $N=x+y+z=142-y-4z$, so $y+4z=142-N$.

To use many stamps we want small denominations. Try $x=20,y=20$ fails to give integer $z$. Search systematically.

Since $y\le20$, to satisfy $2y+5z\ge122$ we need $z$ large. If $z\le16$, then $2y+5z\le40+80=120<122$, impossible. So $z\ge17$.

Try $z=17,y=19$: then $2y+5z=38+85=123$, so $x=142-123=19$, within $0-20$. This gives:
$$19\cdot5+19\cdot10+17\cdot25=95+190+425=710,$$
with $N=19+19+17=55$ stamps. So 55 is achievable.

Can we do 56 or more? Let $N\ge56$.
Then $y+4z=142-N\le86$ and $x=N-y-z\le20$ gives $y+z\ge N-20$.
Subtract: $3z=(y+4z)-(y+z)\le (142-N)-(N-20)=162-2N$.
So $z\le(162-2N)/3$.
Also $y=142-N-4z\le20$ gives $z\ge(122-N)/4$.

For $N=56$: $z\ge\lceil66/4\rceil=17$ and $z\le\lfloor50/3\rfloor=16$, contradiction. For $N>56$ the interval is even emptier; in general need $(122-N)/4\le(162-2N)/3$, i.e. $5N\le282$, so $N\le56.4$. Thus $N\ge57$ impossible, and $N=56$ impossible by the integer contradiction.

Hence 55 is greatest possible.

ANSWER 4: E

Problem 5:
We have $ABA\times CD = CDCD$, where $A,B,C,D$ are different digits, $A\ne0$, $C\ne0$.

Write $CD$ as the 2-digit number $10C+D$.
$CDCD$ as a 4-digit number is:
$$1000C+100D+10C+D =1010C+101D=101(10C+D)=101\cdot CD.$$

So:
$$ABA \times CD = 101\times CD.$$
Since $CD\ne0$, divide:
$$ABA =101.$$

Thus $A=1$, $B=0$, and $A+B=1$.

This is consistent with distinctness: we can choose e.g. $C=2,D=3$, all of $1,0,2,3$ different, and $101\times23=2323$. So valid.

Other sums are ruled out.

ANSWER 5: A

Problem 6:
Total 100: 52 boys, 48 girls. 40 from Jonas, 60 from Clay.

20 of the girls are from Jonas. So girls from Clay = total girls $-$ girls from Jonas:
$$48-20=28.$$

Clay total 60 = boys from Clay + girls from Clay, so:
$$\text{boys from Clay}=60-28=32.$$

Check: boys from Jonas $=52-32=20$, Jonas total $=20+20=40$. Works.

ANSWER 6: B

Problem 7:
Compute $(1901+\cdots+1993)-(101+\cdots+193)$.

Both sums have $1993-1901+1=93$ terms and $193-101+1=93$ terms.

Pair termwise: for $k=0,\dots,92$,
$$(1901+k)-(101+k)=1800.$$

So difference $=93\times1800$.
$93\times18=1674$, so $93\times1800=167{,}400$.

Other choices come from miscounting terms or arithmetic error.

ANSWER 7: A

Problem 8:
Find tens digit of $7^{2011}$, i.e. $7^{2011}\pmod{100}$.

Compute powers mod 100:
$7^1=7$,
$7^2=49$,
$7^3=343\equiv43$,
$7^4\equiv43\times7=301\equiv1$.

Since $7^4\equiv1\pmod{100}$, the last two digits repeat with period 4.

$2011 = 4\cdot502+3$, so:
$$7^{2011}=(7^4)^{502}\cdot7^3\equiv1^{502}\cdot43\equiv43\pmod{100}.$$

Thus the number ends $\dots43$, tens digit $4$.

ANSWER 8: D

Problem 9:
10 different positive whole numbers have mean 10, so sum $=10\times10=100$.

To make one number as large as possible, make the other nine as small as possible while distinct and positive.

Smallest 9 distinct positives are $1,2,3,4,5,6,7,8,9$, sum $=45$.

Then the tenth number $=100-45=55$, which is distinct from the others and valid. Any larger value would force the other nine to sum to $<45$, impossible with distinct positives.

So largest possible is 55.

ANSWER 9: C

Problem 10:
Count student-days (one student working one day).

Allen: $7$ students $\times$ $3$ days $=21$.
Balboa: $4\times5=20$.
Carver: $5\times9=45$.
Total $=21+20+45=86$ student-days.

Total pay $\$774$ for $86$ student-days at equal rate:
$$\text{rate}=774/86=9 \text{ dollars per student-day}.$$

Bal

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that $o$ is an odd whole number and $n$ is any whole number. We must determine which statement about $o^2 + no$ is always true.

**Step-by-Step Solution:**
1. Factor the expression:
   $$o^2 + no = o(o + n)$$
2. Since $o$ is an odd whole number, the parity of the product $o(o + n)$ is entirely determined by the parity of $(o + n)$:
   - If $n$ is **odd**, then $\text{odd} + \text{odd} = \text{even}$, so $o(o+n)$ is $\text{odd} \times \text{even} = \textbf{even}$.
   - If $n$ is **even**, then $\text{odd} + \text{even} = \text{odd}$, so $o(o+n)$ is $\text{odd} \times \text{odd} = \textbf{odd}$.
3. Therefore, the expression $o^2 + no$ is odd if and only if $n$ is even. 
4. This matches the statement: "it is odd only if $n$ is even."

ANSWER 1: E

---

### Problem 2:
**Restatement:** Isabella redeems 6 coupons, one every 10 days. Pete's is closed on Sundays, and none of the 6 dates falls on a Sunday. We need to find the day of the week she redeems her first coupon.

**Step-by-Step Solution:**
1. There are 7 days in a week. Moving forward by 10 days advances the day of the week by:
   $$10 \equiv 3 \pmod 7$$
2. Let $d$ represent the day of the week of the first coupon ($0 = \text{Sunday}, 1 = \text{Monday}, \dots, 6 = \text{Saturday}$).
3. The days of the week for the 6 coupons will be:
   - Coupon 1: $d$
   - Coupon 2: $d + 3$
   - Coupon 3: $d + 6$
   - Coupon 4: $d + 9 \equiv d + 2 \pmod 7$
   - Coupon 5: $d + 12 \equiv d + 5 \pmod 7$
   - Coupon 6: $d + 15 \equiv d + 1 \pmod 7$
4. The six days of the week visited are $d, d+1, d+2, d+3, d+5, d+6$. Notice that the only day of the week omitted from this list is $d+4$.
5. Since no circled date falls on a Sunday, Sunday must be the omitted day:
   $$d + 4 \equiv 0 \pmod 7 \implies d \equiv 3 \pmod 7$$
6. Day 3 corresponds to **Wednesday**.
   - Checking: Wed $\to$ Sat $\to$ Tue $\to$ Fri $\to$ Mon $\to$ Thu. None of these is Sunday.

ANSWER 2: C

---

### Problem 3:
**Restatement:** We are given the telescoping product $\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdots \frac{a}{b} = 9$. We must find the sum $a + b$.

**Step-by-Step Solution:**
1. Each fraction is of the form $\frac{k+1}{k}$, where the numerator of each term cancels with the denominator of the subsequent term:
   $$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{a}{b} = \frac{a}{2}$$
   where $b = a - 1$.
2. We are given that this product equals 9:
   $$\frac{a}{2} = 9 \implies a = 18$$
3. Since $b = a - 1$, we have $b = 17$.
4. The sum of $a$ and $b$ is:
   $$a + b = 18 + 17 = 35$$

ANSWER 3: D

---

### Problem 4:
**Restatement:** Nicolas has 20 stamps each of 5¢, 10¢, and 25¢. We want to find the greatest number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-Step Solution:**
1. Calculate the total value and count of all stamps:
   - $20 \times 5¢ = 100¢$
   - $20 \times 10¢ = 200¢$
   - $20 \times 25¢ = 500¢$
   - Total value $= 100 + 200 + 500 = 800¢$ using 60 stamps.
2. To use the **maximum** number of stamps to make 710¢, Nicolas must **minimize** the number of stamps removed to subtract:
   $$800¢ - 710¢ = 90¢$$
3. To remove the fewest stamps totaling 90¢, we try to use the largest denomination stamps (25¢):
   - With 3 stamps of 25¢ ($75¢$), the remaining amount needed is $15¢$.
   - Since Nicolas has only 5¢, 10¢, and 25¢ stamps, 15¢ cannot be made with a single stamp. It requires at least two stamps ($10¢ + 5¢$).
   - Thus, 90¢ can be formed with $3 \times 25¢ + 1 \times 10¢ + 1 \times 5¢$, which uses $3 + 1 + 1 = 5$ stamps.
   - Can 90¢ be formed using 4 stamps? The maximum value of 4 stamps less than or equal to 90¢ would be $3 \times 25¢ + 10¢ = 85¢ \ne 90¢$. Thus, 4 stamps is impossible.
4. The minimum number of stamps to remove is 5.
5. Therefore, the maximum number of stamps used is:
   $$60 - 5 = 55$$

ANSWER 4: E

---

### Problem 5:
**Restatement:** In the multiplication $ABA \times CD = CDCD$, $A, B, C, D$ are distinct digits. We need to find $A + B$.

**Step-by-Step Solution:**
1. Express $CDCD$ in terms of the two-digit number $CD$:
   $$CDCD = CD \times 100 + CD = CD \times 101$$
2. Substitute this into the equation:
   $$ABA \times CD = 101 \times CD$$
3. Since $CD$ is a non-zero two-digit number, divide both sides by $CD$:
   $$ABA = 101$$
4. Comparing digits, we have $A = 1$ and $B = 0$.
5. Thus:
   $$A + B = 1 + 0 = 1$$

ANSWER 5: A

---

### Problem 6:
**Restatement:** Out of 100 students at a camp, 52 are boys and 48 are girls. Also, 40 are from Jonas Middle School and 60 from Clay Middle School. Given that 20 of the girls are from Jonas, find how many boys are from Clay.

**Step-by-Step Solution:**
1. Find the number of girls from Clay Middle School:
   $$\text{Total girls} - \text{Girls from Jonas} = 48 - 20 = 28\text{ girls from Clay}$$
2. Find the number of boys from Clay Middle School:
   $$\text{Total students from Clay} - \text{Girls from Clay} = 60 - 28 = 32\text{ boys from Clay}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** Compute the value of:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-Step Solution:**
1. Pair each term in the first sum with the corresponding term in the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
2. Each paired difference is equal to:
   $$1900 + k - (100 + k) = 1800$$
3. Find the number of terms: from $k = 1$ to $k = 93$, there are:
   $$93 - 1 + 1 = 93\text{ terms}$$
4. Multiply the difference by the number of terms:
   $$93 \times 1800 = 93 \times 18 \times 100 = 1674 \times 100 = 167,400$$

ANSWER 7: A

---

### Problem 8:
**Restatement:** Find the tens digit of $7^{2011}$.

**Step-by-Step Solution:**
1. Find the pattern of the last two digits of powers of 7 by computing $7^n \pmod{100}$:
   - $7^1 = 07$
   - $7^2 = 49$
   - $7^3 = 343 \equiv 43 \pmod{100}$
   - $7^4 = 43

# Problem 1:
We must decide when $o^2 + no$ is odd or even, given $o$ is odd and $n$ is any whole number.

**Key facts:** odd × odd = odd, so $o^2$ is **odd**. Since $o$ is odd, multiplying by $o$ does not change $n$'s parity, so $no$ has the **same parity as $n$**.

**Case 1: $n$ even.** Then $no$ is even, so $o^2 + no = \text{odd} + \text{even} = \textbf{odd}$.
**Case 2: $n$ odd.** Then $no$ is odd, so $o^2 + no = \text{odd} + \text{odd} = \textbf{even}$.

So the number is odd exactly when $n$ is even. In particular, **if the number is odd, $n$ must be even**.

Ruling out traps: (A) fails since $o=1, n=1$ gives $1+1=2$, even; (B) fails since $n=2$ gives $1+2=3$, odd; (C) and (D) state the exact opposite of the truth. Only (E) matches.

ANSWER 1: E

# Problem 2:
The 6 redemption dates are 10 days apart, so their days of the week shift by $10 \bmod 7 = 3$ each time.

If the first date falls on weekday $W$, the six dates fall on (mod 7):
$$W,\; W+3,\; W+6,\; W+2,\; W+5,\; W+1$$
That is **six of the seven weekdays**, and the only weekday missed is $W+4$.

Since no circled date is a Sunday, the one missed weekday must be Sunday:
$$W + 4 = \text{Sunday} \implies W = \text{Wednesday}.$$

**Check:** Starting Wednesday, the dates fall on Wed, Sat, Tue, Fri, Mon, Thu — no Sunday. (Trap: starting Thursday gives Thu, Sun, ... — fails immediately.)

ANSWER 2: C

# Problem 3:
The product is a chain of fractions (the bars were lost in formatting):
$$\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdots\frac{a}{b} = 9.$$

Each numerator cancels the denominator before it (e.g., $\frac{3}{2}\cdot\frac{4}{3} = \frac{4}{2}$), and each fraction has the form $\frac{k+1}{k}$, so $b = a-1$. Everything telescopes:
$$\text{Product} = \frac{a}{2} = 9 \implies a = 18, \quad b = 17.$$
Then $a + b = 18 + 17 = 35$.

Trap check: $a+b=17$ would mean $a=9$, giving product $\tfrac92$, not $9$; only $35$ fits the pattern.

ANSWER 3: D

# Problem 4:
Work in units of 5¢: we need total value $710¢ = 142$ units, using stamps worth $1, 2, 5$ units (i.e., 5¢, 10¢, 25¢), at most 20 of each. To **maximize the count**, use as many cheap stamps as possible.

**Building a 55-stamp solution:** Take nineteen 1's and nineteen 2's: value $19 + 38 = 57$ units with 38 stamps. Remaining value $142 - 57 = 85 = 17\times 5$: seventeen 5's.
$$19\times 5¢ + 19\times 10¢ + 17\times 25¢ = 95 + 190 + 425 = 710¢ = \$7.10.\; ✓$$
That's $19+19+17 = 55$ stamps, all counts $\le 20$. ✓

**Why 56 is impossible:** with 56 stamps, at least $56 - 40 = 16$ must be 25¢ stamps (only 40 cheap ones exist). Let $z$ = number of 25¢ stamps:
- $z=16$: max value $= 100 + 200 + 400 = 700 < 710$. ✗
- $z=17$: need $5x+10y=285$, $x+y=39 \Rightarrow x=21>20$. ✗
- $z=18$: $x=24>20$. ✗  $z=19$: $x=27>20$. ✗  $z=20$: $x=30>20$. ✗

Traps: 54 comes from $20×5+16×10+18×25 = 710$; 51 from $20×5+11×10+20×25 = 710$ — both valid but not maximal.

ANSWER 4: E

# Problem 5:
Write $CDCD$ as a number:
$$CDCD = 1000C + 100D + 10C + D = 101(10C + D) = 101 \times CD.$$
The given equation is
$$ABA \times CD = CDCD = 101 \times CD.$$
Since $CD \neq 0$ (it's a two-digit number), we can divide both sides by $CD$:
$$ABA = 101.$$
So $A = 1$ and $B = 0$, giving $A + B = 1$.

**Consistency check:** $C, D$ just need to be distinct from $0, 1$ and each other, e.g., $101 \times 23 = 2323$. ✓ (Trap: guessing digits without using the factorization 101 wastes time or gives 9.)

ANSWER 5: A

# Problem 6:
Set up a two-way table with 100 total students.

- Girls: 48 total, and 20 girls are from Jonas $\Rightarrow$ girls from Clay $= 48 - 20 = 28$.
- Jonas has 40 students, of whom 20 are girls $\Rightarrow$ boys from Jonas $= 40 - 20 = 20$.
- Boys: 52 total $\Rightarrow$ boys from Clay $= 52 - 20 = \textbf{32}$.

**Check:** Jonas $20+20=40$ ✓; Clay $28+32=60$ ✓; boys $20+32=52$ ✓; girls $20+28=48$ ✓; total $100$ ✓. (Trap: 20 counts the boys from Jonas, not Clay.)

ANSWER 6: B

# Problem 7:
Both sums have the same number of terms: $1901$ to $1993$ has $1993-1901+1 = 93$ terms, and $101$ to $193$ has $193-101+1 = 93$ terms.

Pair them up: each term in the first sum is exactly $1800$ more than its partner ($1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$). So the difference is
$$93 \times 1800 = 167{,}400.$$

**Check via formulas:** $\frac{93(1901+1993)}{2} - \frac{93(101+193)}{2} = 181{,}071 - 13{,}671 = 167{,}400$. ✓ (Trap: 181,071 is the first sum alone — forgetting to subtract.)

ANSWER 7: A

# Problem 8:
Only the tens digit matters, so track the last two digits of $7^{k}$, which cycle with period 4:
$$7^1 = 07,\quad 7^2 = 49,\quad 7^3 = 343 \to 43,\quad 7^4 = 2401 \to 01,\quad 7^5 \to 07, \ldots$$

Find where $2011$ falls in the cycle:
$$2011 = 4 \times 502 + 3,$$
so $7^{2011}$ ends in the same two digits as $7^3$, namely $\ldots 43$.

The tens digit is $\textbf{4}$. (Trap: $7^{2012}$, a multiple of 4, ends in $01$, tens digit 0 — choice A; that would require the exponent to be divisible by 4.)

ANSWER 8: D

# Problem 9:
The ten **different** positive whole numbers have average 10, so their sum is $10 \times 10 = 100$.

To make one number as large as possible, make the other nine as small as possible. Since they must be **distinct positive** whole numbers, the smallest possible set is $1, 2, 3, \ldots, 9$, with sum
$$1+2+\cdots+9 = 45.$$
So the largest possible value is
$$100 - 45 = 55.$$

**Check:** $1,2,\ldots,9,55$ are ten different positive whole numbers with sum $100$ and average $10$. ✓ (Trap: 91 arises if you ignore "different" and use nine 1's; but the numbers must be distinct.)

ANSWER 9: C

# Problem 10:
Count student-days (the unit being paid):
- Allen: $7 \times 3 = 21$ student-days
- Balboa: $4 \times 5 = 20$ student-days
- Carver: $5 \times 9 = 45$ student-days

Total: $21 + 20 + 45 = 86$ student-days, paid $\$774$ altogether. So the daily rate is
$$774 \div 86 = 9 \text{ dollars per student-day}.$$
Balboa's students worked 20 student-days, so they earned
$$20 \times 9 = 18

Problem 1:  
We need the parity of \(o^2+no\), where \(o\) is odd and \(n\) is any whole number.

Factor:
\[
o^2+no=o(o+n).
\]

Since \(o\) is odd:

- If \(n\) is even, then \(o+n\) is odd, so \(o(o+n)\) is odd.
- If \(n\) is odd, then \(o+n\) is even, so \(o(o+n)\) is even.

Thus the expression is odd exactly when \(n\) is even.

Check the choices:
- A and B are false because the parity depends on \(n\).
- C says “even only if \(n\) is even,” but when the expression is even, \(n\) is actually odd.
- D says “odd only if \(n\) is odd,” but when the expression is odd, \(n\) is even.
- E says “odd only if \(n\) is even,” which is true.

