🔬 Model Lab

Problem 1:
We are asked for the value of $(1-1/2)(1-1/3)\cdots (1-1/10)$.

Write each factor as a single fraction:
$1-1/2 = 1/2$, $1-1/3 = 2/3$, $1-1/4 = 3/4$, ..., $1-1/10 = 9/10$.

So the product is:
$$\frac12 \cdot \frac23 \cdot \frac34 \cdots \frac89 \cdot \frac9{10}.$$

This telescopes: the numerator 2 cancels denominator 2, numerator 3 cancels denominator 3, etc. All cancel except the first denominator 2? Let's see:
$$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10} = \frac{1}{10}.$$
Formally numerator $=1\cdot2\cdots9$, denominator $=2\cdot3\cdots10$, ratio $=1/10$.

Choices B $1/9$, C $1/2$, D $10/11$, E $11/2$ do not result from cancellation.
ANSWER 1: A

Problem 2:
We need the sum of the distinct prime divisors of 2016.

Factor 2016. It is even:
$2016 = 2\times1008 = 2\times2\times504 = 2^2\times504$.
$504 = 2\times252 = 2^2\times252$? Continue systematically:
$2016 /2 =1008$, $/2=504$, $/2=252$, $/2=126$, $/2=63$.
So $2016 = 2^5 \times 63$.
$63 = 7\times9 = 3^2\times7$.

Hence $2016 = 2^5 \cdot 3^2 \cdot 7^1$.
The distinct primes occurring are $2,3,7$.

Their sum is $2+3+7 = 12$.

This rules out A 9, C 16, D 49, E 63 which would come from missing a prime or including powers.
ANSWER 2: B

Problem 3:
In 1998 population is 200, triples every 25 years. When is it about 6000?

Ratio wanted: $6000/200 = 30$. So we need about 30 times growth.

Each 25 years multiply by 3:
After 25 yr: $200\times3=600$.
After 50 yr: $600\times3=1800$.
After 75 yr: $1800\times3=5400$.

5400 is about 6000. So about 3 triplings, i.e. about $3\times25=75$ years after 1998.

$1998+75 = 2073$, closest to 2075 among choices.

2050 is only ~2 triplings, 2100+ is too many.
ANSWER 3: B

Problem 4:
Some $n$ people asked, exactly $74\%$ said yes. Find smallest possible $n$.

Let $k$ be number saying yes. Then $k/n = 74\% = 74/100 = 37/50$.

So $k = 37n/50$ must be an integer. Since 37 and 50 are coprime (gcd=1), $n$ must be a multiple of 50 for $37n/50$ to be integer.

Smallest positive multiple is $n=50$, giving $k=37$.

$n=10,20,25$ give $7.4,14.8,18.5$ yes-votes, impossible. $n=100$ works but is not fewest.
ANSWER 4: D

Problem 5:
True product is $0.075 \times 2.56$. Entering $75 \times 256$ gave $19200$. What is correct with decimals?

$0.075$ has 3 decimal places: $0.075 = 75/1000$.
$2.56$ has 2 decimal places: $2.56 = 256/100$.

So product $= 75\times256 / (1000\times100) = 19200 / 100000$.

$19200/100000 = 0.192$.

In general move decimal 3+2=5 places left. .0192 would be 6 places, 1.92 would be 4 places, etc.
ANSWER 5: B

Problem 6:
Speech must be $1/2$ hour to $3/4$ hour at 150 words per minute. Which word count fits?

Convert hours to minutes: $1/2$ hr $=30$ min, $3/4$ hr $=45$ min.

At 150 wpm:
minimum words $=30\times150 = 4500$.
maximum words $=45\times150 = 6750$.

We need word count between 4500 and 6750 inclusive.

A 2250 is 15 min, B 3000 is 20 min, C 4200 is 28 min, D 4350 is 29 min — all too short. Only E 5650 satisfies $4500 \le 5650 \le 6750$.
ANSWER 6: E

Problem 7:
2 miles each day for 3 days. Speeds: Monday 5 mph, Wednesday 3 mph, Friday 4 mph. Compare total time to time if all 3 days at 4 mph. How many minutes saved?

Time = distance / speed.

Actual:
Monday: $2/5$ hr $=0.4$ hr $=0.4\times60=24$ min.
Wednesday: $2/3$ hr $= (2/3)\times60 =40$ min.
Friday: $2/4=1/2$ hr $=30$ min.
Total $=24+40+30=94$ min.

If always 4 mph: each day $30$ min, total $3\times30=90$ min.