ANSWER 1: E

Problem 2:  
Isabella redeems coupons 10 days apart. Since
\[
10\equiv 3 \pmod 7,
\]
the six circled days are the first day plus
\[
0,3,6,9,12,15
\]
days, which modulo 7 gives offsets
\[
0,3,6,2,5,1.
\]

So the six circled days are all weekdays except the one 4 days after the first day. Since none of the circled days is a Sunday, Sunday must be the un-circled day: 4 days after the first day.

Therefore the first coupon is 4 days before Sunday, which is Wednesday.

Check: Wednesday, Saturday, Tuesday, Friday, Monday, Thursday — no Sunday.

ANSWER 2: C

Problem 3:  
The product is the telescoping product of fractions:
\[
\frac32\cdot \frac43\cdot \frac54\cdot \frac65\cdots \frac{a}{b}=9.
\]

The numerators cancel the next denominators:
\[
\frac32\cdot \frac43=\frac42,
\]
\[
\frac42\cdot \frac54=\frac52,
\]
and continuing, the whole product simplifies to
\[
\frac{a}{2}.
\]

So
\[
\frac{a}{2}=9 \implies a=18.
\]

By the pattern, the denominator is one less than the numerator, so
\[
b=17.
\]

Thus
\[
a+b=18+17=35.
\]

ANSWER 3: D

Problem 4:  
Nicolas has 20 of each type of stamp, so he has
\[
20+20+20=60
\]
stamps total. Their total value is
\[
20(5+10+25)=20(40)=800
\]
cents.

He wants postage of
\[
\$7.10=710
\]
cents, so the stamps he does **not** use must total
\[
800-710=90
\]
cents.

To maximize the number of stamps he uses, we minimize the number of stamps he leaves unused.

Can 4 unused stamps total 90 cents? Let \(x,y,z\) be the numbers of 25-, 10-, and 5-cent unused stamps. Then
\[
x+y+z=4
\]
and
\[
25x+10y+5z=90.
\]

Divide by 5:
\[
5x+2y+z=18.
\]

Subtract \(x+y+z=4\):
\[
4x+y=14.
\]

Testing possible \(x=0,1,2,3,4\):

- \(x=0\Rightarrow y=14\), impossible.
- \(x=1\Rightarrow y=10\), impossible.
- \(x=2\Rightarrow y=6\), impossible.
- \(x=3\Rightarrow y=2\), then \(z=4-3-2=-1\), impossible.
- \(x=4\Rightarrow y=-2\), impossible.

So 4 unused stamps cannot total 90 cents.

Can 5 unused stamps total 90 cents? Yes:
\[
25+25+25+10+5=90.
\]

Then he uses
\[
60-5=55
\]
stamps. For example, he could use 19 five-cent stamps, 19 ten-cent stamps, and 17 twenty-five-cent stamps:
\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So the greatest possible number of stamps is 55. Choice D, 54, is possible but not greatest.

ANSWER 4: E

Problem 5:  
We are told
\[
ABA\times CD=CDCD,
\]
where \(A,B,C,D\) are different digits.

Write the numbers in place-value form:
\[
ABA=101A+10B,
\]
\[
CD=10C+D,
\]
\[
CDCD=1000C+100D+10C+D=101(10C+D).
\]

So
\[
(101A+10B)(10C+D)=101(10C+D).
\]

Since \(CD\) is a two-digit number, \(10C+D\neq 0\), so we can divide both sides by \(10C+D\):
\[
101A+10B=101.
\]

The only digit solution is
\[
A=1,\quad B=0.
\]

Therefore
\[
A+B=1+0=1.
\]

ANSWER 5: A

Problem 6:  
There are 48 girls total, and 20 of the girls are from Jonas Middle School. So the number of girls from Clay Middle School is
\[
48-20=28.
\]

There are 60 students total from Clay Middle School, so the number of boys from Clay is
\[
60-28=32.
\]

ANSWER 6: B

Problem 7:  
Pair the terms:
\[
(1901+1902+\cdots+1993)-(101+102+\cdots+193)
\]
\[
=(1901-101)+(1902-102)+\cdots+(1993-193).
\]

Each pair is
\[
1800.
\]

Now count the number of terms:
\[
193-101+1=93.
\]

So the value is
\[
93\cdot 1800=167400.
\]

ANSWER 7: A

Problem 8:  
We need the tens digit of \(7^{2011}\), so we need the last two digits.

Powers of 7 modulo 100 cycle:
\[
7^1=07,
\]
\[
7^2=49,
\]
\[
7^3=343\equiv 43,
\]
\[
7^4=2401\equiv 01.
\]

Then the cycle repeats:
\[
07,\ 49,\ 43,\ 01.
\]

Since
\[
2011\equiv 3 \pmod 4,
\]
we have
\[
7^{2011}\equiv 7^3\equiv 43 \pmod{100}.
\]

The last two digits are 43, so the tens digit is 4.

ANSWER 8: D

Problem 9:  
The average of 10 different positive whole numbers is 10, so their sum is
\[
10\cdot 10=100.
\]

To make one number as large as possible, make the other 9 numbers as small as possible while still being different positive whole numbers. The smallest 9 distinct positive whole numbers are
\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is
\[
1+2+\cdots+9=45.
\]

So the largest number can be
\[
100-45=55.
\]

This is possible: \(1,2,3,4,5,6,7,8,9,55\) are all different positive whole numbers.

ANSWER 9: C

Problem 10:  
First count total student-days worked:

- Allen: \(7\cdot 3=21\) student-days
- Balboa: \(4\cdot 5=20\) student-days
- Carver: \(5\cdot 9=45\) student-days

Total:
\[
21+20+45=86
\]
student-days.

The total payment is \$774, so the daily rate is
\[
774\div 86=9
\]
dollars per student-day.

Balboa students worked 20 student-days, so they earned
\[
20\cdot 9=180
\]
dollars.

ANSWER 10: C

# Problem 1:
We need a statement always true about o² + no, where o is odd.

Factor: o² + no = o(o + n).

Since o is odd, the parity of the product is the parity of (o + n).
- If n is even, o + n is odd → product = odd·odd = **odd**.
- If n is odd, o + n is even → product = odd·even = **even**.

So the expression is odd exactly when n is even.

Ruling out others: A fails (n odd gives even), B fails (n even gives odd), C says "even only if n is even" — but it's even when n is *odd*, false. D says "odd only if n is odd" — false, it's odd when n is even. E: "odd only if n is even" — true.

ANSWER 1: E

# Problem 2:
The redemption days are 0, 10, 20, 30, 40, 50 days after the first.

Reduce mod 7 (days of week shift): 0, 3, 6, 2, 5, 1.

These residues are {0, 1, 2, 3, 5, 6} — every residue except **4**. So the only day of the week never hit is the day 4 days after the starting day. For no coupon to land on Sunday, Sunday must be that missed day:

start + 4 ≡ Sunday.

Counting back 4 days from Sunday: Saturday, Friday, Thursday, **Wednesday**.

Check: Wed → +10 Sat → +20 Tue → +30 Fri → +40 Mon → +50 Thu. No Sundays. ✓

ANSWER 2: C

# Problem 3:
The product is (3/2)·(4/3)·(5/4)·(6/5)·…·(a/b), where each numerator is one more than its denominator, so b = a − 1.

Telescoping: everything cancels except the last numerator and first denominator:
a/2 = 9 → a = 18, so b = 17.

a + b = 18 + 17 = 35.

(Choice E, 37, would come from mistakenly taking a = 19.)

ANSWER 3: D

# Problem 4:
Total value of all 60 stamps: 20(5 + 10 + 25) = 20·40 = 800 cents.

We need 710 cents, so we must *remove* stamps worth exactly 800 − 710 = 90 cents, using as **few** stamps as possible (to keep the most stamps).

Can 90 be made with 4 or fewer stamps? Max with 4 stamps is 100; possible 4-stamp totals near 90: 100, 85, 80, 75, 70… none equal 90. Three or fewer is at most 75. So 4 or fewer is impossible.

With 5 stamps: 25 + 25 + 25 + 10 + 5 = 90 ✓

So remove 5 stamps: 60 − 5 = 55 stamps.

Check: 17(25) + 19(10) + 19(5) = 425 + 190 + 95 = 710 cents, using 17 + 19 + 19 = 55 stamps.

ANSWER 4: E

# Problem 5:
CDCD as a number = CD·100 + CD = CD·101.

So ABA × CD = 101 × CD, and since CD ≠ 0, we get ABA = 101.

Thus A = 1, B = 0 (and indeed A, B, C, D distinct works, e.g. 101 × 23 = 2323).

A + B = 1 + 0 = 1.

ANSWER 5: A

# Problem 6:
Jonas total = 40, and 20 of those were girls, so boys from Jonas = 40 − 20 = 20.

Total boys = 52, so boys from Clay = 52 − 20 = 32.

(Check: girls from Clay = 48 − 20 = 28; 32 + 28 = 60 = Clay total ✓.)

Choice C (40) is the Jonas total, and E (52) is all boys — traps.

ANSWER 6: B

# Problem 7:
Pair each term: 1901 − 101 = 1800, 1902 − 102 = 1800, …, each pair differs by 1800.

Number of terms: 1993 − 1901 + 1 = 93.

Total = 93 × 1800 = 167,400.

ANSWER 7: A

# Problem 8:
Look at 7ⁿ mod 100:
- 7¹ = 07
- 7² = 49
- 7³ = 343 → 43
- 7⁴ = 2401 → 01
- then the pattern repeats with period 4.

2011 ÷ 4 = 502 remainder 3, so 7²⁰¹¹ ≡ 7³ ≡ 43 (mod 100).

Last two digits are 43, so the tens digit is 4.

ANSWER 8: D

# Problem 9:
Sum of the 10 numbers = 10 × 10 = 100.

To make one number as large as possible, make the other nine as small as possible while distinct and positive: 1 + 2 + … + 9 = 45.

Largest = 100 − 45 = 55.

(90 or 91 would require repeats or a zero, which isn't allowed.)

ANSWER 9: C

# Problem 10:
Total student-days:
- Allen: 7 × 3 = 21
- Balboa: 4 × 5 = 20
- Carver: 5 × 9 = 45

Total = 86 student-days.

Daily wage = 774 ÷ 86 = $9.00 per student-day.

Balboa earnings = 20 × 9 = $180.00.

(A is one day's pay for one student; E is Carver's total.)

ANSWER 10: C

Problem 1:

We need determine the parity of

\[
o^2+no.
\]

Factor out \(o\):

\[
o^2+no=o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on \(o+n\).

- If \(n\) is even, then odd \(+\) even \(=\) odd, so \(o+n\) is odd. Thus odd \(\cdot\) odd is odd.
- If \(n\) is odd, then odd \(+\) odd \(=\) even, so the product is even.

Therefore, the expression is odd only when \(n\) is even.

- A is wrong because the expression is not always odd.
- B is wrong because it is odd when \(n\) is even.
- C reverses the condition for being even.
- D reverses the condition for being odd.

ANSWER 1: E

Problem 2:

Isabella uses a coupon every 10 days. Since

\[
10\equiv 3\pmod 7,
\]

each coupon date moves forward by 3 days of the week.

If the first coupon is redeemed on day \(d\), the six redemption days are:

\[
d,\ d+3,\ d+6,\ d+2,\ d+5,\ d+1.
\]

These are six different days of the week. The only day missing is

\[
d+4.
\]

Since none of the six coupon dates is Sunday, the missing day must be Sunday. Thus,

\[
d+4=\text{Sunday}.
\]

Four days before Sunday is Wednesday.

Indeed, beginning on Wednesday gives:

\[
\text{Wednesday, Saturday, Tuesday, Friday, Monday, Thursday}.
\]

Sunday is the only missing day.

ANSWER 2: C

Problem 3:

The product is

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9.
\]

Almost every numerator cancels with a denominator:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab
=\frac a2.
\]

So

\[
\frac a2=9.
\]

Therefore,

\[
a=18.
\]

The final fraction must be

\[
\frac{18}{17},
\]

so \(b=17\). Therefore,

\[
a+b=18+17=35.
\]

Choices 11, 13, and 17 are too small, while 37 would correspond to a different ending fraction.

ANSWER 3: D

Problem 4:

Let

- \(x\) = number of 5-cent stamps,
- \(y\) = number of 10-cent stamps,
- \(z\) = number of 25-cent stamps.

In units of 5 cents, the postage total is

\[
\frac{710}{5}=142.
\]

Thus,

\[
x+2y+5z=142.
\]

We want to maximize

\[
x+y+z.
\]

Since there are at most 20 five-cent stamps, \(x\leq 20\). Therefore,

\[
2y+5z=142-x\geq 142-20=122.
\]

Since \(y\leq20\), if \(z\leq16\), then

\[
2y+5z\leq 2(20)+5(16)=40+80=120,
\]

which is not enough. So \(z\geq17\).

If \(z=17\), then

\[
2y+5(17)\geq122,
\]

so

\[
2y+85\geq122,
\]

\[
2y\geq37,
\]

and hence \(y\geq19\).

Now use

\[
x+y+z=142-y-4z.
\]

Since \(y\geq19\) and \(z\geq17\),

\[
x+y+z\leq142-19-4(17)
=142-19-68
=55.
\]

This maximum is attainable with

\[
x=19,\qquad y=19,\qquad z=17.
\]

Check:

\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So 55 stamps can be used. Choices 45, 46, 51, and 54 are possible totals to consider, but 55 is the greatest possible.

ANSWER 4: E

Problem 5:

We are told

\[
ABA\times CD=CDCD.
\]

The number \(ABA\) is

\[
100A+10B+A=101A+10B.
\]

The number \(CDCD\) is

\[
1000C+100D+10C+D.
\]

Group the \(C\)'s and \(D\)'s:

\[
CDCD=1010C+101D=101(10C+D)=101\cdot CD.
\]

Thus the equation becomes

\[
ABA\cdot CD=101\cdot CD.
\]

Since \(CD\) is a two-digit number and is not zero, divide both sides by \(CD\):

\[
ABA=101.
\]

Therefore,

\[
A=1,\qquad B=0.
\]

So

\[
A+B=1+0=1.
\]

For example, \(101\times23=2323\), and the digits \(1,0,2,3\) are all different.

ANSWER 5: A

Problem 6:

There are 40 students from Jonas Middle School, and 20 of those are girls.

Therefore the number of boys from Jonas is

\[
40-20=20.
\]

There are 52 boys altogether. Therefore the number of boys from Clay Middle School is

\[
52-20=32.
\]

The other choices do not correctly account for the 20 Jonas girls and the total of 52 boys.

ANSWER 6: B

Problem 7:

There are

\[
1993-1901+1=93
\]

terms in the first sum.

There are also

\[
193-101+1=93
\]

terms in the second sum.

Pair corresponding terms:

\[
1901-101=1800,
\]

\[
1902-102=1800,
\]

and so on, through

\[
1993-193=1800.
\]

Thus there are 93 differences of 1800:

\[
93\cdot1800=167400.
\]

Therefore the value is

\[
167,\!400.
\]

ANSWER 7: A

Problem 8:

We need the tens digit of

\[
7^{2011}.
\]

It is enough to examine powers of 7 modulo 100:

\[
7^1=7,
\]

\[
7^2=49,
\]

\[
7^3=343,
\]

whose last two digits are 43, and

\[
7^4=2401,
\]

whose last two digits are 01.

Thus the last two digits repeat every 4 powers because

\[
7^4\equiv1\pmod{100}.
\]

Now,

\[
2011\div4
\]

leaves remainder 3, since

\[
2011=4(502)+3.
\]

Therefore,

\[
7^{2011}
\]

has the same last two digits as

\[
7^3=343,
\]

which are 43. The tens digit is 4.

ANSWER 8: D

Problem 9:

The average of 10 different positive whole numbers is 10, so their total sum is

\[
10\cdot10=100.
\]

To make one number as large as possible, make the other nine numbers as small as possible.

The nine smallest different positive whole numbers are

\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is

\[
1+2+3+4+5+6+7+8+9=45.
\]

So the largest possible tenth number is

\[
100-45=55.
\]

This works because the ten numbers can be

\[
1,2,3,4,5,6,7,8,9,55,
\]

which are all different and have total 100.

A value such as 90 or 91 would leave too small a total for nine distinct positive whole numbers.

ANSWER 9: C

Problem 10:

Compute the total number of student-days worked.

Allen students worked:

\[
7\cdot3=21
\]

student-days.

Balboa students worked:

\[
4\cdot5=20
\]

student-days.

Carver students worked:

\[
5\cdot9=45
\]

student-days.

So the total number of student-days is

\[
21+20+45=86.
\]

The total pay was \$774, so the pay per student per day was

\[
\frac{774}{86}=9.
\]

Balboa students worked 20 student-days, so they earned

\[
20\cdot9=180.
\]

Thus the Balboa students earned \$180.00 altogether.

ANSWER 10: C

Problem 1:
We are asked about the parity of $o^2+no$ where $o$ is odd and $n$ is any whole number.

Factor:
$$o^2+no = o(o+n).$$

$o$ is odd, so $o^2 = o\cdot o$ is odd times odd = odd.

Hence parity is odd + $n\cdot o$.
$n\cdot o$ has the same parity as $n$, because $o$ is odd: if $n$ is even, $n o$ is even; if $n$ is odd, $n o$ is odd.

So:
- if $n$ even: odd + even = odd.
- if $n$ odd: odd + odd = even.

Thus the number is odd iff $n$ is even, and even iff $n$ is odd.

Check choices:
A. always odd - false, e.g. $o=1,n=1$ gives $1+1=2$ even.
B. always even - false, e.g. $o=1,n=2$ gives $1+2=3$ odd.
C. it is even only if $n$ is even - means even $\implies$ $n$ even. False, even happens when $n$ is odd.
D. it is odd only if $n$ is odd - means odd $\implies$ $n$ odd. False, odd happens when $n$ is even.
E. it is odd only if $n$ is even - means odd $\implies$ $n$ even. True by above.

ANSWER 1: E

Problem 2:
Isabella redeems on days $d, d+10, d+20, d+30, d+40, d+50$.
No date is a Sunday.

Modulo 7, $10\equiv 3$. So successive dates advance by 3 weekdays.

Starting from residue $r$, the residues are:
$$r,\ r+3,\ r+6,\ r+9=r+2,\ r+12=r+5,\ r+15=r+1 \pmod 7.$$
These 6 are all distinct, because adding 3 mod 7 cycles through all 7 residues. Continuing one more step would give $r+18=r+4$.

Thus the 6 dates occupy 6 distinct days of the week, missing exactly the 7th residue $r+4$.

Since none of the 6 is Sunday, the missing one must be Sunday. So:
$$r+4 \equiv \text{Sunday}.$$

Hence $r$ is 4 days before Sunday, i.e. Wednesday.

Check: Wed $\to$ Wed+3=Sat $\to$ +3=Tue $\to$ +3=Fri $\to$ +3=Mon $\to$ +3=Thu. That is Wed,Sat,Tue,Fri,Mon,Thu - no Sunday. Any other start would include Sunday.

ANSWER 2: C

Problem 3:
We interpret $3\ 2\cdot 4\ 3\cdots$ as fractions:
$$\frac32 \cdot \frac43 \cdot \frac54 \cdot \frac65 \cdots \frac{a}{b}=9.$$

The pattern is numerator $3,4,5,6,\dots$ and denominator $2,3,4,5,\dots$, i.e. term $k/(k-1)$. The last term is $a/b$, so we must have $b=a-1$ for the pattern to continue.

Then the product telescopes:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}
= \frac{a}{2},$$
because all intermediate $3,4,\dots,a-1$ cancel.

Set equal to 9:
$$\frac{a}{2}=9 \implies a=18.$$
Then $b=a-1=17$.

Sum $a+b=18+17=35$.

This rules out A,B,C,E.

ANSWER 3: D

Problem 4:
Let $x,y,z$ be numbers of 5-cent, 10-cent, 25-cent stamps used. $0\le x,y,z\le 20$ integers.
Value $710$ cents:
$$5x+10y+25z=710.$$
Divide by 5:
$$x+2y+5z=142. \tag{1}$$
Maximize $N=x+y+z$.

From (1), $x=142-2y-5z$, so $0\le x\le20$ gives:
$$122\le 2y+5z\le 142.$$
Also $N=x+y+z=142-y-4z$, so $y+4z=142-N$.

To use many stamps we want small denominations. Try $x=20,y=20$ fails to give integer $z$. Search systematically.

Since $y\le20$, to satisfy $2y+5z\ge122$ we need $z$ large. If $z\le16$, then $2y+5z\le40+80=120<122$, impossible. So $z\ge17$.

Try $z=17,y=19$: then $2y+5z=38+85=123$, so $x=142-123=19$, within $0-20$. This gives:
$$19\cdot5+19\cdot10+17\cdot25=95+190+425=710,$$
with $N=19+19+17=55$ stamps. So 55 is achievable.

Can we do 56 or more? Let $N\ge56$.
Then $y+4z=142-N\le86$ and $x=N-y-z\le20$ gives $y+z\ge N-20$.
Subtract: $3z=(y+4z)-(y+z)\le (142-N)-(N-20)=162-2N$.
So $z\le(162-2N)/3$.
Also $y=142-N-4z\le20$ gives $z\ge(122-N)/4$.

For $N=56$: $z\ge\lceil66/4\rceil=17$ and $z\le\lfloor50/3\rfloor=16$, contradiction. For $N>56$ the interval is even emptier; in general need $(122-N)/4\le(162-2N)/3$, i.e. $5N\le282$, so $N\le56.4$. Thus $N\ge57$ impossible, and $N=56$ impossible by the integer contradiction.

Hence 55 is greatest possible.

ANSWER 4: E

Problem 5:
We have $ABA\times CD = CDCD$, where $A,B,C,D$ are different digits, $A\ne0$, $C\ne0$.

Write $CD$ as the 2-digit number $10C+D$.
$CDCD$ as a 4-digit number is:
$$1000C+100D+10C+D =1010C+101D=101(10C+D)=101\cdot CD.$$

So:
$$ABA \times CD = 101\times CD.$$
Since $CD\ne0$, divide:
$$ABA =101.$$

Thus $A=1$, $B=0$, and $A+B=1$.

This is consistent with distinctness: we can choose e.g. $C=2,D=3$, all of $1,0,2,3$ different, and $101\times23=2323$. So valid.

Other sums are ruled out.

ANSWER 5: A

Problem 6:
Total 100: 52 boys, 48 girls. 40 from Jonas, 60 from Clay.

20 of the girls are from Jonas. So girls from Clay = total girls $-$ girls from Jonas:
$$48-20=28.$$

Clay total 60 = boys from Clay + girls from Clay, so:
$$\text{boys from Clay}=60-28=32.$$

Check: boys from Jonas $=52-32=20$, Jonas total $=20+20=40$. Works.

ANSWER 6: B

Problem 7:
Compute $(1901+\cdots+1993)-(101+\cdots+193)$.

Both sums have $1993-1901+1=93$ terms and $193-101+1=93$ terms.

Pair termwise: for $k=0,\dots,92$,
$$(1901+k)-(101+k)=1800.$$

So difference $=93\times1800$.
$93\times18=1674$, so $93\times1800=167{,}400$.

Other choices come from miscounting terms or arithmetic error.

ANSWER 7: A

Problem 8:
Find tens digit of $7^{2011}$, i.e. $7^{2011}\pmod{100}$.

Compute powers mod 100:
$7^1=7$,
$7^2=49$,
$7^3=343\equiv43$,
$7^4\equiv43\times7=301\equiv1$.

Since $7^4\equiv1\pmod{100}$, the last two digits repeat with period 4.

$2011 = 4\cdot502+3$, so:
$$7^{2011}=(7^4)^{502}\cdot7^3\equiv1^{502}\cdot43\equiv43\pmod{100}.$$

Thus the number ends $\dots43$, tens digit $4$.

ANSWER 8: D

Problem 9:
10 different positive whole numbers have mean 10, so sum $=10\times10=100$.

To make one number as large as possible, make the other nine as small as possible while distinct and positive.

Smallest 9 distinct positives are $1,2,3,4,5,6,7,8,9$, sum $=45$.

Then the tenth number $=100-45=55$, which is distinct from the others and valid. Any larger value would force the other nine to sum to $<45$, impossible with distinct positives.

So largest possible is 55.

ANSWER 9: C

Problem 10:
Count student-days (one student working one day).

Allen: $7$ students $\times$ $3$ days $=21$.
Balboa: $4\times5=20$.
Carver: $5\times9=45$.
Total $=21+20+45=86$ student-days.

Total pay $\$774$ for $86$ student-days at equal rate:
$$\text{rate}=774/86=9 \text{ dollars per student-day}.$$

Bal

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that $o$ is an odd whole number and $n$ is any whole number. We must determine which statement about $o^2 + no$ is always true.

**Step-by-Step Solution:**
1. Factor the expression:
   $$o^2 + no = o(o + n)$$
2. Since $o$ is an odd whole number, the parity of the product $o(o + n)$ is entirely determined by the parity of $(o + n)$:
   - If $n$ is **odd**, then $\text{odd} + \text{odd} = \text{even}$, so $o(o+n)$ is $\text{odd} \times \text{even} = \textbf{even}$.
   - If $n$ is **even**, then $\text{odd} + \text{even} = \text{odd}$, so $o(o+n)$ is $\text{odd} \times \text{odd} = \textbf{odd}$.
3. Therefore, the expression $o^2 + no$ is odd if and only if $n$ is even. 
4. This matches the statement: "it is odd only if $n$ is even."

ANSWER 1: E

---

### Problem 2:
**Restatement:** Isabella redeems 6 coupons, one every 10 days. Pete's is closed on Sundays, and none of the 6 dates falls on a Sunday. We need to find the day of the week she redeems her first coupon.

**Step-by-Step Solution:**
1. There are 7 days in a week. Moving forward by 10 days advances the day of the week by:
   $$10 \equiv 3 \pmod 7$$
2. Let $d$ represent the day of the week of the first coupon ($0 = \text{Sunday}, 1 = \text{Monday}, \dots, 6 = \text{Saturday}$).
3. The days of the week for the 6 coupons will be:
   - Coupon 1: $d$
   - Coupon 2: $d + 3$
   - Coupon 3: $d + 6$
   - Coupon 4: $d + 9 \equiv d + 2 \pmod 7$
   - Coupon 5: $d + 12 \equiv d + 5 \pmod 7$
   - Coupon 6: $d + 15 \equiv d + 1 \pmod 7$
4. The six days of the week visited are $d, d+1, d+2, d+3, d+5, d+6$. Notice that the only day of the week omitted from this list is $d+4$.
5. Since no circled date falls on a Sunday, Sunday must be the omitted day:
   $$d + 4 \equiv 0 \pmod 7 \implies d \equiv 3 \pmod 7$$
6. Day 3 corresponds to **Wednesday**.
   - Checking: Wed $\to$ Sat $\to$ Tue $\to$ Fri $\to$ Mon $\to$ Thu. None of these is Sunday.

ANSWER 2: C

---

### Problem 3:
**Restatement:** We are given the telescoping product $\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdots \frac{a}{b} = 9$. We must find the sum $a + b$.

**Step-by-Step Solution:**
1. Each fraction is of the form $\frac{k+1}{k}$, where the numerator of each term cancels with the denominator of the subsequent term:
   $$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{a}{b} = \frac{a}{2}$$
   where $b = a - 1$.
2. We are given that this product equals 9:
   $$\frac{a}{2} = 9 \implies a = 18$$
3. Since $b = a - 1$, we have $b = 17$.
4. The sum of $a$ and $b$ is:
   $$a + b = 18 + 17 = 35$$

ANSWER 3: D

---

### Problem 4:
**Restatement:** Nicolas has 20 stamps each of 5¢, 10¢, and 25¢. We want to find the greatest number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-Step Solution:**
1. Calculate the total value and count of all stamps:
   - $20 \times 5¢ = 100¢$
   - $20 \times 10¢ = 200¢$
   - $20 \times 25¢ = 500¢$
   - Total value $= 100 + 200 + 500 = 800¢$ using 60 stamps.
2. To use the **maximum** number of stamps to make 710¢, Nicolas must **minimize** the number of stamps removed to subtract:
   $$800¢ - 710¢ = 90¢$$
3. To remove the fewest stamps totaling 90¢, we try to use the largest denomination stamps (25¢):
   - With 3 stamps of 25¢ ($75¢$), the remaining amount needed is $15¢$.
   - Since Nicolas has only 5¢, 10¢, and 25¢ stamps, 15¢ cannot be made with a single stamp. It requires at least two stamps ($10¢ + 5¢$).
   - Thus, 90¢ can be formed with $3 \times 25¢ + 1 \times 10¢ + 1 \times 5¢$, which uses $3 + 1 + 1 = 5$ stamps.
   - Can 90¢ be formed using 4 stamps? The maximum value of 4 stamps less than or equal to 90¢ would be $3 \times 25¢ + 10¢ = 85¢ \ne 90¢$. Thus, 4 stamps is impossible.
4. The minimum number of stamps to remove is 5.
5. Therefore, the maximum number of stamps used is:
   $$60 - 5 = 55$$

ANSWER 4: E

---

### Problem 5:
**Restatement:** In the multiplication $ABA \times CD = CDCD$, $A, B, C, D$ are distinct digits. We need to find $A + B$.

**Step-by-Step Solution:**
1. Express $CDCD$ in terms of the two-digit number $CD$:
   $$CDCD = CD \times 100 + CD = CD \times 101$$
2. Substitute this into the equation:
   $$ABA \times CD = 101 \times CD$$
3. Since $CD$ is a non-zero two-digit number, divide both sides by $CD$:
   $$ABA = 101$$
4. Comparing digits, we have $A = 1$ and $B = 0$.
5. Thus:
   $$A + B = 1 + 0 = 1$$

ANSWER 5: A

---

### Problem 6:
**Restatement:** Out of 100 students at a camp, 52 are boys and 48 are girls. Also, 40 are from Jonas Middle School and 60 from Clay Middle School. Given that 20 of the girls are from Jonas, find how many boys are from Clay.

**Step-by-Step Solution:**
1. Find the number of girls from Clay Middle School:
   $$\text{Total girls} - \text{Girls from Jonas} = 48 - 20 = 28\text{ girls from Clay}$$
2. Find the number of boys from Clay Middle School:
   $$\text{Total students from Clay} - \text{Girls from Clay} = 60 - 28 = 32\text{ boys from Clay}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** Compute the value of:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-Step Solution:**
1. Pair each term in the first sum with the corresponding term in the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
2. Each paired difference is equal to:
   $$1900 + k - (100 + k) = 1800$$
3. Find the number of terms: from $k = 1$ to $k = 93$, there are:
   $$93 - 1 + 1 = 93\text{ terms}$$
4. Multiply the difference by the number of terms:
   $$93 \times 1800 = 93 \times 18 \times 100 = 1674 \times 100 = 167,400$$

ANSWER 7: A

---

### Problem 8:
**Restatement:** Find the tens digit of $7^{2011}$.

**Step-by-Step Solution:**
1. Find the pattern of the last two digits of powers of 7 by computing $7^n \pmod{100}$:
   - $7^1 = 07$
   - $7^2 = 49$
   - $7^3 = 343 \equiv 43 \pmod{100}$
   - $7^4 = 43

# Problem 1:
We must decide when $o^2 + no$ is odd or even, given $o$ is odd and $n$ is any whole number.

**Key facts:** odd × odd = odd, so $o^2$ is **odd**. Since $o$ is odd, multiplying by $o$ does not change $n$'s parity, so $no$ has the **same parity as $n$**.

**Case 1: $n$ even.** Then $no$ is even, so $o^2 + no = \text{odd} + \text{even} = \textbf{odd}$.
**Case 2: $n$ odd.** Then $no$ is odd, so $o^2 + no = \text{odd} + \text{odd} = \textbf{even}$.

So the number is odd exactly when $n$ is even. In particular, **if the number is odd, $n$ must be even**.

Ruling out traps: (A) fails since $o=1, n=1$ gives $1+1=2$, even; (B) fails since $n=2$ gives $1+2=3$, odd; (C) and (D) state the exact opposite of the truth. Only (E) matches.

ANSWER 1: E

# Problem 2:
The 6 redemption dates are 10 days apart, so their days of the week shift by $10 \bmod 7 = 3$ each time.

If the first date falls on weekday $W$, the six dates fall on (mod 7):
$$W,\; W+3,\; W+6,\; W+2,\; W+5,\; W+1$$
That is **six of the seven weekdays**, and the only weekday missed is $W+4$.

Since no circled date is a Sunday, the one missed weekday must be Sunday:
$$W + 4 = \text{Sunday} \implies W = \text{Wednesday}.$$

**Check:** Starting Wednesday, the dates fall on Wed, Sat, Tue, Fri, Mon, Thu — no Sunday. (Trap: starting Thursday gives Thu, Sun, ... — fails immediately.)

ANSWER 2: C

# Problem 3:
The product is a chain of fractions (the bars were lost in formatting):
$$\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdots\frac{a}{b} = 9.$$

Each numerator cancels the denominator before it (e.g., $\frac{3}{2}\cdot\frac{4}{3} = \frac{4}{2}$), and each fraction has the form $\frac{k+1}{k}$, so $b = a-1$. Everything telescopes:
$$\text{Product} = \frac{a}{2} = 9 \implies a = 18, \quad b = 17.$$
Then $a + b = 18 + 17 = 35$.

Trap check: $a+b=17$ would mean $a=9$, giving product $\tfrac92$, not $9$; only $35$ fits the pattern.

ANSWER 3: D

# Problem 4:
Work in units of 5¢: we need total value $710¢ = 142$ units, using stamps worth $1, 2, 5$ units (i.e., 5¢, 10¢, 25¢), at most 20 of each. To **maximize the count**, use as many cheap stamps as possible.

**Building a 55-stamp solution:** Take nineteen 1's and nineteen 2's: value $19 + 38 = 57$ units with 38 stamps. Remaining value $142 - 57 = 85 = 17\times 5$: seventeen 5's.
$$19\times 5¢ + 19\times 10¢ + 17\times 25¢ = 95 + 190 + 425 = 710¢ = \$7.10.\; ✓$$
That's $19+19+17 = 55$ stamps, all counts $\le 20$. ✓

**Why 56 is impossible:** with 56 stamps, at least $56 - 40 = 16$ must be 25¢ stamps (only 40 cheap ones exist). Let $z$ = number of 25¢ stamps:
- $z=16$: max value $= 100 + 200 + 400 = 700 < 710$. ✗
- $z=17$: need $5x+10y=285$, $x+y=39 \Rightarrow x=21>20$. ✗
- $z=18$: $x=24>20$. ✗  $z=19$: $x=27>20$. ✗  $z=20$: $x=30>20$. ✗

Traps: 54 comes from $20×5+16×10+18×25 = 710$; 51 from $20×5+11×10+20×25 = 710$ — both valid but not maximal.

ANSWER 4: E

# Problem 5:
Write $CDCD$ as a number:
$$CDCD = 1000C + 100D + 10C + D = 101(10C + D) = 101 \times CD.$$
The given equation is
$$ABA \times CD = CDCD = 101 \times CD.$$
Since $CD \neq 0$ (it's a two-digit number), we can divide both sides by $CD$:
$$ABA = 101.$$
So $A = 1$ and $B = 0$, giving $A + B = 1$.

**Consistency check:** $C, D$ just need to be distinct from $0, 1$ and each other, e.g., $101 \times 23 = 2323$. ✓ (Trap: guessing digits without using the factorization 101 wastes time or gives 9.)

ANSWER 5: A

# Problem 6:
Set up a two-way table with 100 total students.

- Girls: 48 total, and 20 girls are from Jonas $\Rightarrow$ girls from Clay $= 48 - 20 = 28$.
- Jonas has 40 students, of whom 20 are girls $\Rightarrow$ boys from Jonas $= 40 - 20 = 20$.
- Boys: 52 total $\Rightarrow$ boys from Clay $= 52 - 20 = \textbf{32}$.

**Check:** Jonas $20+20=40$ ✓; Clay $28+32=60$ ✓; boys $20+32=52$ ✓; girls $20+28=48$ ✓; total $100$ ✓. (Trap: 20 counts the boys from Jonas, not Clay.)

ANSWER 6: B

# Problem 7:
Both sums have the same number of terms: $1901$ to $1993$ has $1993-1901+1 = 93$ terms, and $101$ to $193$ has $193-101+1 = 93$ terms.

Pair them up: each term in the first sum is exactly $1800$ more than its partner ($1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$). So the difference is
$$93 \times 1800 = 167{,}400.$$

**Check via formulas:** $\frac{93(1901+1993)}{2} - \frac{93(101+193)}{2} = 181{,}071 - 13{,}671 = 167{,}400$. ✓ (Trap: 181,071 is the first sum alone — forgetting to subtract.)

ANSWER 7: A

# Problem 8:
Only the tens digit matters, so track the last two digits of $7^{k}$, which cycle with period 4:
$$7^1 = 07,\quad 7^2 = 49,\quad 7^3 = 343 \to 43,\quad 7^4 = 2401 \to 01,\quad 7^5 \to 07, \ldots$$

Find where $2011$ falls in the cycle:
$$2011 = 4 \times 502 + 3,$$
so $7^{2011}$ ends in the same two digits as $7^3$, namely $\ldots 43$.

The tens digit is $\textbf{4}$. (Trap: $7^{2012}$, a multiple of 4, ends in $01$, tens digit 0 — choice A; that would require the exponent to be divisible by 4.)

ANSWER 8: D

# Problem 9:
The ten **different** positive whole numbers have average 10, so their sum is $10 \times 10 = 100$.

To make one number as large as possible, make the other nine as small as possible. Since they must be **distinct positive** whole numbers, the smallest possible set is $1, 2, 3, \ldots, 9$, with sum
$$1+2+\cdots+9 = 45.$$
So the largest possible value is
$$100 - 45 = 55.$$

**Check:** $1,2,\ldots,9,55$ are ten different positive whole numbers with sum $100$ and average $10$. ✓ (Trap: 91 arises if you ignore "different" and use nine 1's; but the numbers must be distinct.)

ANSWER 9: C

# Problem 10:
Count student-days (the unit being paid):
- Allen: $7 \times 3 = 21$ student-days
- Balboa: $4 \times 5 = 20$ student-days
- Carver: $5 \times 9 = 45$ student-days

Total: $21 + 20 + 45 = 86$ student-days, paid $\$774$ altogether. So the daily rate is
$$774 \div 86 = 9 \text{ dollars per student-day}.$$
Balboa's students worked 20 student-days, so they earned
$$20 \times 9 = 18

Problem 1:  
We need the parity of \(o^2+no\), where \(o\) is odd and \(n\) is any whole number.