Saving $=94-90=4$ minutes.
ANSWER 7: D

Problem 8:
Which integer cannot be sum of four consecutive odd integers?

Let the four be $n, n+2, n+4, n+6$ with $n$ odd.
Sum $=4n+12 =4(n+3)$.

Since $n$ is odd, $n+3$ is even, say $n+3=2m$. Then sum $=4\cdot2m=8m$, a multiple of 8. So any such sum must be divisible by 8.

Test:
16: $16/8=2$ possible; indeed $1+3+5+7=16$.
40: $40/8=5$ possible; indeed $7+9+11+13=40$.
72: $72/8=9$ possible; e.g. $15+17+19+21=72$.
200: $200/8=25$ possible; e.g. $47+49+51+53=200$.
100: $100/8=12.5$ not integer, so impossible.

Thus 100 cannot be written that way.
ANSWER 8: D

Problem 9:
Blake averaged 78 over four tests. Jenny vs Blake: +10, -10, +20, +20. By how much does Jenny's average exceed Blake's?

Only differences matter. Total points difference over 4 tests:
$+10-10+20+20 = +40$.

Average is total/4, so Jenny's average is $40/4=10$ points higher.

Blake's 78 is irrelevant to the difference. Choices 15,20,25,40 confuse total with average.
ANSWER 9: A

Problem 10:
Compute $(.2)^3/(.02)^2$.

$(.2)^3 = 0.2\times0.2\times0.2 = 0.04\times0.2 = 0.008$.
$(.02)^2 = 0.02\times0.02 = 0.0004$.

Ratio $=0.008/0.0004 = 8/0.4 = 80/4 =20$.

Alternatively with powers of 10: $(2\times10^{-1})^3=8\times10^{-3}$, $(2\times10^{-2})^2=4\times10^{-4}$, ratio $=2\times10^{1}=20$.

So not .2, 2, 10, or 15.
ANSWER 10: E

Problem 1:
We are asked to evaluate the product of the 9 factors:
$$\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{10}\right)$$

First, simplify each factor:
$$1 - \frac{1}{2} = \frac{1}{2}$$
$$1 - \frac{1}{3} = \frac{2}{3}$$
$$1 - \frac{1}{4} = \frac{3}{4}$$
$$\vdots$$
$$1 - \frac{1}{10} = \frac{9}{10}$$

Now, multiply the fractions together:
$$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{9}{10}$$

Notice that this is a telescoping product where the denominator of each fraction cancels with the numerator of the following fraction:
$$\frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \cdots \times \frac{\cancel{9}}{10} = \frac{1}{10}$$

ANSWER 1: A

---

Problem 2:
We are asked to find the sum of the distinct prime integer divisors of $2016$.

First, find the prime factorization of $2016$:
- $2016 = 2 \times 1008$
- $1008 = 2 \times 504$
- $504 = 2 \times 252$
- $252 = 2 \times 126$
- $126 = 2 \times 63$
- $63 = 9 \times 7 = 3^2 \times 7$

Thus, the prime factorization of $2016$ is:
$$2016 = 2^5 \times 3^2 \times 7$$

The distinct prime divisors are $2$, $3$, and $7$. 
Their sum is:
$$2 + 3 + 7 = 12$$

ANSWER 2: B

---

Problem 3:
We are given that the population of Nisos Isles is $200$ in the year $1998$, and it triples every $25$ years. We want to find the year in which the population reaches approximately $6000$.

Let $n$ be the number of $25$-year periods that pass after $1998$. The population can be represented as:
$$P = 200 \times 3^n$$

We want $P \approx 6000$:
$$200 \times 3^n \approx 6000 \implies 3^n \approx \frac{6000}{200} = 30$$

Let's test powers of $3$:
- For $n = 1$: $3^1 = 3 \implies P = 600$
- For $n = 2$: $3^2 = 9 \implies P = 1800$
- For $n = 3$: $3^3 = 27 \implies P = 5400$
- For $n = 4$: $3^4 = 81 \implies P = 16200$

$5400$ is very close to $6000$, so about $n = 3$ periods of $25$ years are needed:
$$3 \times 25 = 75\text{ years}$$

Adding $75$ years to $1998$:
$$1998 + 75 = 2073 \approx 2075$$

ANSWER 3: B

---

Problem 4:
We are given that exactly $74\%$ of the people surveyed answered "yes." We need to find the minimum number of people who could have been surveyed.