Factor:
\[
o^2+no=o(o+n).
\]

Since \(o\) is odd:

- If \(n\) is even, then \(o+n\) is odd, so \(o(o+n)\) is odd.
- If \(n\) is odd, then \(o+n\) is even, so \(o(o+n)\) is even.

Thus the expression is odd exactly when \(n\) is even.

Check the choices:
- A and B are false because the parity depends on \(n\).
- C says “even only if \(n\) is even,” but when the expression is even, \(n\) is actually odd.
- D says “odd only if \(n\) is odd,” but when the expression is odd, \(n\) is even.
- E says “odd only if \(n\) is even,” which is true.

ANSWER 1: E

Problem 2:  
Isabella redeems coupons 10 days apart. Since
\[
10\equiv 3 \pmod 7,
\]
the six circled days are the first day plus
\[
0,3,6,9,12,15
\]
days, which modulo 7 gives offsets
\[
0,3,6,2,5,1.
\]

So the six circled days are all weekdays except the one 4 days after the first day. Since none of the circled days is a Sunday, Sunday must be the un-circled day: 4 days after the first day.

Therefore the first coupon is 4 days before Sunday, which is Wednesday.

Check: Wednesday, Saturday, Tuesday, Friday, Monday, Thursday — no Sunday.

ANSWER 2: C

Problem 3:  
The product is the telescoping product of fractions:
\[
\frac32\cdot \frac43\cdot \frac54\cdot \frac65\cdots \frac{a}{b}=9.
\]

The numerators cancel the next denominators:
\[
\frac32\cdot \frac43=\frac42,
\]
\[
\frac42\cdot \frac54=\frac52,
\]
and continuing, the whole product simplifies to
\[
\frac{a}{2}.
\]

So
\[
\frac{a}{2}=9 \implies a=18.
\]

By the pattern, the denominator is one less than the numerator, so
\[
b=17.
\]

Thus
\[
a+b=18+17=35.
\]

ANSWER 3: D

Problem 4:  
Nicolas has 20 of each type of stamp, so he has
\[
20+20+20=60
\]
stamps total. Their total value is
\[
20(5+10+25)=20(40)=800
\]
cents.

He wants postage of
\[
\$7.10=710
\]
cents, so the stamps he does **not** use must total
\[
800-710=90
\]
cents.

To maximize the number of stamps he uses, we minimize the number of stamps he leaves unused.

Can 4 unused stamps total 90 cents? Let \(x,y,z\) be the numbers of 25-, 10-, and 5-cent unused stamps. Then
\[
x+y+z=4
\]
and
\[
25x+10y+5z=90.
\]

Divide by 5:
\[
5x+2y+z=18.
\]

Subtract \(x+y+z=4\):
\[
4x+y=14.
\]

Testing possible \(x=0,1,2,3,4\):

- \(x=0\Rightarrow y=14\), impossible.
- \(x=1\Rightarrow y=10\), impossible.
- \(x=2\Rightarrow y=6\), impossible.
- \(x=3\Rightarrow y=2\), then \(z=4-3-2=-1\), impossible.
- \(x=4\Rightarrow y=-2\), impossible.

So 4 unused stamps cannot total 90 cents.

Can 5 unused stamps total 90 cents? Yes:
\[
25+25+25+10+5=90.
\]

Then he uses
\[
60-5=55
\]
stamps. For example, he could use 19 five-cent stamps, 19 ten-cent stamps, and 17 twenty-five-cent stamps:
\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So the greatest possible number of stamps is 55. Choice D, 54, is possible but not greatest.

ANSWER 4: E

Problem 5:  
We are told
\[
ABA\times CD=CDCD,
\]
where \(A,B,C,D\) are different digits.

Write the numbers in place-value form:
\[
ABA=101A+10B,
\]
\[
CD=10C+D,
\]
\[
CDCD=1000C+100D+10C+D=101(10C+D).
\]

So
\[
(101A+10B)(10C+D)=101(10C+D).
\]

Since \(CD\) is a two-digit number, \(10C+D\neq 0\), so we can divide both sides by \(10C+D\):
\[
101A+10B=101.
\]

The only digit solution is
\[
A=1,\quad B=0.
\]

Therefore
\[
A+B=1+0=1.
\]

ANSWER 5: A

Problem 6:  
There are 48 girls total, and 20 of the girls are from Jonas Middle School. So the number of girls from Clay Middle School is
\[
48-20=28.
\]

There are 60 students total from Clay Middle School, so the number of boys from Clay is
\[
60-28=32.
\]

ANSWER 6: B

Problem 7:  
Pair the terms:
\[
(1901+1902+\cdots+1993)-(101+102+\cdots+193)
\]
\[
=(1901-101)+(1902-102)+\cdots+(1993-193).
\]

Each pair is
\[
1800.
\]

Now count the number of terms:
\[
193-101+1=93.
\]

So the value is
\[
93\cdot 1800=167400.
\]

ANSWER 7: A

Problem 8:  
We need the tens digit of \(7^{2011}\), so we need the last two digits.

Powers of 7 modulo 100 cycle:
\[
7^1=07,
\]
\[
7^2=49,
\]
\[
7^3=343\equiv 43,
\]
\[
7^4=2401\equiv 01.
\]

Then the cycle repeats:
\[
07,\ 49,\ 43,\ 01.
\]

Since
\[
2011\equiv 3 \pmod 4,
\]
we have
\[
7^{2011}\equiv 7^3\equiv 43 \pmod{100}.
\]

The last two digits are 43, so the tens digit is 4.

ANSWER 8: D

Problem 9:  
The average of 10 different positive whole numbers is 10, so their sum is
\[
10\cdot 10=100.
\]

To make one number as large as possible, make the other 9 numbers as small as possible while still being different positive whole numbers. The smallest 9 distinct positive whole numbers are
\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is
\[
1+2+\cdots+9=45.
\]

So the largest number can be
\[
100-45=55.
\]

This is possible: \(1,2,3,4,5,6,7,8,9,55\) are all different positive whole numbers.

ANSWER 9: C

Problem 10:  
First count total student-days worked:

- Allen: \(7\cdot 3=21\) student-days
- Balboa: \(4\cdot 5=20\) student-days
- Carver: \(5\cdot 9=45\) student-days

Total:
\[
21+20+45=86
\]
student-days.

The total payment is \$774, so the daily rate is
\[
774\div 86=9
\]
dollars per student-day.

Balboa students worked 20 student-days, so they earned
\[
20\cdot 9=180
\]
dollars.

ANSWER 10: C

# Problem 1:
We need a statement always true about o² + no, where o is odd.

Factor: o² + no = o(o + n).

Since o is odd, the parity of the product is the parity of (o + n).
- If n is even, o + n is odd → product = odd·odd = **odd**.
- If n is odd, o + n is even → product = odd·even = **even**.

So the expression is odd exactly when n is even.

Ruling out others: A fails (n odd gives even), B fails (n even gives odd), C says "even only if n is even" — but it's even when n is *odd*, false. D says "odd only if n is odd" — false, it's odd when n is even. E: "odd only if n is even" — true.

ANSWER 1: E

# Problem 2:
The redemption days are 0, 10, 20, 30, 40, 50 days after the first.

Reduce mod 7 (days of week shift): 0, 3, 6, 2, 5, 1.

These residues are {0, 1, 2, 3, 5, 6} — every residue except **4**. So the only day of the week never hit is the day 4 days after the starting day. For no coupon to land on Sunday, Sunday must be that missed day:

start + 4 ≡ Sunday.

Counting back 4 days from Sunday: Saturday, Friday, Thursday, **Wednesday**.

Check: Wed → +10 Sat → +20 Tue → +30 Fri → +40 Mon → +50 Thu. No Sundays. ✓

ANSWER 2: C

# Problem 3:
The product is (3/2)·(4/3)·(5/4)·(6/5)·…·(a/b), where each numerator is one more than its denominator, so b = a − 1.

Telescoping: everything cancels except the last numerator and first denominator:
a/2 = 9 → a = 18, so b = 17.

a + b = 18 + 17 = 35.

(Choice E, 37, would come from mistakenly taking a = 19.)

ANSWER 3: D

# Problem 4:
Total value of all 60 stamps: 20(5 + 10 + 25) = 20·40 = 800 cents.

We need 710 cents, so we must *remove* stamps worth exactly 800 − 710 = 90 cents, using as **few** stamps as possible (to keep the most stamps).

Can 90 be made with 4 or fewer stamps? Max with 4 stamps is 100; possible 4-stamp totals near 90: 100, 85, 80, 75, 70… none equal 90. Three or fewer is at most 75. So 4 or fewer is impossible.

With 5 stamps: 25 + 25 + 25 + 10 + 5 = 90 ✓

So remove 5 stamps: 60 − 5 = 55 stamps.

Check: 17(25) + 19(10) + 19(5) = 425 + 190 + 95 = 710 cents, using 17 + 19 + 19 = 55 stamps.

ANSWER 4: E

# Problem 5:
CDCD as a number = CD·100 + CD = CD·101.

So ABA × CD = 101 × CD, and since CD ≠ 0, we get ABA = 101.

Thus A = 1, B = 0 (and indeed A, B, C, D distinct works, e.g. 101 × 23 = 2323).

A + B = 1 + 0 = 1.

ANSWER 5: A

# Problem 6:
Jonas total = 40, and 20 of those were girls, so boys from Jonas = 40 − 20 = 20.

Total boys = 52, so boys from Clay = 52 − 20 = 32.

(Check: girls from Clay = 48 − 20 = 28; 32 + 28 = 60 = Clay total ✓.)

Choice C (40) is the Jonas total, and E (52) is all boys — traps.

ANSWER 6: B

# Problem 7:
Pair each term: 1901 − 101 = 1800, 1902 − 102 = 1800, …, each pair differs by 1800.

Number of terms: 1993 − 1901 + 1 = 93.

Total = 93 × 1800 = 167,400.

ANSWER 7: A

# Problem 8:
Look at 7ⁿ mod 100:
- 7¹ = 07
- 7² = 49
- 7³ = 343 → 43
- 7⁴ = 2401 → 01
- then the pattern repeats with period 4.

2011 ÷ 4 = 502 remainder 3, so 7²⁰¹¹ ≡ 7³ ≡ 43 (mod 100).

Last two digits are 43, so the tens digit is 4.

ANSWER 8: D

# Problem 9:
Sum of the 10 numbers = 10 × 10 = 100.

To make one number as large as possible, make the other nine as small as possible while distinct and positive: 1 + 2 + … + 9 = 45.

Largest = 100 − 45 = 55.

(90 or 91 would require repeats or a zero, which isn't allowed.)

ANSWER 9: C

# Problem 10:
Total student-days:
- Allen: 7 × 3 = 21
- Balboa: 4 × 5 = 20
- Carver: 5 × 9 = 45

Total = 86 student-days.

Daily wage = 774 ÷ 86 = $9.00 per student-day.

Balboa earnings = 20 × 9 = $180.00.

(A is one day's pay for one student; E is Carver's total.)

ANSWER 10: C

Problem 1:

We need determine the parity of

\[
o^2+no.
\]

Factor out \(o\):

\[
o^2+no=o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on \(o+n\).

- If \(n\) is even, then odd \(+\) even \(=\) odd, so \(o+n\) is odd. Thus odd \(\cdot\) odd is odd.
- If \(n\) is odd, then odd \(+\) odd \(=\) even, so the product is even.

Therefore, the expression is odd only when \(n\) is even.

- A is wrong because the expression is not always odd.
- B is wrong because it is odd when \(n\) is even.
- C reverses the condition for being even.
- D reverses the condition for being odd.

ANSWER 1: E

Problem 2:

Isabella uses a coupon every 10 days. Since

\[
10\equiv 3\pmod 7,
\]

each coupon date moves forward by 3 days of the week.

If the first coupon is redeemed on day \(d\), the six redemption days are:

\[
d,\ d+3,\ d+6,\ d+2,\ d+5,\ d+1.
\]

These are six different days of the week. The only day missing is

\[
d+4.
\]

Since none of the six coupon dates is Sunday, the missing day must be Sunday. Thus,

\[
d+4=\text{Sunday}.
\]

Four days before Sunday is Wednesday.

Indeed, beginning on Wednesday gives:

\[
\text{Wednesday, Saturday, Tuesday, Friday, Monday, Thursday}.
\]

Sunday is the only missing day.

ANSWER 2: C

Problem 3:

The product is

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9.
\]

Almost every numerator cancels with a denominator:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab
=\frac a2.
\]

So

\[
\frac a2=9.
\]

Therefore,

\[
a=18.
\]

The final fraction must be

\[
\frac{18}{17},
\]

so \(b=17\). Therefore,

\[
a+b=18+17=35.
\]

Choices 11, 13, and 17 are too small, while 37 would correspond to a different ending fraction.

ANSWER 3: D

Problem 4:

Let

- \(x\) = number of 5-cent stamps,
- \(y\) = number of 10-cent stamps,
- \(z\) = number of 25-cent stamps.

In units of 5 cents, the postage total is

\[
\frac{710}{5}=142.
\]

Thus,

\[
x+2y+5z=142.
\]

We want to maximize

\[
x+y+z.
\]

Since there are at most 20 five-cent stamps, \(x\leq 20\). Therefore,

\[
2y+5z=142-x\geq 142-20=122.
\]

Since \(y\leq20\), if \(z\leq16\), then

\[
2y+5z\leq 2(20)+5(16)=40+80=120,
\]

which is not enough. So \(z\geq17\).

If \(z=17\), then

\[
2y+5(17)\geq122,
\]

so

\[
2y+85\geq122,
\]

\[
2y\geq37,
\]

and hence \(y\geq19\).

Now use

\[
x+y+z=142-y-4z.
\]

Since \(y\geq19\) and \(z\geq17\),

\[
x+y+z\leq142-19-4(17)
=142-19-68
=55.
\]

This maximum is attainable with

\[
x=19,\qquad y=19,\qquad z=17.
\]

Check:

\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So 55 stamps can be used. Choices 45, 46, 51, and 54 are possible totals to consider, but 55 is the greatest possible.

ANSWER 4: E

Problem 5:

We are told

\[
ABA\times CD=CDCD.
\]

The number \(ABA\) is

\[
100A+10B+A=101A+10B.
\]

The number \(CDCD\) is

\[
1000C+100D+10C+D.
\]

Group the \(C\)'s and \(D\)'s:

\[
CDCD=1010C+101D=101(10C+D)=101\cdot CD.
\]

Thus the equation becomes

\[
ABA\cdot CD=101\cdot CD.
\]

Since \(CD\) is a two-digit number and is not zero, divide both sides by \(CD\):

\[
ABA=101.
\]

Therefore,

\[
A=1,\qquad B=0.
\]

So

\[
A+B=1+0=1.
\]

For example, \(101\times23=2323\), and the digits \(1,0,2,3\) are all different.

ANSWER 5: A

Problem 6:

There are 40 students from Jonas Middle School, and 20 of those are girls.

Therefore the number of boys from Jonas is

\[
40-20=20.
\]

There are 52 boys altogether. Therefore the number of boys from Clay Middle School is

\[
52-20=32.
\]

The other choices do not correctly account for the 20 Jonas girls and the total of 52 boys.

ANSWER 6: B

Problem 7:

There are

\[
1993-1901+1=93
\]

terms in the first sum.

There are also

\[
193-101+1=93
\]

terms in the second sum.

Pair corresponding terms:

\[
1901-101=1800,
\]

\[
1902-102=1800,
\]

and so on, through

\[
1993-193=1800.
\]

Thus there are 93 differences of 1800:

\[
93\cdot1800=167400.
\]

Therefore the value is

\[
167,\!400.
\]

ANSWER 7: A

Problem 8:

We need the tens digit of

\[
7^{2011}.
\]

It is enough to examine powers of 7 modulo 100:

\[
7^1=7,
\]

\[
7^2=49,
\]

\[
7^3=343,
\]

whose last two digits are 43, and

\[
7^4=2401,
\]

whose last two digits are 01.

Thus the last two digits repeat every 4 powers because

\[
7^4\equiv1\pmod{100}.
\]

Now,

\[
2011\div4
\]

leaves remainder 3, since

\[
2011=4(502)+3.
\]

Therefore,

\[
7^{2011}
\]

has the same last two digits as

\[
7^3=343,
\]

which are 43. The tens digit is 4.

ANSWER 8: D

Problem 9:

The average of 10 different positive whole numbers is 10, so their total sum is

\[
10\cdot10=100.
\]

To make one number as large as possible, make the other nine numbers as small as possible.

The nine smallest different positive whole numbers are

\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is

\[
1+2+3+4+5+6+7+8+9=45.
\]

So the largest possible tenth number is

\[
100-45=55.
\]

This works because the ten numbers can be

\[
1,2,3,4,5,6,7,8,9,55,
\]

which are all different and have total 100.

A value such as 90 or 91 would leave too small a total for nine distinct positive whole numbers.

ANSWER 9: C

Problem 10:

Compute the total number of student-days worked.

Allen students worked:

\[
7\cdot3=21
\]

student-days.

Balboa students worked:

\[
4\cdot5=20
\]

student-days.

Carver students worked:

\[
5\cdot9=45
\]

student-days.

So the total number of student-days is

\[
21+20+45=86.
\]

The total pay was \$774, so the pay per student per day was

\[
\frac{774}{86}=9.
\]

Balboa students worked 20 student-days, so they earned

\[
20\cdot9=180.
\]

Thus the Balboa students earned \$180.00 altogether.

ANSWER 10: C

Problem 1:
We are asked about the parity of $o^2+no$ where $o$ is odd and $n$ is any whole number.

Factor:
$$o^2+no = o(o+n).$$

$o$ is odd, so $o^2 = o\cdot o$ is odd times odd = odd.

Hence parity is odd + $n\cdot o$.
$n\cdot o$ has the same parity as $n$, because $o$ is odd: if $n$ is even, $n o$ is even; if $n$ is odd, $n o$ is odd.

So:
- if $n$ even: odd + even = odd.
- if $n$ odd: odd + odd = even.

Thus the number is odd iff $n$ is even, and even iff $n$ is odd.

Check choices:
A. always odd - false, e.g. $o=1,n=1$ gives $1+1=2$ even.
B. always even - false, e.g. $o=1,n=2$ gives $1+2=3$ odd.
C. it is even only if $n$ is even - means even $\implies$ $n$ even. False, even happens when $n$ is odd.
D. it is odd only if $n$ is odd - means odd $\implies$ $n$ odd. False, odd happens when $n$ is even.
E. it is odd only if $n$ is even - means odd $\implies$ $n$ even. True by above.

ANSWER 1: E

Problem 2:
Isabella redeems on days $d, d+10, d+20, d+30, d+40, d+50$.
No date is a Sunday.

Modulo 7, $10\equiv 3$. So successive dates advance by 3 weekdays.

Starting from residue $r$, the residues are:
$$r,\ r+3,\ r+6,\ r+9=r+2,\ r+12=r+5,\ r+15=r+1 \pmod 7.$$
These 6 are all distinct, because adding 3 mod 7 cycles through all 7 residues. Continuing one more step would give $r+18=r+4$.

Thus the 6 dates occupy 6 distinct days of the week, missing exactly the 7th residue $r+4$.

Since none of the 6 is Sunday, the missing one must be Sunday. So:
$$r+4 \equiv \text{Sunday}.$$

Hence $r$ is 4 days before Sunday, i.e. Wednesday.

Check: Wed $\to$ Wed+3=Sat $\to$ +3=Tue $\to$ +3=Fri $\to$ +3=Mon $\to$ +3=Thu. That is Wed,Sat,Tue,Fri,Mon,Thu - no Sunday. Any other start would include Sunday.

ANSWER 2: C

Problem 3:
We interpret $3\ 2\cdot 4\ 3\cdots$ as fractions:
$$\frac32 \cdot \frac43 \cdot \frac54 \cdot \frac65 \cdots \frac{a}{b}=9.$$

The pattern is numerator $3,4,5,6,\dots$ and denominator $2,3,4,5,\dots$, i.e. term $k/(k-1)$. The last term is $a/b$, so we must have $b=a-1$ for the pattern to continue.

Then the product telescopes:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}
= \frac{a}{2},$$
because all intermediate $3,4,\dots,a-1$ cancel.

Set equal to 9:
$$\frac{a}{2}=9 \implies a=18.$$
Then $b=a-1=17$.

Sum $a+b=18+17=35$.

This rules out A,B,C,E.

ANSWER 3: D

Problem 4:
Let $x,y,z$ be numbers of 5-cent, 10-cent, 25-cent stamps used. $0\le x,y,z\le 20$ integers.
Value $710$ cents:
$$5x+10y+25z=710.$$
Divide by 5:
$$x+2y+5z=142. \tag{1}$$
Maximize $N=x+y+z$.

From (1), $x=142-2y-5z$, so $0\le x\le20$ gives:
$$122\le 2y+5z\le 142.$$
Also $N=x+y+z=142-y-4z$, so $y+4z=142-N$.

To use many stamps we want small denominations. Try $x=20,y=20$ fails to give integer $z$. Search systematically.

Since $y\le20$, to satisfy $2y+5z\ge122$ we need $z$ large. If $z\le16$, then $2y+5z\le40+80=120<122$, impossible. So $z\ge17$.

Try $z=17,y=19$: then $2y+5z=38+85=123$, so $x=142-123=19$, within $0-20$. This gives:
$$19\cdot5+19\cdot10+17\cdot25=95+190+425=710,$$
with $N=19+19+17=55$ stamps. So 55 is achievable.

Can we do 56 or more? Let $N\ge56$.
Then $y+4z=142-N\le86$ and $x=N-y-z\le20$ gives $y+z\ge N-20$.
Subtract: $3z=(y+4z)-(y+z)\le (142-N)-(N-20)=162-2N$.
So $z\le(162-2N)/3$.
Also $y=142-N-4z\le20$ gives $z\ge(122-N)/4$.

For $N=56$: $z\ge\lceil66/4\rceil=17$ and $z\le\lfloor50/3\rfloor=16$, contradiction. For $N>56$ the interval is even emptier; in general need $(122-N)/4\le(162-2N)/3$, i.e. $5N\le282$, so $N\le56.4$. Thus $N\ge57$ impossible, and $N=56$ impossible by the integer contradiction.

Hence 55 is greatest possible.

ANSWER 4: E

Problem 5:
We have $ABA\times CD = CDCD$, where $A,B,C,D$ are different digits, $A\ne0$, $C\ne0$.

Write $CD$ as the 2-digit number $10C+D$.
$CDCD$ as a 4-digit number is:
$$1000C+100D+10C+D =1010C+101D=101(10C+D)=101\cdot CD.$$

So:
$$ABA \times CD = 101\times CD.$$
Since $CD\ne0$, divide:
$$ABA =101.$$

Thus $A=1$, $B=0$, and $A+B=1$.

This is consistent with distinctness: we can choose e.g. $C=2,D=3$, all of $1,0,2,3$ different, and $101\times23=2323$. So valid.

Other sums are ruled out.

ANSWER 5: A

Problem 6:
Total 100: 52 boys, 48 girls. 40 from Jonas, 60 from Clay.

20 of the girls are from Jonas. So girls from Clay = total girls $-$ girls from Jonas:
$$48-20=28.$$

Clay total 60 = boys from Clay + girls from Clay, so:
$$\text{boys from Clay}=60-28=32.$$

Check: boys from Jonas $=52-32=20$, Jonas total $=20+20=40$. Works.

ANSWER 6: B

Problem 7:
Compute $(1901+\cdots+1993)-(101+\cdots+193)$.

Both sums have $1993-1901+1=93$ terms and $193-101+1=93$ terms.

Pair termwise: for $k=0,\dots,92$,
$$(1901+k)-(101+k)=1800.$$

So difference $=93\times1800$.
$93\times18=1674$, so $93\times1800=167{,}400$.

Other choices come from miscounting terms or arithmetic error.

ANSWER 7: A

Problem 8:
Find tens digit of $7^{2011}$, i.e. $7^{2011}\pmod{100}$.

Compute powers mod 100:
$7^1=7$,
$7^2=49$,
$7^3=343\equiv43$,
$7^4\equiv43\times7=301\equiv1$.

Since $7^4\equiv1\pmod{100}$, the last two digits repeat with period 4.

$2011 = 4\cdot502+3$, so:
$$7^{2011}=(7^4)^{502}\cdot7^3\equiv1^{502}\cdot43\equiv43\pmod{100}.$$

Thus the number ends $\dots43$, tens digit $4$.

ANSWER 8: D

Problem 9:
10 different positive whole numbers have mean 10, so sum $=10\times10=100$.

To make one number as large as possible, make the other nine as small as possible while distinct and positive.

Smallest 9 distinct positives are $1,2,3,4,5,6,7,8,9$, sum $=45$.

Then the tenth number $=100-45=55$, which is distinct from the others and valid. Any larger value would force the other nine to sum to $<45$, impossible with distinct positives.

So largest possible is 55.

ANSWER 9: C

Problem 10:
Count student-days (one student working one day).

Allen: $7$ students $\times$ $3$ days $=21$.
Balboa: $4\times5=20$.
Carver: $5\times9=45$.
Total $=21+20+45=86$ student-days.

Total pay $\$774$ for $86$ student-days at equal rate:
$$\text{rate}=774/86=9 \text{ dollars per student-day}.$$

Bal

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that $o$ is an odd whole number and $n$ is any whole number. We must determine which statement about $o^2 + no$ is always true.

**Step-by-Step Solution:**
1. Factor the expression:
   $$o^2 + no = o(o + n)$$
2. Since $o$ is an odd whole number, the parity of the product $o(o + n)$ is entirely determined by the parity of $(o + n)$:
   - If $n$ is **odd**, then $\text{odd} + \text{odd} = \text{even}$, so $o(o+n)$ is $\text{odd} \times \text{even} = \textbf{even}$.
   - If $n$ is **even**, then $\text{odd} + \text{even} = \text{odd}$, so $o(o+n)$ is $\text{odd} \times \text{odd} = \textbf{odd}$.
3. Therefore, the expression $o^2 + no$ is odd if and only if $n$ is even. 
4. This matches the statement: "it is odd only if $n$ is even."

ANSWER 1: E

---

### Problem 2:
**Restatement:** Isabella redeems 6 coupons, one every 10 days. Pete's is closed on Sundays, and none of the 6 dates falls on a Sunday. We need to find the day of the week she redeems her first coupon.

**Step-by-Step Solution:**
1. There are 7 days in a week. Moving forward by 10 days advances the day of the week by:
   $$10 \equiv 3 \pmod 7$$
2. Let $d$ represent the day of the week of the first coupon ($0 = \text{Sunday}, 1 = \text{Monday}, \dots, 6 = \text{Saturday}$).
3. The days of the week for the 6 coupons will be:
   - Coupon 1: $d$
   - Coupon 2: $d + 3$
   - Coupon 3: $d + 6$
   - Coupon 4: $d + 9 \equiv d + 2 \pmod 7$
   - Coupon 5: $d + 12 \equiv d + 5 \pmod 7$
   - Coupon 6: $d + 15 \equiv d + 1 \pmod 7$
4. The six days of the week visited are $d, d+1, d+2, d+3, d+5, d+6$. Notice that the only day of the week omitted from this list is $d+4$.
5. Since no circled date falls on a Sunday, Sunday must be the omitted day:
   $$d + 4 \equiv 0 \pmod 7 \implies d \equiv 3 \pmod 7$$
6. Day 3 corresponds to **Wednesday**.
   - Checking: Wed $\to$ Sat $\to$ Tue $\to$ Fri $\to$ Mon $\to$ Thu. None of these is Sunday.

ANSWER 2: C

---

### Problem 3:
**Restatement:** We are given the telescoping product $\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdots \frac{a}{b} = 9$. We must find the sum $a + b$.

**Step-by-Step Solution:**
1. Each fraction is of the form $\frac{k+1}{k}$, where the numerator of each term cancels with the denominator of the subsequent term:
   $$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{a}{b} = \frac{a}{2}$$
   where $b = a - 1$.
2. We are given that this product equals 9:
   $$\frac{a}{2} = 9 \implies a = 18$$
3. Since $b = a - 1$, we have $b = 17$.
4. The sum of $a$ and $b$ is:
   $$a + b = 18 + 17 = 35$$

ANSWER 3: D

---

### Problem 4:
**Restatement:** Nicolas has 20 stamps each of 5¢, 10¢, and 25¢. We want to find the greatest number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-Step Solution:**
1. Calculate the total value and count of all stamps:
   - $20 \times 5¢ = 100¢$
   - $20 \times 10¢ = 200¢$
   - $20 \times 25¢ = 500¢$
   - Total value $= 100 + 200 + 500 = 800¢$ using 60 stamps.
2. To use the **maximum** number of stamps to make 710¢, Nicolas must **minimize** the number of stamps removed to subtract:
   $$800¢ - 710¢ = 90¢$$
3. To remove the fewest stamps totaling 90¢, we try to use the largest denomination stamps (25¢):
   - With 3 stamps of 25¢ ($75¢$), the remaining amount needed is $15¢$.
   - Since Nicolas has only 5¢, 10¢, and 25¢ stamps, 15¢ cannot be made with a single stamp. It requires at least two stamps ($10¢ + 5¢$).
   - Thus, 90¢ can be formed with $3 \times 25¢ + 1 \times 10¢ + 1 \times 5¢$, which uses $3 + 1 + 1 = 5$ stamps.
   - Can 90¢ be formed using 4 stamps? The maximum value of 4 stamps less than or equal to 90¢ would be $3 \times 25¢ + 10¢ = 85¢ \ne 90¢$. Thus, 4 stamps is impossible.
4. The minimum number of stamps to remove is 5.
5. Therefore, the maximum number of stamps used is:
   $$60 - 5 = 55$$

ANSWER 4: E

---

### Problem 5:
**Restatement:** In the multiplication $ABA \times CD = CDCD$, $A, B, C, D$ are distinct digits. We need to find $A + B$.

**Step-by-Step Solution:**
1. Express $CDCD$ in terms of the two-digit number $CD$:
   $$CDCD = CD \times 100 + CD = CD \times 101$$
2. Substitute this into the equation:
   $$ABA \times CD = 101 \times CD$$
3. Since $CD$ is a non-zero two-digit number, divide both sides by $CD$:
   $$ABA = 101$$
4. Comparing digits, we have $A = 1$ and $B = 0$.
5. Thus:
   $$A + B = 1 + 0 = 1$$

ANSWER 5: A

---

### Problem 6:
**Restatement:** Out of 100 students at a camp, 52 are boys and 48 are girls. Also, 40 are from Jonas Middle School and 60 from Clay Middle School. Given that 20 of the girls are from Jonas, find how many boys are from Clay.

**Step-by-Step Solution:**
1. Find the number of girls from Clay Middle School:
   $$\text{Total girls} - \text{Girls from Jonas} = 48 - 20 = 28\text{ girls from Clay}$$
2. Find the number of boys from Clay Middle School:
   $$\text{Total students from Clay} - \text{Girls from Clay} = 60 - 28 = 32\text{ boys from Clay}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** Compute the value of:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-Step Solution:**
1. Pair each term in the first sum with the corresponding term in the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
2. Each paired difference is equal to:
   $$1900 + k - (100 + k) = 1800$$
3. Find the number of terms: from $k = 1$ to $k = 93$, there are:
   $$93 - 1 + 1 = 93\text{ terms}$$
4. Multiply the difference by the number of terms:
   $$93 \times 1800 = 93 \times 18 \times 100 = 1674 \times 100 = 167,400$$

ANSWER 7: A

---

### Problem 8:
**Restatement:** Find the tens digit of $7^{2011}$.

**Step-by-Step Solution:**
1. Find the pattern of the last two digits of powers of 7 by computing $7^n \pmod{100}$:
   - $7^1 = 07$
   - $7^2 = 49$
   - $7^3 = 343 \equiv 43 \pmod{100}$
   - $7^4 = 43

# Problem 1:
We must decide when $o^2 + no$ is odd or even, given $o$ is odd and $n$ is any whole number.

**Key facts:** odd × odd = odd, so $o^2$ is **odd**. Since $o$ is odd, multiplying by $o$ does not change $n$'s parity, so $no$ has the **same parity as $n$**.

**Case 1: $n$ even.** Then $no$ is even, so $o^2 + no = \text{odd} + \text{even} = \textbf{odd}$.
**Case 2: $n$ odd.** Then $no$ is odd, so $o^2 + no = \text{odd} + \text{odd} = \textbf{even}$.

So the number is odd exactly when $n$ is even. In particular, **if the number is odd, $n$ must be even**.

Ruling out traps: (A) fails since $o=1, n=1$ gives $1+1=2$, even; (B) fails since $n=2$ gives $1+2=3$, odd; (C) and (D) state the exact opposite of the truth. Only (E) matches.

ANSWER 1: E

# Problem 2:
The 6 redemption dates are 10 days apart, so their days of the week shift by $10 \bmod 7 = 3$ each time.

If the first date falls on weekday $W$, the six dates fall on (mod 7):
$$W,\; W+3,\; W+6,\; W+2,\; W+5,\; W+1$$
That is **six of the seven weekdays**, and the only weekday missed is $W+4$.

Since no circled date is a Sunday, the one missed weekday must be Sunday:
$$W + 4 = \text{Sunday} \implies W = \text{Wednesday}.$$

**Check:** Starting Wednesday, the dates fall on Wed, Sat, Tue, Fri, Mon, Thu — no Sunday. (Trap: starting Thursday gives Thu, Sun, ... — fails immediately.)

ANSWER 2: C

# Problem 3:
The product is a chain of fractions (the bars were lost in formatting):
$$\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdots\frac{a}{b} = 9.$$

Each numerator cancels the denominator before it (e.g., $\frac{3}{2}\cdot\frac{4}{3} = \frac{4}{2}$), and each fraction has the form $\frac{k+1}{k}$, so $b = a-1$. Everything telescopes:
$$\text{Product} = \frac{a}{2} = 9 \implies a = 18, \quad b = 17.$$
Then $a + b = 18 + 17 = 35$.

Trap check: $a+b=17$ would mean $a=9$, giving product $\tfrac92$, not $9$; only $35$ fits the pattern.

ANSWER 3: D

# Problem 4:
Work in units of 5¢: we need total value $710¢ = 142$ units, using stamps worth $1, 2, 5$ units (i.e., 5¢, 10¢, 25¢), at most 20 of each. To **maximize the count**, use as many cheap stamps as possible.

**Building a 55-stamp solution:** Take nineteen 1's and nineteen 2's: value $19 + 38 = 57$ units with 38 stamps. Remaining value $142 - 57 = 85 = 17\times 5$: seventeen 5's.
$$19\times 5¢ + 19\times 10¢ + 17\times 25¢ = 95 + 190 + 425 = 710¢ = \$7.10.\; ✓$$
That's $19+19+17 = 55$ stamps, all counts $\le 20$. ✓

**Why 56 is impossible:** with 56 stamps, at least $56 - 40 = 16$ must be 25¢ stamps (only 40 cheap ones exist). Let $z$ = number of 25¢ stamps:
- $z=16$: max value $= 100 + 200 + 400 = 700 < 710$. ✗
- $z=17$: need $5x+10y=285$, $x+y=39 \Rightarrow x=21>20$. ✗
- $z=18$: $x=24>20$. ✗  $z=19$: $x=27>20$. ✗  $z=20$: $x=30>20$. ✗

Traps: 54 comes from $20×5+16×10+18×25 = 710$; 51 from $20×5+11×10+20×25 = 710$ — both valid but not maximal.

ANSWER 4: E

# Problem 5:
Write $CDCD$ as a number:
$$CDCD = 1000C + 100D + 10C + D = 101(10C + D) = 101 \times CD.$$
The given equation is
$$ABA \times CD = CDCD = 101 \times CD.$$
Since $CD \neq 0$ (it's a two-digit number), we can divide both sides by $CD$:
$$ABA = 101.$$
So $A = 1$ and $B = 0$, giving $A + B = 1$.

**Consistency check:** $C, D$ just need to be distinct from $0, 1$ and each other, e.g., $101 \times 23 = 2323$. ✓ (Trap: guessing digits without using the factorization 101 wastes time or gives 9.)

ANSWER 5: A

# Problem 6:
Set up a two-way table with 100 total students.

- Girls: 48 total, and 20 girls are from Jonas $\Rightarrow$ girls from Clay $= 48 - 20 = 28$.
- Jonas has 40 students, of whom 20 are girls $\Rightarrow$ boys from Jonas $= 40 - 20 = 20$.
- Boys: 52 total $\Rightarrow$ boys from Clay $= 52 - 20 = \textbf{32}$.

**Check:** Jonas $20+20=40$ ✓; Clay $28+32=60$ ✓; boys $20+32=52$ ✓; girls $20+28=48$ ✓; total $100$ ✓. (Trap: 20 counts the boys from Jonas, not Clay.)

ANSWER 6: B

# Problem 7:
Both sums have the same number of terms: $1901$ to $1993$ has $1993-1901+1 = 93$ terms, and $101$ to $193$ has $193-101+1 = 93$ terms.

Pair them up: each term in the first sum is exactly $1800$ more than its partner ($1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$). So the difference is
$$93 \times 1800 = 167{,}400.$$

**Check via formulas:** $\frac{93(1901+1993)}{2} - \frac{93(101+193)}{2} = 181{,}071 - 13{,}671 = 167{,}400$. ✓ (Trap: 181,071 is the first sum alone — forgetting to subtract.)

ANSWER 7: A

# Problem 8:
Only the tens digit matters, so track the last two digits of $7^{k}$, which cycle with period 4:
$$7^1 = 07,\quad 7^2 = 49,\quad 7^3 = 343 \to 43,\quad 7^4 = 2401 \to 01,\quad 7^5 \to 07, \ldots$$

Find where $2011$ falls in the cycle:
$$2011 = 4 \times 502 + 3,$$
so $7^{2011}$ ends in the same two digits as $7^3$, namely $\ldots 43$.

The tens digit is $\textbf{4}$. (Trap: $7^{2012}$, a multiple of 4, ends in $01$, tens digit 0 — choice A; that would require the exponent to be divisible by 4.)