Let $N$ be the total number of people surveyed. The number of people who answered "yes" is:
$$74\% \times N = \frac{74}{100} N = \frac{37}{50} N$$

Since the number of people who answered "yes" must be a whole number, $\frac{37}{50} N$ must be an integer. 
Since $37$ and $50$ share no common factors other than $1$ (they are coprime), $N$ must be a multiple of $50$. 

The smallest positive integer multiple of $50$ is $50$. 
With $N = 50$, exactly $\frac{37}{50} \times 50 = 37$ people answered "yes."

ANSWER 4: D

---

Problem 5:
Betty is multiplying $0.075$ by $2.56$.
She calculated $75 \times 256 = 19200$.

Count the total number of digits to the right of the decimal point in the original numbers:
- $0.075$ has $3$ decimal places.
- $2.56$ has $2$ decimal places.

The product must therefore have $3 + 2 = 5$ decimal places:
$$19200 \times 10^{-5} = 0.19200 = 0.192$$

ANSWER 5: B

---

Problem 6:
Julie's speech must last between one-half hour ($30$ minutes) and three-quarters of an hour ($45$ minutes). Her speaking rate is $150$ words per minute.

Calculate the minimum and maximum acceptable word counts:
- Minimum words: $30\text{ minutes} \times 150\text{ words/minute} = 4500\text{ words}$
- Maximum words: $45\text{ minutes} \times 150\text{ words/minute} = 6750\text{ words}$

An appropriate word count must lie in the range $[4500, 6750]$. 
Looking at the choices:
- A. $2250$ (too low)
- B. $3000$ (too low)
- C. $4200$ (too low)
- D. $4350$ (too low)
- E. $5650$ is between $4500$ and $6750$ ($\frac{5650}{150} \approx 37.7$ minutes).

ANSWER 6: E

---

Problem 7:
Ted's grandfather covered $2$ miles on each of the $3$ days. 
First, find the time he actually spent each day using $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
- **Monday:** $\frac{2\text{ miles}}{5\text{ mph}} = \frac{2}{5}\text{ hour} = \frac{2}{5} \times 60 = 24\text{ minutes}$.
- **Wednesday:** $\frac{2\text{ miles}}{3\text{ mph}} = \frac{2}{3}\text{ hour} = \frac{2}{3} \times 60 = 40\text{ minutes}$.
- **Friday:** $\frac{2\text{ miles}}{4\text{ mph}} = \frac{2}{4}\text{ hour} = \frac{1}{2} \times 60 = 30\text{ minutes}$.

Total actual time spent:
$$24 + 40 + 30 = 94\text{ minutes}$$

If he had walked at $4\text{ mph}$ every day, the time spent each day would be $\frac{2}{4}\text{ hour} = 30\text{ minutes}$, so for all $3$ days:
$$3 \times 30 = 90\text{ minutes}$$

He would have spent:
$$94 - 90 = 4\text{ minutes less}$$

ANSWER 7: D

---

Problem 8:
Let four consecutive odd integers be represented as:
$$n, \quad n+2, \quad n+4, \quad n+6$$
where $n$ is an odd integer.

Their sum is:
$$S = n + (n+2) + (n+4) + (n+6) = 4n + 12 = 4(n+3)$$

Since $n$ is an odd integer, $n+3$ must be an even integer, which can be written as $2k$ for some integer $k$.
Substituting this in gives:
$$S = 4(2k) = 8k$$

Therefore, the sum of four consecutive odd integers must be a multiple of $8$.
Let's check the given options:
- A. $16 = 8 \times 2$ ($1 + 3 + 5 + 7 = 16$)
- B. $40 = 8 \times 5$ ($7 + 9 + 11 + 13 = 40$)
- C. $72 = 8 \times 9$ ($15 + 17 + 19 + 21 = 72$)
- D. $100$: not divisible by $8$ ($100 / 8 = 12.5$)
- E. $200 = 8 \times 25$ ($47 + 49 + 51 + 53 = 200$)

Hence, $100$ cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

---

Problem 9:
Let Blake's scores on the four tests be $B_1, B_2, B_3, B_4$.
Jenny's scores are:
- First test: $B_1 + 10$
- Second test: $B_2 - 10$
- Third test: $B_3 + 20$
- Fourth test: $B_4 + 20$

The sum of Jenny's scores minus the sum of Blake's scores is:
$$(B_1 + 10 + B_2 - 10 + B_3 + 20 + B_4 + 20) - (B_1 + B_2 + B_3 + B_4) = 10 - 10 + 20 + 20 = 40\text{ points}$$

Since there are $4$ tests, the difference between Jenny's average and Blake's aver