ANSWER 8: D

# Problem 9:
The ten **different** positive whole numbers have average 10, so their sum is $10 \times 10 = 100$.

To make one number as large as possible, make the other nine as small as possible. Since they must be **distinct positive** whole numbers, the smallest possible set is $1, 2, 3, \ldots, 9$, with sum
$$1+2+\cdots+9 = 45.$$
So the largest possible value is
$$100 - 45 = 55.$$

**Check:** $1,2,\ldots,9,55$ are ten different positive whole numbers with sum $100$ and average $10$. ✓ (Trap: 91 arises if you ignore "different" and use nine 1's; but the numbers must be distinct.)

ANSWER 9: C

# Problem 10:
Count student-days (the unit being paid):
- Allen: $7 \times 3 = 21$ student-days
- Balboa: $4 \times 5 = 20$ student-days
- Carver: $5 \times 9 = 45$ student-days

Total: $21 + 20 + 45 = 86$ student-days, paid $\$774$ altogether. So the daily rate is
$$774 \div 86 = 9 \text{ dollars per student-day}.$$
Balboa's students worked 20 student-days, so they earned
$$20 \times 9 = 18

Problem 1:  
We need the parity of \(o^2+no\), where \(o\) is odd and \(n\) is any whole number.

Factor:
\[
o^2+no=o(o+n).
\]

Since \(o\) is odd:

- If \(n\) is even, then \(o+n\) is odd, so \(o(o+n)\) is odd.
- If \(n\) is odd, then \(o+n\) is even, so \(o(o+n)\) is even.

Thus the expression is odd exactly when \(n\) is even.

Check the choices:
- A and B are false because the parity depends on \(n\).
- C says “even only if \(n\) is even,” but when the expression is even, \(n\) is actually odd.
- D says “odd only if \(n\) is odd,” but when the expression is odd, \(n\) is even.
- E says “odd only if \(n\) is even,” which is true.

ANSWER 1: E

Problem 2:  
Isabella redeems coupons 10 days apart. Since
\[
10\equiv 3 \pmod 7,
\]
the six circled days are the first day plus
\[
0,3,6,9,12,15
\]
days, which modulo 7 gives offsets
\[
0,3,6,2,5,1.
\]

So the six circled days are all weekdays except the one 4 days after the first day. Since none of the circled days is a Sunday, Sunday must be the un-circled day: 4 days after the first day.

Therefore the first coupon is 4 days before Sunday, which is Wednesday.

Check: Wednesday, Saturday, Tuesday, Friday, Monday, Thursday — no Sunday.

ANSWER 2: C

Problem 3:  
The product is the telescoping product of fractions:
\[
\frac32\cdot \frac43\cdot \frac54\cdot \frac65\cdots \frac{a}{b}=9.
\]

The numerators cancel the next denominators:
\[
\frac32\cdot \frac43=\frac42,
\]
\[
\frac42\cdot \frac54=\frac52,
\]
and continuing, the whole product simplifies to
\[
\frac{a}{2}.
\]

So
\[
\frac{a}{2}=9 \implies a=18.
\]

By the pattern, the denominator is one less than the numerator, so
\[
b=17.
\]

Thus
\[
a+b=18+17=35.
\]

ANSWER 3: D

Problem 4:  
Nicolas has 20 of each type of stamp, so he has
\[
20+20+20=60
\]
stamps total. Their total value is
\[
20(5+10+25)=20(40)=800
\]
cents.

He wants postage of
\[
\$7.10=710
\]
cents, so the stamps he does **not** use must total
\[
800-710=90
\]
cents.

To maximize the number of stamps he uses, we minimize the number of stamps he leaves unused.

Can 4 unused stamps total 90 cents? Let \(x,y,z\) be the numbers of 25-, 10-, and 5-cent unused stamps. Then
\[
x+y+z=4
\]
and
\[
25x+10y+5z=90.
\]

Divide by 5:
\[
5x+2y+z=18.
\]

Subtract \(x+y+z=4\):
\[
4x+y=14.
\]

Testing possible \(x=0,1,2,3,4\):

- \(x=0\Rightarrow y=14\), impossible.
- \(x=1\Rightarrow y=10\), impossible.
- \(x=2\Rightarrow y=6\), impossible.
- \(x=3\Rightarrow y=2\), then \(z=4-3-2=-1\), impossible.
- \(x=4\Rightarrow y=-2\), impossible.

So 4 unused stamps cannot total 90 cents.

Can 5 unused stamps total 90 cents? Yes:
\[
25+25+25+10+5=90.
\]

Then he uses
\[
60-5=55
\]
stamps. For example, he could use 19 five-cent stamps, 19 ten-cent stamps, and 17 twenty-five-cent stamps:
\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So the greatest possible number of stamps is 55. Choice D, 54, is possible but not greatest.

ANSWER 4: E

Problem 5:  
We are told
\[
ABA\times CD=CDCD,
\]
where \(A,B,C,D\) are different digits.

Write the numbers in place-value form:
\[
ABA=101A+10B,
\]
\[
CD=10C+D,
\]
\[
CDCD=1000C+100D+10C+D=101(10C+D).
\]

So
\[
(101A+10B)(10C+D)=101(10C+D).
\]

Since \(CD\) is a two-digit number, \(10C+D\neq 0\), so we can divide both sides by \(10C+D\):
\[
101A+10B=101.
\]

The only digit solution is
\[
A=1,\quad B=0.
\]

Therefore
\[
A+B=1+0=1.
\]

ANSWER 5: A

Problem 6:  
There are 48 girls total, and 20 of the girls are from Jonas Middle School. So the number of girls from Clay Middle School is
\[
48-20=28.
\]

There are 60 students total from Clay Middle School, so the number of boys from Clay is
\[
60-28=32.
\]

ANSWER 6: B

Problem 7:  
Pair the terms:
\[
(1901+1902+\cdots+1993)-(101+102+\cdots+193)
\]
\[
=(1901-101)+(1902-102)+\cdots+(1993-193).
\]

Each pair is
\[
1800.
\]

Now count the number of terms:
\[
193-101+1=93.
\]

So the value is
\[
93\cdot 1800=167400.
\]

ANSWER 7: A

Problem 8:  
We need the tens digit of \(7^{2011}\), so we need the last two digits.

Powers of 7 modulo 100 cycle:
\[
7^1=07,
\]
\[
7^2=49,
\]
\[
7^3=343\equiv 43,
\]
\[
7^4=2401\equiv 01.
\]

Then the cycle repeats:
\[
07,\ 49,\ 43,\ 01.
\]

Since
\[
2011\equiv 3 \pmod 4,
\]
we have
\[
7^{2011}\equiv 7^3\equiv 43 \pmod{100}.
\]

The last two digits are 43, so the tens digit is 4.

ANSWER 8: D

Problem 9:  
The average of 10 different positive whole numbers is 10, so their sum is
\[
10\cdot 10=100.
\]

To make one number as large as possible, make the other 9 numbers as small as possible while still being different positive whole numbers. The smallest 9 distinct positive whole numbers are
\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is
\[
1+2+\cdots+9=45.
\]

So the largest number can be
\[
100-45=55.
\]

This is possible: \(1,2,3,4,5,6,7,8,9,55\) are all different positive whole numbers.

ANSWER 9: C

Problem 10:  
First count total student-days worked:

- Allen: \(7\cdot 3=21\) student-days
- Balboa: \(4\cdot 5=20\) student-days
- Carver: \(5\cdot 9=45\) student-days

Total:
\[
21+20+45=86
\]
student-days.

The total payment is \$774, so the daily rate is
\[
774\div 86=9
\]
dollars per student-day.

Balboa students worked 20 student-days, so they earned
\[
20\cdot 9=180
\]
dollars.

ANSWER 10: C

# Problem 1:
We need a statement always true about o² + no, where o is odd.

Factor: o² + no = o(o + n).

Since o is odd, the parity of the product is the parity of (o + n).
- If n is even, o + n is odd → product = odd·odd = **odd**.
- If n is odd, o + n is even → product = odd·even = **even**.

So the expression is odd exactly when n is even.

Ruling out others: A fails (n odd gives even), B fails (n even gives odd), C says "even only if n is even" — but it's even when n is *odd*, false. D says "odd only if n is odd" — false, it's odd when n is even. E: "odd only if n is even" — true.

ANSWER 1: E

# Problem 2:
The redemption days are 0, 10, 20, 30, 40, 50 days after the first.

Reduce mod 7 (days of week shift): 0, 3, 6, 2, 5, 1.

These residues are {0, 1, 2, 3, 5, 6} — every residue except **4**. So the only day of the week never hit is the day 4 days after the starting day. For no coupon to land on Sunday, Sunday must be that missed day:

start + 4 ≡ Sunday.

Counting back 4 days from Sunday: Saturday, Friday, Thursday, **Wednesday**.

Check: Wed → +10 Sat → +20 Tue → +30 Fri → +40 Mon → +50 Thu. No Sundays. ✓

ANSWER 2: C

# Problem 3:
The product is (3/2)·(4/3)·(5/4)·(6/5)·…·(a/b), where each numerator is one more than its denominator, so b = a − 1.

Telescoping: everything cancels except the last numerator and first denominator:
a/2 = 9 → a = 18, so b = 17.

a + b = 18 + 17 = 35.

(Choice E, 37, would come from mistakenly taking a = 19.)

ANSWER 3: D

# Problem 4:
Total value of all 60 stamps: 20(5 + 10 + 25) = 20·40 = 800 cents.

We need 710 cents, so we must *remove* stamps worth exactly 800 − 710 = 90 cents, using as **few** stamps as possible (to keep the most stamps).

Can 90 be made with 4 or fewer stamps? Max with 4 stamps is 100; possible 4-stamp totals near 90: 100, 85, 80, 75, 70… none equal 90. Three or fewer is at most 75. So 4 or fewer is impossible.

With 5 stamps: 25 + 25 + 25 + 10 + 5 = 90 ✓

So remove 5 stamps: 60 − 5 = 55 stamps.

Check: 17(25) + 19(10) + 19(5) = 425 + 190 + 95 = 710 cents, using 17 + 19 + 19 = 55 stamps.

ANSWER 4: E

# Problem 5:
CDCD as a number = CD·100 + CD = CD·101.

So ABA × CD = 101 × CD, and since CD ≠ 0, we get ABA = 101.

Thus A = 1, B = 0 (and indeed A, B, C, D distinct works, e.g. 101 × 23 = 2323).

A + B = 1 + 0 = 1.

ANSWER 5: A

# Problem 6:
Jonas total = 40, and 20 of those were girls, so boys from Jonas = 40 − 20 = 20.

Total boys = 52, so boys from Clay = 52 − 20 = 32.

(Check: girls from Clay = 48 − 20 = 28; 32 + 28 = 60 = Clay total ✓.)

Choice C (40) is the Jonas total, and E (52) is all boys — traps.

ANSWER 6: B

# Problem 7:
Pair each term: 1901 − 101 = 1800, 1902 − 102 = 1800, …, each pair differs by 1800.

Number of terms: 1993 − 1901 + 1 = 93.

Total = 93 × 1800 = 167,400.

ANSWER 7: A

# Problem 8:
Look at 7ⁿ mod 100:
- 7¹ = 07
- 7² = 49
- 7³ = 343 → 43
- 7⁴ = 2401 → 01
- then the pattern repeats with period 4.

2011 ÷ 4 = 502 remainder 3, so 7²⁰¹¹ ≡ 7³ ≡ 43 (mod 100).

Last two digits are 43, so the tens digit is 4.

ANSWER 8: D

# Problem 9:
Sum of the 10 numbers = 10 × 10 = 100.

To make one number as large as possible, make the other nine as small as possible while distinct and positive: 1 + 2 + … + 9 = 45.

Largest = 100 − 45 = 55.

(90 or 91 would require repeats or a zero, which isn't allowed.)

ANSWER 9: C

# Problem 10:
Total student-days:
- Allen: 7 × 3 = 21
- Balboa: 4 × 5 = 20
- Carver: 5 × 9 = 45

Total = 86 student-days.

Daily wage = 774 ÷ 86 = $9.00 per student-day.

Balboa earnings = 20 × 9 = $180.00.

(A is one day's pay for one student; E is Carver's total.)

ANSWER 10: C

Problem 1:

We need determine the parity of

\[
o^2+no.
\]

Factor out \(o\):

\[
o^2+no=o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on \(o+n\).

- If \(n\) is even, then odd \(+\) even \(=\) odd, so \(o+n\) is odd. Thus odd \(\cdot\) odd is odd.
- If \(n\) is odd, then odd \(+\) odd \(=\) even, so the product is even.

Therefore, the expression is odd only when \(n\) is even.

- A is wrong because the expression is not always odd.
- B is wrong because it is odd when \(n\) is even.
- C reverses the condition for being even.
- D reverses the condition for being odd.

ANSWER 1: E

Problem 2:

Isabella uses a coupon every 10 days. Since

\[
10\equiv 3\pmod 7,
\]

each coupon date moves forward by 3 days of the week.

If the first coupon is redeemed on day \(d\), the six redemption days are:

\[
d,\ d+3,\ d+6,\ d+2,\ d+5,\ d+1.
\]

These are six different days of the week. The only day missing is

\[
d+4.
\]

Since none of the six coupon dates is Sunday, the missing day must be Sunday. Thus,

\[
d+4=\text{Sunday}.
\]

Four days before Sunday is Wednesday.

Indeed, beginning on Wednesday gives:

\[
\text{Wednesday, Saturday, Tuesday, Friday, Monday, Thursday}.
\]

Sunday is the only missing day.

ANSWER 2: C

Problem 3:

The product is

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9.
\]

Almost every numerator cancels with a denominator:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab
=\frac a2.
\]

So

\[
\frac a2=9.
\]

Therefore,

\[
a=18.
\]

The final fraction must be

\[
\frac{18}{17},
\]

so \(b=17\). Therefore,

\[
a+b=18+17=35.
\]

Choices 11, 13, and 17 are too small, while 37 would correspond to a different ending fraction.

ANSWER 3: D

Problem 4:

Let

- \(x\) = number of 5-cent stamps,
- \(y\) = number of 10-cent stamps,
- \(z\) = number of 25-cent stamps.

In units of 5 cents, the postage total is

\[
\frac{710}{5}=142.
\]

Thus,

\[
x+2y+5z=142.
\]

We want to maximize

\[
x+y+z.
\]

Since there are at most 20 five-cent stamps, \(x\leq 20\). Therefore,

\[
2y+5z=142-x\geq 142-20=122.
\]

Since \(y\leq20\), if \(z\leq16\), then

\[
2y+5z\leq 2(20)+5(16)=40+80=120,
\]

which is not enough. So \(z\geq17\).

If \(z=17\), then

\[
2y+5(17)\geq122,
\]

so

\[
2y+85\geq122,
\]

\[
2y\geq37,
\]

and hence \(y\geq19\).

Now use

\[
x+y+z=142-y-4z.
\]

Since \(y\geq19\) and \(z\geq17\),

\[
x+y+z\leq142-19-4(17)
=142-19-68
=55.
\]

This maximum is attainable with

\[
x=19,\qquad y=19,\qquad z=17.
\]

Check:

\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So 55 stamps can be used. Choices 45, 46, 51, and 54 are possible totals to consider, but 55 is the greatest possible.

ANSWER 4: E

Problem 5:

We are told

\[
ABA\times CD=CDCD.
\]

The number \(ABA\) is

\[
100A+10B+A=101A+10B.
\]

The number \(CDCD\) is

\[
1000C+100D+10C+D.
\]

Group the \(C\)'s and \(D\)'s:

\[
CDCD=1010C+101D=101(10C+D)=101\cdot CD.
\]

Thus the equation becomes

\[
ABA\cdot CD=101\cdot CD.
\]

Since \(CD\) is a two-digit number and is not zero, divide both sides by \(CD\):

\[
ABA=101.
\]

Therefore,

\[
A=1,\qquad B=0.
\]

So

\[
A+B=1+0=1.
\]

For example, \(101\times23=2323\), and the digits \(1,0,2,3\) are all different.

ANSWER 5: A

Problem 6:

There are 40 students from Jonas Middle School, and 20 of those are girls.

Therefore the number of boys from Jonas is

\[
40-20=20.
\]

There are 52 boys altogether. Therefore the number of boys from Clay Middle School is

\[
52-20=32.
\]

The other choices do not correctly account for the 20 Jonas girls and the total of 52 boys.

ANSWER 6: B

Problem 7:

There are

\[
1993-1901+1=93
\]

terms in the first sum.

There are also

\[
193-101+1=93
\]

terms in the second sum.

Pair corresponding terms:

\[
1901-101=1800,
\]

\[
1902-102=1800,
\]

and so on, through

\[
1993-193=1800.
\]

Thus there are 93 differences of 1800:

\[
93\cdot1800=167400.
\]

Therefore the value is

\[
167,\!400.
\]

ANSWER 7: A

Problem 8:

We need the tens digit of

\[
7^{2011}.
\]

It is enough to examine powers of 7 modulo 100:

\[
7^1=7,
\]

\[
7^2=49,
\]

\[
7^3=343,
\]

whose last two digits are 43, and

\[
7^4=2401,
\]

whose last two digits are 01.

Thus the last two digits repeat every 4 powers because

\[
7^4\equiv1\pmod{100}.
\]

Now,

\[
2011\div4
\]

leaves remainder 3, since

\[
2011=4(502)+3.
\]

Therefore,

\[
7^{2011}
\]

has the same last two digits as

\[
7^3=343,
\]

which are 43. The tens digit is 4.

ANSWER 8: D

Problem 9:

The average of 10 different positive whole numbers is 10, so their total sum is

\[
10\cdot10=100.
\]

To make one number as large as possible, make the other nine numbers as small as possible.

The nine smallest different positive whole numbers are

\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is

\[
1+2+3+4+5+6+7+8+9=45.
\]

So the largest possible tenth number is

\[
100-45=55.
\]

This works because the ten numbers can be

\[
1,2,3,4,5,6,7,8,9,55,
\]

which are all different and have total 100.

A value such as 90 or 91 would leave too small a total for nine distinct positive whole numbers.

ANSWER 9: C

Problem 10:

Compute the total number of student-days worked.

Allen students worked:

\[
7\cdot3=21
\]

student-days.

Balboa students worked:

\[
4\cdot5=20
\]

student-days.

Carver students worked:

\[
5\cdot9=45
\]

student-days.

So the total number of student-days is

\[
21+20+45=86.
\]

The total pay was \$774, so the pay per student per day was

\[
\frac{774}{86}=9.
\]

Balboa students worked 20 student-days, so they earned

\[
20\cdot9=180.
\]

Thus the Balboa students earned \$180.00 altogether.

ANSWER 10: C

Problem 1:
We are asked about the parity of $o^2+no$ where $o$ is odd and $n$ is any whole number.

Factor:
$$o^2+no = o(o+n).$$

$o$ is odd, so $o^2 = o\cdot o$ is odd times odd = odd.

Hence parity is odd + $n\cdot o$.
$n\cdot o$ has the same parity as $n$, because $o$ is odd: if $n$ is even, $n o$ is even; if $n$ is odd, $n o$ is odd.

So:
- if $n$ even: odd + even = odd.
- if $n$ odd: odd + odd = even.

Thus the number is odd iff $n$ is even, and even iff $n$ is odd.

Check choices:
A. always odd - false, e.g. $o=1,n=1$ gives $1+1=2$ even.
B. always even - false, e.g. $o=1,n=2$ gives $1+2=3$ odd.
C. it is even only if $n$ is even - means even $\implies$ $n$ even. False, even happens when $n$ is odd.
D. it is odd only if $n$ is odd - means odd $\implies$ $n$ odd. False, odd happens when $n$ is even.
E. it is odd only if $n$ is even - means odd $\implies$ $n$ even. True by above.

ANSWER 1: E

Problem 2:
Isabella redeems on days $d, d+10, d+20, d+30, d+40, d+50$.
No date is a Sunday.

Modulo 7, $10\equiv 3$. So successive dates advance by 3 weekdays.

Starting from residue $r$, the residues are:
$$r,\ r+3,\ r+6,\ r+9=r+2,\ r+12=r+5,\ r+15=r+1 \pmod 7.$$
These 6 are all distinct, because adding 3 mod 7 cycles through all 7 residues. Continuing one more step would give $r+18=r+4$.

Thus the 6 dates occupy 6 distinct days of the week, missing exactly the 7th residue $r+4$.

Since none of the 6 is Sunday, the missing one must be Sunday. So:
$$r+4 \equiv \text{Sunday}.$$

Hence $r$ is 4 days before Sunday, i.e. Wednesday.

Check: Wed $\to$ Wed+3=Sat $\to$ +3=Tue $\to$ +3=Fri $\to$ +3=Mon $\to$ +3=Thu. That is Wed,Sat,Tue,Fri,Mon,Thu - no Sunday. Any other start would include Sunday.

ANSWER 2: C

Problem 3:
We interpret $3\ 2\cdot 4\ 3\cdots$ as fractions:
$$\frac32 \cdot \frac43 \cdot \frac54 \cdot \frac65 \cdots \frac{a}{b}=9.$$

The pattern is numerator $3,4,5,6,\dots$ and denominator $2,3,4,5,\dots$, i.e. term $k/(k-1)$. The last term is $a/b$, so we must have $b=a-1$ for the pattern to continue.

Then the product telescopes:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}
= \frac{a}{2},$$
because all intermediate $3,4,\dots,a-1$ cancel.

Set equal to 9:
$$\frac{a}{2}=9 \implies a=18.$$
Then $b=a-1=17$.

Sum $a+b=18+17=35$.

This rules out A,B,C,E.

ANSWER 3: D

Problem 4:
Let $x,y,z$ be numbers of 5-cent, 10-cent, 25-cent stamps used. $0\le x,y,z\le 20$ integers.
Value $710$ cents:
$$5x+10y+25z=710.$$
Divide by 5:
$$x+2y+5z=142. \tag{1}$$
Maximize $N=x+y+z$.

From (1), $x=142-2y-5z$, so $0\le x\le20$ gives:
$$122\le 2y+5z\le 142.$$
Also $N=x+y+z=142-y-4z$, so $y+4z=142-N$.

To use many stamps we want small denominations. Try $x=20,y=20$ fails to give integer $z$. Search systematically.

Since $y\le20$, to satisfy $2y+5z\ge122$ we need $z$ large. If $z\le16$, then $2y+5z\le40+80=120<122$, impossible. So $z\ge17$.

Try $z=17,y=19$: then $2y+5z=38+85=123$, so $x=142-123=19$, within $0-20$. This gives:
$$19\cdot5+19\cdot10+17\cdot25=95+190+425=710,$$
with $N=19+19+17=55$ stamps. So 55 is achievable.

Can we do 56 or more? Let $N\ge56$.
Then $y+4z=142-N\le86$ and $x=N-y-z\le20$ gives $y+z\ge N-20$.
Subtract: $3z=(y+4z)-(y+z)\le (142-N)-(N-20)=162-2N$.
So $z\le(162-2N)/3$.
Also $y=142-N-4z\le20$ gives $z\ge(122-N)/4$.

For $N=56$: $z\ge\lceil66/4\rceil=17$ and $z\le\lfloor50/3\rfloor=16$, contradiction. For $N>56$ the interval is even emptier; in general need $(122-N)/4\le(162-2N)/3$, i.e. $5N\le282$, so $N\le56.4$. Thus $N\ge57$ impossible, and $N=56$ impossible by the integer contradiction.

Hence 55 is greatest possible.

ANSWER 4: E

Problem 5:
We have $ABA\times CD = CDCD$, where $A,B,C,D$ are different digits, $A\ne0$, $C\ne0$.

Write $CD$ as the 2-digit number $10C+D$.
$CDCD$ as a 4-digit number is:
$$1000C+100D+10C+D =1010C+101D=101(10C+D)=101\cdot CD.$$

So:
$$ABA \times CD = 101\times CD.$$
Since $CD\ne0$, divide:
$$ABA =101.$$

Thus $A=1$, $B=0$, and $A+B=1$.

This is consistent with distinctness: we can choose e.g. $C=2,D=3$, all of $1,0,2,3$ different, and $101\times23=2323$. So valid.

Other sums are ruled out.

ANSWER 5: A

Problem 6:
Total 100: 52 boys, 48 girls. 40 from Jonas, 60 from Clay.

20 of the girls are from Jonas. So girls from Clay = total girls $-$ girls from Jonas:
$$48-20=28.$$

Clay total 60 = boys from Clay + girls from Clay, so:
$$\text{boys from Clay}=60-28=32.$$

Check: boys from Jonas $=52-32=20$, Jonas total $=20+20=40$. Works.

ANSWER 6: B

Problem 7:
Compute $(1901+\cdots+1993)-(101+\cdots+193)$.

Both sums have $1993-1901+1=93$ terms and $193-101+1=93$ terms.

Pair termwise: for $k=0,\dots,92$,
$$(1901+k)-(101+k)=1800.$$

So difference $=93\times1800$.
$93\times18=1674$, so $93\times1800=167{,}400$.

Other choices come from miscounting terms or arithmetic error.

ANSWER 7: A

Problem 8:
Find tens digit of $7^{2011}$, i.e. $7^{2011}\pmod{100}$.

Compute powers mod 100:
$7^1=7$,
$7^2=49$,
$7^3=343\equiv43$,
$7^4\equiv43\times7=301\equiv1$.

Since $7^4\equiv1\pmod{100}$, the last two digits repeat with period 4.

$2011 = 4\cdot502+3$, so:
$$7^{2011}=(7^4)^{502}\cdot7^3\equiv1^{502}\cdot43\equiv43\pmod{100}.$$

Thus the number ends $\dots43$, tens digit $4$.

ANSWER 8: D

Problem 9:
10 different positive whole numbers have mean 10, so sum $=10\times10=100$.

To make one number as large as possible, make the other nine as small as possible while distinct and positive.

Smallest 9 distinct positives are $1,2,3,4,5,6,7,8,9$, sum $=45$.

Then the tenth number $=100-45=55$, which is distinct from the others and valid. Any larger value would force the other nine to sum to $<45$, impossible with distinct positives.

So largest possible is 55.

ANSWER 9: C

Problem 10:
Count student-days (one student working one day).

Allen: $7$ students $\times$ $3$ days $=21$.
Balboa: $4\times5=20$.
Carver: $5\times9=45$.
Total $=21+20+45=86$ student-days.

Total pay $\$774$ for $86$ student-days at equal rate:
$$\text{rate}=774/86=9 \text{ dollars per student-day}.$$

Bal

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that $o$ is an odd whole number and $n$ is any whole number. We must determine which statement about $o^2 + no$ is always true.

**Step-by-Step Solution:**
1. Factor the expression:
   $$o^2 + no = o(o + n)$$
2. Since $o$ is an odd whole number, the parity of the product $o(o + n)$ is entirely determined by the parity of $(o + n)$:
   - If $n$ is **odd**, then $\text{odd} + \text{odd} = \text{even}$, so $o(o+n)$ is $\text{odd} \times \text{even} = \textbf{even}$.
   - If $n$ is **even**, then $\text{odd} + \text{even} = \text{odd}$, so $o(o+n)$ is $\text{odd} \times \text{odd} = \textbf{odd}$.
3. Therefore, the expression $o^2 + no$ is odd if and only if $n$ is even. 
4. This matches the statement: "it is odd only if $n$ is even."

ANSWER 1: E

---

### Problem 2:
**Restatement:** Isabella redeems 6 coupons, one every 10 days. Pete's is closed on Sundays, and none of the 6 dates falls on a Sunday. We need to find the day of the week she redeems her first coupon.

**Step-by-Step Solution:**
1. There are 7 days in a week. Moving forward by 10 days advances the day of the week by:
   $$10 \equiv 3 \pmod 7$$
2. Let $d$ represent the day of the week of the first coupon ($0 = \text{Sunday}, 1 = \text{Monday}, \dots, 6 = \text{Saturday}$).
3. The days of the week for the 6 coupons will be:
   - Coupon 1: $d$
   - Coupon 2: $d + 3$
   - Coupon 3: $d + 6$
   - Coupon 4: $d + 9 \equiv d + 2 \pmod 7$
   - Coupon 5: $d + 12 \equiv d + 5 \pmod 7$
   - Coupon 6: $d + 15 \equiv d + 1 \pmod 7$
4. The six days of the week visited are $d, d+1, d+2, d+3, d+5, d+6$. Notice that the only day of the week omitted from this list is $d+4$.
5. Since no circled date falls on a Sunday, Sunday must be the omitted day:
   $$d + 4 \equiv 0 \pmod 7 \implies d \equiv 3 \pmod 7$$
6. Day 3 corresponds to **Wednesday**.
   - Checking: Wed $\to$ Sat $\to$ Tue $\to$ Fri $\to$ Mon $\to$ Thu. None of these is Sunday.

ANSWER 2: C

---

### Problem 3:
**Restatement:** We are given the telescoping product $\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdots \frac{a}{b} = 9$. We must find the sum $a + b$.

**Step-by-Step Solution:**
1. Each fraction is of the form $\frac{k+1}{k}$, where the numerator of each term cancels with the denominator of the subsequent term:
   $$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{a}{b} = \frac{a}{2}$$
   where $b = a - 1$.
2. We are given that this product equals 9:
   $$\frac{a}{2} = 9 \implies a = 18$$
3. Since $b = a - 1$, we have $b = 17$.
4. The sum of $a$ and $b$ is:
   $$a + b = 18 + 17 = 35$$

ANSWER 3: D

---

### Problem 4:
**Restatement:** Nicolas has 20 stamps each of 5¢, 10¢, and 25¢. We want to find the greatest number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-Step Solution:**
1. Calculate the total value and count of all stamps:
   - $20 \times 5¢ = 100¢$
   - $20 \times 10¢ = 200¢$
   - $20 \times 25¢ = 500¢$
   - Total value $= 100 + 200 + 500 = 800¢$ using 60 stamps.
2. To use the **maximum** number of stamps to make 710¢, Nicolas must **minimize** the number of stamps removed to subtract:
   $$800¢ - 710¢ = 90¢$$
3. To remove the fewest stamps totaling 90¢, we try to use the largest denomination stamps (25¢):
   - With 3 stamps of 25¢ ($75¢$), the remaining amount needed is $15¢$.
   - Since Nicolas has only 5¢, 10¢, and 25¢ stamps, 15¢ cannot be made with a single stamp. It requires at least two stamps ($10¢ + 5¢$).
   - Thus, 90¢ can be formed with $3 \times 25¢ + 1 \times 10¢ + 1 \times 5¢$, which uses $3 + 1 + 1 = 5$ stamps.
   - Can 90¢ be formed using 4 stamps? The maximum value of 4 stamps less than or equal to 90¢ would be $3 \times 25¢ + 10¢ = 85¢ \ne 90¢$. Thus, 4 stamps is impossible.
4. The minimum number of stamps to remove is 5.
5. Therefore, the maximum number of stamps used is:
   $$60 - 5 = 55$$

ANSWER 4: E

---

### Problem 5:
**Restatement:** In the multiplication $ABA \times CD = CDCD$, $A, B, C, D$ are distinct digits. We need to find $A + B$.

**Step-by-Step Solution:**
1. Express $CDCD$ in terms of the two-digit number $CD$:
   $$CDCD = CD \times 100 + CD = CD \times 101$$
2. Substitute this into the equation:
   $$ABA \times CD = 101 \times CD$$
3. Since $CD$ is a non-zero two-digit number, divide both sides by $CD$:
   $$ABA = 101$$
4. Comparing digits, we have $A = 1$ and $B = 0$.
5. Thus:
   $$A + B = 1 + 0 = 1$$

ANSWER 5: A

---

### Problem 6:
**Restatement:** Out of 100 students at a camp, 52 are boys and 48 are girls. Also, 40 are from Jonas Middle School and 60 from Clay Middle School. Given that 20 of the girls are from Jonas, find how many boys are from Clay.

**Step-by-Step Solution:**
1. Find the number of girls from Clay Middle School:
   $$\text{Total girls} - \text{Girls from Jonas} = 48 - 20 = 28\text{ girls from Clay}$$
2. Find the number of boys from Clay Middle School:
   $$\text{Total students from Clay} - \text{Girls from Clay} = 60 - 28 = 32\text{ boys from Clay}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** Compute the value of:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-Step Solution:**
1. Pair each term in the first sum with the corresponding term in the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
2. Each paired difference is equal to:
   $$1900 + k - (100 + k) = 1800$$
3. Find the number of terms: from $k = 1$ to $k = 93$, there are:
   $$93 - 1 + 1 = 93\text{ terms}$$
4. Multiply the difference by the number of terms:
   $$93 \times 1800 = 93 \times 18 \times 100 = 1674 \times 100 = 167,400$$

ANSWER 7: A

---

### Problem 8:
**Restatement:** Find the tens digit of $7^{2011}$.

**Step-by-Step Solution:**
1. Find the pattern of the last two digits of powers of 7 by computing $7^n \pmod{100}$:
   - $7^1 = 07$
   - $7^2 = 49$
   - $7^3 = 343 \equiv 43 \pmod{100}$
   - $7^4 = 43

# Problem 1:
We must decide when $o^2 + no$ is odd or even, given $o$ is odd and $n$ is any whole number.

**Key facts:** odd × odd = odd, so $o^2$ is **odd**. Since $o$ is odd, multiplying by $o$ does not change $n$'s parity, so $no$ has the **same parity as $n$**.

**Case 1: $n$ even.** Then $no$ is even, so $o^2 + no = \text{odd} + \text{even} = \textbf{odd}$.
**Case 2: $n$ odd.** Then $no$ is odd, so $o^2 + no = \text{odd} + \text{odd} = \textbf{even}$.

So the number is odd exactly when $n$ is even. In particular, **if the number is odd, $n$ must be even**.

Ruling out traps: (A) fails since $o=1, n=1$ gives $1+1=2$, even; (B) fails since $n=2$ gives $1+2=3$, odd; (C) and (D) state the exact opposite of the truth. Only (E) matches.

ANSWER 1: E

# Problem 2:
The 6 redemption dates are 10 days apart, so their days of the week shift by $10 \bmod 7 = 3$ each time.

If the first date falls on weekday $W$, the six dates fall on (mod 7):
$$W,\; W+3,\; W+6,\; W+2,\; W+5,\; W+1$$
That is **six of the seven weekdays**, and the only weekday missed is $W+4$.

Since no circled date is a Sunday, the one missed weekday must be Sunday:
$$W + 4 = \text{Sunday} \implies W = \text{Wednesday}.$$

**Check:** Starting Wednesday, the dates fall on Wed, Sat, Tue, Fri, Mon, Thu — no Sunday. (Trap: starting Thursday gives Thu, Sun, ... — fails immediately.)

ANSWER 2: C

# Problem 3:
The product is a chain of fractions (the bars were lost in formatting):
$$\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdots\frac{a}{b} = 9.$$

Each numerator cancels the denominator before it (e.g., $\frac{3}{2}\cdot\frac{4}{3} = \frac{4}{2}$), and each fraction has the form $\frac{k+1}{k}$, so $b = a-1$. Everything telescopes:
$$\text{Product} = \frac{a}{2} = 9 \implies a = 18, \quad b = 17.$$
Then $a + b = 18 + 17 = 35$.

Trap check: $a+b=17$ would mean $a=9$, giving product $\tfrac92$, not $9$; only $35$ fits the pattern.

ANSWER 3: D

# Problem 4:
Work in units of 5¢: we need total value $710¢ = 142$ units, using stamps worth $1, 2, 5$ units (i.e., 5¢, 10¢, 25¢), at most 20 of each. To **maximize the count**, use as many cheap stamps as possible.

**Building a 55-stamp solution:** Take nineteen 1's and nineteen 2's: value $19 + 38 = 57$ units with 38 stamps. Remaining value $142 - 57 = 85 = 17\times 5$: seventeen 5's.
$$19\times 5¢ + 19\times 10¢ + 17\times 25¢ = 95 + 190 + 425 = 710¢ = \$7.10.\; ✓$$
That's $19+19+17 = 55$ stamps, all counts $\le 20$. ✓

**Why 56 is impossible:** with 56 stamps, at least $56 - 40 = 16$ must be 25¢ stamps (only 40 cheap ones exist). Let $z$ = number of 25¢ stamps:
- $z=16$: max value $= 100 + 200 + 400 = 700 < 710$. ✗
- $z=17$: need $5x+10y=285$, $x+y=39 \Rightarrow x=21>20$. ✗
- $z=18$: $x=24>20$. ✗  $z=19$: $x=27>20$. ✗  $z=20$: $x=30>20$. ✗

Traps: 54 comes from $20×5+16×10+18×25 = 710$; 51 from $20×5+11×10+20×25 = 710$ — both valid but not maximal.

ANSWER 4: E

# Problem 5:
Write $CDCD$ as a number:
$$CDCD = 1000C + 100D + 10C + D = 101(10C + D) = 101 \times CD.$$
The given equation is
$$ABA \times CD = CDCD = 101 \times CD.$$
Since $CD \neq 0$ (it's a two-digit number), we can divide both sides by $CD$:
$$ABA = 101.$$
So $A = 1$ and $B = 0$, giving $A + B = 1$.

**Consistency check:** $C, D$ just need to be distinct from $0, 1$ and each other, e.g., $101 \times 23 = 2323$. ✓ (Trap: guessing digits without using the factorization 101 wastes time or gives 9.)

ANSWER 5: A

# Problem 6:
Set up a two-way table with 100 total students.

- Girls: 48 total, and 20 girls are from Jonas $\Rightarrow$ girls from Clay $= 48 - 20 = 28$.
- Jonas has 40 students, of whom 20 are girls $\Rightarrow$ boys from Jonas $= 40 - 20 = 20$.
- Boys: 52 total $\Rightarrow$ boys from Clay $= 52 - 20 = \textbf{32}$.

**Check:** Jonas $20+20=40$ ✓; Clay $28+32=60$ ✓; boys $20+32=52$ ✓; girls $20+28=48$ ✓; total $100$ ✓. (Trap: 20 counts the boys from Jonas, not Clay.)

ANSWER 6: B

# Problem 7:
Both sums have the same number of terms: $1901$ to $1993$ has $1993-1901+1 = 93$ terms, and $101$ to $193$ has $193-101+1 = 93$ terms.

Pair them up: each term in the first sum is exactly $1800$ more than its partner ($1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$). So the difference is
$$93 \times 1800 = 167{,}400.$$

**Check via formulas:** $\frac{93(1901+1993)}{2} - \frac{93(101+193)}{2} = 181{,}071 - 13{,}671 = 167{,}400$. ✓ (Trap: 181,071 is the first sum alone — forgetting to subtract.)