# Problem 1:
We need the product of the 9 factors (1 − 1/n) for n = 2, 3, 4, …, 10.

Rewrite each factor: 1 − 1/n = (n−1)/n. So the product becomes

(1/2)(2/3)(3/4)(4/5)(5/6)(6/7)(7/8)(8/9)(9/10)

This **telescopes**: each numerator cancels the denominator before it (2 cancels 2, 3 cancels 3, …, 9 cancels 9), leaving only the first numerator and the last denominator:

Product = 1/10

Ruling out traps: every factor is less than 1, so the product must be *smaller* than any single factor — this eliminates D (10/11) and E (11/2) immediately. Choice C (1/2) would be the answer if only the first factor were used, and B (1/9) would result from miscounting the factors (stopping at 1 − 1/9). Since n runs from 2 to 10 inclusive, there are exactly 9 factors and the answer is 1/10.

ANSWER 1: A

---

# Problem 2:
Factor 2016 into primes by repeated division:

2016 = 2 × 1008 = 2² × 504 = 2³ × 252 = 2⁴ × 126 = 2⁵ × 63, and 63 = 3² × 7.

So 2016 = 2⁵ × 3² × 7.

The **distinct** prime divisors are just 2, 3, and 7 (the exponents don't matter — we count each prime once).

Sum = 2 + 3 + 7 = **12**

Ruling out traps: A (9) would come from forgetting the prime 3; E (63) is the odd part 3² × 7 of 2016, not a sum of primes; adding primes *with repetition* gives 2+2+2+2+2+3+3+7 = 23, which isn't even a choice. The correct distinct-prime sum is 12.

ANSWER 2: B

---

# Problem 3:
Starting at 200 people in 1998, the population triples every 25 years:

- 1998: 200
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16 200

We want about 6000. Check how many triplings are needed: 6000/200 = 30, and 3³ = 27 ≈ 30 (while 3⁴ = 81 is way too big). So about **three triplings ≈ 75 years** after 1998, i.e., around 2073.

Comparing to the choices: at 2073 the population is 5400 (off by only 600), while at 2098 it is 16 200 (off by over 10 000). The closest estimate offered is **2075**.

Ruling out traps: 2050 and 2100 bracket the answer but are farther from the true crossing point than 2075; 2125 and 2150 correspond to populations of tens of thousands.

ANSWER 3: B

---

# Problem 4:
If N people were asked and exactly 74% said yes, then 0.74N must be a whole number of people:

0.74N = (74/100)N = (37/50)N

Since 37 is prime and shares no factor with 50, the fraction 37/50 is fully reduced. For (37/50)N to be an integer, **50 must divide N**. The smallest such N is 50.

Check: with 50 people, 0.74 × 50 = 37 people said yes — exactly 74%. ✓

Ruling out the smaller choices: 10 people → 7.4 yes-votes; 20 → 14.8; 25 → 18.5 — none of these is a whole number of people, so each is impossible. Hence 50 is the fewest.

ANSWER 4: D

---

# Problem 5:
First, confirm the calculator display: entering 075 × 256 (no decimal points) gives 75 × 256 = 19 200. ✓ That matches.

Now place the decimal correctly. The rule for multiplying decimals: the product has as many decimal places as the two factors combined.

- 0.075 has **3** decimal places
- 2.56 has **2** decimal places
- Product has 3 + 2 = **5** decimal places

Start from 19 200 and move the decimal point 5 places to the left:

19 200 → 0.19200 = **0.192**

Ruling out traps: 192 (E) and 19.2 (D) come from shifting the point only 2 or 3 places; 0.0192 (A) from shifting 6 places. A quick sanity check: 0.075 × 2.56 should be a bit less than 0.1 × 2.56 = 0.256, and 0.192 fits perfectly.

ANSWER 5: B

---

# Problem 6:
Convert the time limits to minutes, then to words:

- Minimum time: 1/2 hour = 30 minutes → 150 × 30 = **4500 words**
- Maximum time: 3/4 hour = 45 minutes → 150 × 45 = **6750 words**

So the speech needs between 4500 and 6750 words.

Check each choice: 2250 words = 15 min (too short); 3000 = 20 min (too short); 4200 = 28 min (too short); 4350 = 29 min (just under a half hour — too short); 5650 = 5650/150 ≈ 37.7 minutes, which lies nicely between 30 and 45 minutes. ✓

Only 5650 works.