ANSWER 7: A

# Problem 8:
Only the tens digit matters, so track the last two digits of $7^{k}$, which cycle with period 4:
$$7^1 = 07,\quad 7^2 = 49,\quad 7^3 = 343 \to 43,\quad 7^4 = 2401 \to 01,\quad 7^5 \to 07, \ldots$$

Find where $2011$ falls in the cycle:
$$2011 = 4 \times 502 + 3,$$
so $7^{2011}$ ends in the same two digits as $7^3$, namely $\ldots 43$.

The tens digit is $\textbf{4}$. (Trap: $7^{2012}$, a multiple of 4, ends in $01$, tens digit 0 — choice A; that would require the exponent to be divisible by 4.)

ANSWER 8: D

# Problem 9:
The ten **different** positive whole numbers have average 10, so their sum is $10 \times 10 = 100$.

To make one number as large as possible, make the other nine as small as possible. Since they must be **distinct positive** whole numbers, the smallest possible set is $1, 2, 3, \ldots, 9$, with sum
$$1+2+\cdots+9 = 45.$$
So the largest possible value is
$$100 - 45 = 55.$$

**Check:** $1,2,\ldots,9,55$ are ten different positive whole numbers with sum $100$ and average $10$. ✓ (Trap: 91 arises if you ignore "different" and use nine 1's; but the numbers must be distinct.)

ANSWER 9: C

# Problem 10:
Count student-days (the unit being paid):
- Allen: $7 \times 3 = 21$ student-days
- Balboa: $4 \times 5 = 20$ student-days
- Carver: $5 \times 9 = 45$ student-days

Total: $21 + 20 + 45 = 86$ student-days, paid $\$774$ altogether. So the daily rate is
$$774 \div 86 = 9 \text{ dollars per student-day}.$$
Balboa's students worked 20 student-days, so they earned
$$20 \times 9 = 18

Problem 1:  
We need the parity of \(o^2+no\), where \(o\) is odd and \(n\) is any whole number.

Factor:
\[
o^2+no=o(o+n).
\]

Since \(o\) is odd:

- If \(n\) is even, then \(o+n\) is odd, so \(o(o+n)\) is odd.
- If \(n\) is odd, then \(o+n\) is even, so \(o(o+n)\) is even.

Thus the expression is odd exactly when \(n\) is even.

Check the choices:
- A and B are false because the parity depends on \(n\).
- C says “even only if \(n\) is even,” but when the expression is even, \(n\) is actually odd.
- D says “odd only if \(n\) is odd,” but when the expression is odd, \(n\) is even.
- E says “odd only if \(n\) is even,” which is true.

ANSWER 1: E

Problem 2:  
Isabella redeems coupons 10 days apart. Since
\[
10\equiv 3 \pmod 7,
\]
the six circled days are the first day plus
\[
0,3,6,9,12,15
\]
days, which modulo 7 gives offsets
\[
0,3,6,2,5,1.
\]

So the six circled days are all weekdays except the one 4 days after the first day. Since none of the circled days is a Sunday, Sunday must be the un-circled day: 4 days after the first day.

Therefore the first coupon is 4 days before Sunday, which is Wednesday.

Check: Wednesday, Saturday, Tuesday, Friday, Monday, Thursday — no Sunday.

ANSWER 2: C

Problem 3:  
The product is the telescoping product of fractions:
\[
\frac32\cdot \frac43\cdot \frac54\cdot \frac65\cdots \frac{a}{b}=9.
\]

The numerators cancel the next denominators:
\[
\frac32\cdot \frac43=\frac42,
\]
\[
\frac42\cdot \frac54=\frac52,
\]
and continuing, the whole product simplifies to
\[
\frac{a}{2}.
\]

So
\[
\frac{a}{2}=9 \implies a=18.
\]

By the pattern, the denominator is one less than the numerator, so
\[
b=17.
\]

Thus
\[
a+b=18+17=35.
\]

ANSWER 3: D

Problem 4:  
Nicolas has 20 of each type of stamp, so he has
\[
20+20+20=60
\]
stamps total. Their total value is
\[
20(5+10+25)=20(40)=800
\]
cents.

He wants postage of
\[
\$7.10=710
\]
cents, so the stamps he does **not** use must total
\[
800-710=90
\]
cents.

To maximize the number of stamps he uses, we minimize the number of stamps he leaves unused.

Can 4 unused stamps total 90 cents? Let \(x,y,z\) be the numbers of 25-, 10-, and 5-cent unused stamps. Then
\[
x+y+z=4
\]
and
\[
25x+10y+5z=90.
\]

Divide by 5:
\[
5x+2y+z=18.
\]

Subtract \(x+y+z=4\):
\[
4x+y=14.
\]

Testing possible \(x=0,1,2,3,4\):

- \(x=0\Rightarrow y=14\), impossible.
- \(x=1\Rightarrow y=10\), impossible.
- \(x=2\Rightarrow y=6\), impossible.
- \(x=3\Rightarrow y=2\), then \(z=4-3-2=-1\), impossible.
- \(x=4\Rightarrow y=-2\), impossible.

So 4 unused stamps cannot total 90 cents.

Can 5 unused stamps total 90 cents? Yes:
\[
25+25+25+10+5=90.
\]

Then he uses
\[
60-5=55
\]
stamps. For example, he could use 19 five-cent stamps, 19 ten-cent stamps, and 17 twenty-five-cent stamps:
\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So the greatest possible number of stamps is 55. Choice D, 54, is possible but not greatest.

ANSWER 4: E

Problem 5:  
We are told
\[
ABA\times CD=CDCD,
\]
where \(A,B,C,D\) are different digits.

Write the numbers in place-value form:
\[
ABA=101A+10B,
\]
\[
CD=10C+D,
\]
\[
CDCD=1000C+100D+10C+D=101(10C+D).
\]

So
\[
(101A+10B)(10C+D)=101(10C+D).
\]

Since \(CD\) is a two-digit number, \(10C+D\neq 0\), so we can divide both sides by \(10C+D\):
\[
101A+10B=101.
\]

The only digit solution is
\[
A=1,\quad B=0.
\]

Therefore
\[
A+B=1+0=1.
\]

ANSWER 5: A

Problem 6:  
There are 48 girls total, and 20 of the girls are from Jonas Middle School. So the number of girls from Clay Middle School is
\[
48-20=28.
\]

There are 60 students total from Clay Middle School, so the number of boys from Clay is
\[
60-28=32.
\]

ANSWER 6: B

Problem 7:  
Pair the terms:
\[
(1901+1902+\cdots+1993)-(101+102+\cdots+193)
\]
\[
=(1901-101)+(1902-102)+\cdots+(1993-193).
\]

Each pair is
\[
1800.
\]

Now count the number of terms:
\[
193-101+1=93.
\]

So the value is
\[
93\cdot 1800=167400.
\]

ANSWER 7: A

Problem 8:  
We need the tens digit of \(7^{2011}\), so we need the last two digits.

Powers of 7 modulo 100 cycle:
\[
7^1=07,
\]
\[
7^2=49,
\]
\[
7^3=343\equiv 43,
\]
\[
7^4=2401\equiv 01.
\]

Then the cycle repeats:
\[
07,\ 49,\ 43,\ 01.
\]

Since
\[
2011\equiv 3 \pmod 4,
\]
we have
\[
7^{2011}\equiv 7^3\equiv 43 \pmod{100}.
\]

The last two digits are 43, so the tens digit is 4.

ANSWER 8: D

Problem 9:  
The average of 10 different positive whole numbers is 10, so their sum is
\[
10\cdot 10=100.
\]

To make one number as large as possible, make the other 9 numbers as small as possible while still being different positive whole numbers. The smallest 9 distinct positive whole numbers are
\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is
\[
1+2+\cdots+9=45.
\]

So the largest number can be
\[
100-45=55.
\]

This is possible: \(1,2,3,4,5,6,7,8,9,55\) are all different positive whole numbers.

ANSWER 9: C

Problem 10:  
First count total student-days worked:

- Allen: \(7\cdot 3=21\) student-days
- Balboa: \(4\cdot 5=20\) student-days
- Carver: \(5\cdot 9=45\) student-days

Total:
\[
21+20+45=86
\]
student-days.

The total payment is \$774, so the daily rate is
\[
774\div 86=9
\]
dollars per student-day.

Balboa students worked 20 student-days, so they earned
\[
20\cdot 9=180
\]
dollars.

ANSWER 10: C

# Problem 1:
We need a statement always true about o² + no, where o is odd.

Factor: o² + no = o(o + n).

Since o is odd, the parity of the product is the parity of (o + n).
- If n is even, o + n is odd → product = odd·odd = **odd**.
- If n is odd, o + n is even → product = odd·even = **even**.

So the expression is odd exactly when n is even.

Ruling out others: A fails (n odd gives even), B fails (n even gives odd), C says "even only if n is even" — but it's even when n is *odd*, false. D says "odd only if n is odd" — false, it's odd when n is even. E: "odd only if n is even" — true.

ANSWER 1: E

# Problem 2:
The redemption days are 0, 10, 20, 30, 40, 50 days after the first.

Reduce mod 7 (days of week shift): 0, 3, 6, 2, 5, 1.

These residues are {0, 1, 2, 3, 5, 6} — every residue except **4**. So the only day of the week never hit is the day 4 days after the starting day. For no coupon to land on Sunday, Sunday must be that missed day:

start + 4 ≡ Sunday.

Counting back 4 days from Sunday: Saturday, Friday, Thursday, **Wednesday**.

Check: Wed → +10 Sat → +20 Tue → +30 Fri → +40 Mon → +50 Thu. No Sundays. ✓

ANSWER 2: C

# Problem 3:
The product is (3/2)·(4/3)·(5/4)·(6/5)·…·(a/b), where each numerator is one more than its denominator, so b = a − 1.

Telescoping: everything cancels except the last numerator and first denominator:
a/2 = 9 → a = 18, so b = 17.

a + b = 18 + 17 = 35.

(Choice E, 37, would come from mistakenly taking a = 19.)

ANSWER 3: D

# Problem 4:
Total value of all 60 stamps: 20(5 + 10 + 25) = 20·40 = 800 cents.

We need 710 cents, so we must *remove* stamps worth exactly 800 − 710 = 90 cents, using as **few** stamps as possible (to keep the most stamps).

Can 90 be made with 4 or fewer stamps? Max with 4 stamps is 100; possible 4-stamp totals near 90: 100, 85, 80, 75, 70… none equal 90. Three or fewer is at most 75. So 4 or fewer is impossible.

With 5 stamps: 25 + 25 + 25 + 10 + 5 = 90 ✓

So remove 5 stamps: 60 − 5 = 55 stamps.

Check: 17(25) + 19(10) + 19(5) = 425 + 190 + 95 = 710 cents, using 17 + 19 + 19 = 55 stamps.

ANSWER 4: E

# Problem 5:
CDCD as a number = CD·100 + CD = CD·101.

So ABA × CD = 101 × CD, and since CD ≠ 0, we get ABA = 101.

Thus A = 1, B = 0 (and indeed A, B, C, D distinct works, e.g. 101 × 23 = 2323).

A + B = 1 + 0 = 1.

ANSWER 5: A

# Problem 6:
Jonas total = 40, and 20 of those were girls, so boys from Jonas = 40 − 20 = 20.

Total boys = 52, so boys from Clay = 52 − 20 = 32.

(Check: girls from Clay = 48 − 20 = 28; 32 + 28 = 60 = Clay total ✓.)

Choice C (40) is the Jonas total, and E (52) is all boys — traps.

ANSWER 6: B

# Problem 7:
Pair each term: 1901 − 101 = 1800, 1902 − 102 = 1800, …, each pair differs by 1800.

Number of terms: 1993 − 1901 + 1 = 93.

Total = 93 × 1800 = 167,400.

ANSWER 7: A

# Problem 8:
Look at 7ⁿ mod 100:
- 7¹ = 07
- 7² = 49
- 7³ = 343 → 43
- 7⁴ = 2401 → 01
- then the pattern repeats with period 4.

2011 ÷ 4 = 502 remainder 3, so 7²⁰¹¹ ≡ 7³ ≡ 43 (mod 100).

Last two digits are 43, so the tens digit is 4.

ANSWER 8: D

# Problem 9:
Sum of the 10 numbers = 10 × 10 = 100.

To make one number as large as possible, make the other nine as small as possible while distinct and positive: 1 + 2 + … + 9 = 45.

Largest = 100 − 45 = 55.

(90 or 91 would require repeats or a zero, which isn't allowed.)

ANSWER 9: C

# Problem 10:
Total student-days:
- Allen: 7 × 3 = 21
- Balboa: 4 × 5 = 20
- Carver: 5 × 9 = 45

Total = 86 student-days.

Daily wage = 774 ÷ 86 = $9.00 per student-day.

Balboa earnings = 20 × 9 = $180.00.

(A is one day's pay for one student; E is Carver's total.)

ANSWER 10: C

Problem 1:

We need determine the parity of

\[
o^2+no.
\]

Factor out \(o\):

\[
o^2+no=o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on \(o+n\).

- If \(n\) is even, then odd \(+\) even \(=\) odd, so \(o+n\) is odd. Thus odd \(\cdot\) odd is odd.
- If \(n\) is odd, then odd \(+\) odd \(=\) even, so the product is even.

Therefore, the expression is odd only when \(n\) is even.

- A is wrong because the expression is not always odd.
- B is wrong because it is odd when \(n\) is even.
- C reverses the condition for being even.
- D reverses the condition for being odd.

ANSWER 1: E

Problem 2:

Isabella uses a coupon every 10 days. Since

\[
10\equiv 3\pmod 7,
\]

each coupon date moves forward by 3 days of the week.

If the first coupon is redeemed on day \(d\), the six redemption days are:

\[
d,\ d+3,\ d+6,\ d+2,\ d+5,\ d+1.
\]

These are six different days of the week. The only day missing is

\[
d+4.
\]

Since none of the six coupon dates is Sunday, the missing day must be Sunday. Thus,

\[
d+4=\text{Sunday}.
\]

Four days before Sunday is Wednesday.

Indeed, beginning on Wednesday gives:

\[
\text{Wednesday, Saturday, Tuesday, Friday, Monday, Thursday}.
\]

Sunday is the only missing day.

ANSWER 2: C

Problem 3:

The product is

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9.
\]

Almost every numerator cancels with a denominator:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab
=\frac a2.
\]

So

\[
\frac a2=9.
\]

Therefore,

\[
a=18.
\]

The final fraction must be

\[
\frac{18}{17},
\]

so \(b=17\). Therefore,

\[
a+b=18+17=35.
\]

Choices 11, 13, and 17 are too small, while 37 would correspond to a different ending fraction.

ANSWER 3: D

Problem 4:

Let

- \(x\) = number of 5-cent stamps,
- \(y\) = number of 10-cent stamps,
- \(z\) = number of 25-cent stamps.

In units of 5 cents, the postage total is

\[
\frac{710}{5}=142.
\]

Thus,

\[
x+2y+5z=142.
\]

We want to maximize

\[
x+y+z.
\]

Since there are at most 20 five-cent stamps, \(x\leq 20\). Therefore,

\[
2y+5z=142-x\geq 142-20=122.
\]

Since \(y\leq20\), if \(z\leq16\), then

\[
2y+5z\leq 2(20)+5(16)=40+80=120,
\]

which is not enough. So \(z\geq17\).

If \(z=17\), then

\[
2y+5(17)\geq122,
\]

so

\[
2y+85\geq122,
\]

\[
2y\geq37,
\]

and hence \(y\geq19\).

Now use

\[
x+y+z=142-y-4z.
\]

Since \(y\geq19\) and \(z\geq17\),

\[
x+y+z\leq142-19-4(17)
=142-19-68
=55.
\]

This maximum is attainable with

\[
x=19,\qquad y=19,\qquad z=17.
\]

Check:

\[
19(5)+19(10)+17(25)=95+190+425=710.
\]

So 55 stamps can be used. Choices 45, 46, 51, and 54 are possible totals to consider, but 55 is the greatest possible.

ANSWER 4: E

Problem 5:

We are told

\[
ABA\times CD=CDCD.
\]

The number \(ABA\) is

\[
100A+10B+A=101A+10B.
\]

The number \(CDCD\) is

\[
1000C+100D+10C+D.
\]

Group the \(C\)'s and \(D\)'s:

\[
CDCD=1010C+101D=101(10C+D)=101\cdot CD.
\]

Thus the equation becomes

\[
ABA\cdot CD=101\cdot CD.
\]

Since \(CD\) is a two-digit number and is not zero, divide both sides by \(CD\):

\[
ABA=101.
\]

Therefore,

\[
A=1,\qquad B=0.
\]

So

\[
A+B=1+0=1.
\]

For example, \(101\times23=2323\), and the digits \(1,0,2,3\) are all different.

ANSWER 5: A

Problem 6:

There are 40 students from Jonas Middle School, and 20 of those are girls.

Therefore the number of boys from Jonas is

\[
40-20=20.
\]

There are 52 boys altogether. Therefore the number of boys from Clay Middle School is

\[
52-20=32.
\]

The other choices do not correctly account for the 20 Jonas girls and the total of 52 boys.

ANSWER 6: B

Problem 7:

There are

\[
1993-1901+1=93
\]

terms in the first sum.

There are also

\[
193-101+1=93
\]

terms in the second sum.

Pair corresponding terms:

\[
1901-101=1800,
\]

\[
1902-102=1800,
\]

and so on, through

\[
1993-193=1800.
\]

Thus there are 93 differences of 1800:

\[
93\cdot1800=167400.
\]

Therefore the value is

\[
167,\!400.
\]

ANSWER 7: A

Problem 8:

We need the tens digit of

\[
7^{2011}.
\]

It is enough to examine powers of 7 modulo 100:

\[
7^1=7,
\]

\[
7^2=49,
\]

\[
7^3=343,
\]

whose last two digits are 43, and

\[
7^4=2401,
\]

whose last two digits are 01.

Thus the last two digits repeat every 4 powers because

\[
7^4\equiv1\pmod{100}.
\]

Now,

\[
2011\div4
\]

leaves remainder 3, since

\[
2011=4(502)+3.
\]

Therefore,

\[
7^{2011}
\]

has the same last two digits as

\[
7^3=343,
\]

which are 43. The tens digit is 4.

ANSWER 8: D

Problem 9:

The average of 10 different positive whole numbers is 10, so their total sum is

\[
10\cdot10=100.
\]

To make one number as large as possible, make the other nine numbers as small as possible.

The nine smallest different positive whole numbers are

\[
1,2,3,4,5,6,7,8,9.
\]

Their sum is

\[
1+2+3+4+5+6+7+8+9=45.
\]

So the largest possible tenth number is

\[
100-45=55.
\]

This works because the ten numbers can be

\[
1,2,3,4,5,6,7,8,9,55,
\]

which are all different and have total 100.

A value such as 90 or 91 would leave too small a total for nine distinct positive whole numbers.

ANSWER 9: C

Problem 10:

Compute the total number of student-days worked.

Allen students worked:

\[
7\cdot3=21
\]

student-days.

Balboa students worked:

\[
4\cdot5=20
\]

student-days.

Carver students worked:

\[
5\cdot9=45
\]

student-days.

So the total number of student-days is

\[
21+20+45=86.
\]

The total pay was \$774, so the pay per student per day was

\[
\frac{774}{86}=9.
\]

Balboa students worked 20 student-days, so they earned

\[
20\cdot9=180.
\]

Thus the Balboa students earned \$180.00 altogether.

ANSWER 10: C