ANSWER 6: E

---

# Problem 7:
Use time = distance ÷ speed, and convert to minutes (1 hour = 60 min):

- Monday: 2 miles at 5 mph → 2/5 hour = 24 minutes
- Wednesday: 2 miles at 3 mph → 2/3 hour = 40 minutes
- Friday: 2 miles at 4 mph → 2/4 = 1/2 hour = 30 minutes

Actual total: 24 + 40 + 30 = **94 minutes**.

If he had always gone 4 mph: total distance 6 miles at 4 mph → 6/4 = 1.5 hours = 90 minutes.

Difference: 94 − 90 = **4 minutes less**.

Ruling out traps: note Monday would actually take *longer* at 4 mph (30 vs. 24 min), so you can't just assume each day saves time — Wednesday's big savings (40 → 30) must be combined with Monday's loss (24 → 30). Net saving = 10 − 6 = 4 minutes, matching the totals.

ANSWER 7: D

---

# Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6, where a is odd. Their sum is

a + (a+2) + (a+4) + (a+6) = 4a + 12 = 4(a + 3).

Since a is odd, a + 3 is **even**, so the sum equals 4 × (an even number), i.e., every such sum is **divisible by 8**.

Test divisibility by 8 for each choice:

- 16 ÷ 8 = 2 ✓ (indeed 1 + 3 + 5 + 7 = 16)
- 40 ÷ 8 = 5 ✓ (7 + 9 + 11 + 13 = 40)
- 72 ÷ 8 = 9 ✓ (15 + 17 + 19 + 21 = 72)
- 100 ÷ 8 = 12.5 ✗ — **not divisible by 8**
- 200 ÷ 8 = 25 ✓ (47 + 49 + 51 + 53 = 200)

So 100 cannot be written as such a sum. (Check directly: 100 = 4(a+3) forces a + 3 = 25, so a = 22 — but 22 is even, a contradiction.)

ANSWER 8: D

---

# Problem 9:
Blake's total over four tests: 4 × 78 = 312 points.

Jenny's scores compared to Blake's, test by test: +10, −10, +20, +20.

Total difference: +10 − 10 + 20 + 20 = **+40 points**.

Jenny's total: 312 + 40 = 352, so her average is 352 ÷ 4 = 88.

Jenny's average exceeds Blake's by 88 − 78 = **10 points**.

(Equivalently: the *average* difference is the total difference divided by the number of tests: 40 ÷ 4 = 10.)

Ruling out traps: E (40) is the total point difference — it forgets to divide by the 4 tests; C (20) might come from averaging only the +20's and ignoring the +10/−10 pair; B and D have no valid basis.

ANSWER 9: A

---

# Problem 10:
Compute the numerator and denominator separately:

- (0.2)³ = 

Problem 1:  
The product is  
\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots\left(1-\frac1{10}\right).
\]
Rewrite each factor:
\[
\frac12\cdot \frac23\cdot \frac34\cdot \frac45\cdots \frac9{10}.
\]
Everything cancels except the first numerator and the last denominator:
\[
\frac1{10}.
\]
So the answer is A.

ANSWER 1: A

Problem 2:  
We need the sum of the distinct prime factors of 2016. Factor:
\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot 252=2^4\cdot 126=2^5\cdot 63.
\]
Also,
\[
63=3^2\cdot 7.
\]
Thus
\[
2016=2^5\cdot 3^2\cdot 7.
\]
The distinct prime divisors are 2, 3, and 7. Their sum is
\[
2+3+7=12.
\]

ANSWER 2: B

Problem 3:  
The population triples every 25 years. Starting with 200 in 1998, after \(t\) years the population is approximately
\[
200\cdot 3^{t/25}.
\]
We want this to be about 6000:
\[
200\cdot 3^{t/25}=6000.
\]
Divide by 200:
\[
3^{t/25}=30.
\]
Now,
\[
3^3=27
\]
and
\[
3^4=81.
\]
So \(t/25\) is a little more than 3, meaning \(t\) is a little more than 75 years.  
\[
1998+75=2073,
\]
so the closest choice is 2075.

ANSWER 3: B

Problem 4:  
Exactly 74% answered “yes.” Write 74% as a fraction in lowest terms:
\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]
For the number of “yes” answers to be an integer, the total number of people asked must be a multiple of 50. The smallest such number among the choices is 50.  
Check: \(\frac{37}{50}=74\%\).

ANSWER 4: D

Problem 5:  
Betty forgot the decimal points, so she effectively multiplied
\[
75\cdot 256=19200.
\]
Now restore the decimal places.  
\(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places, so the correct product must have
\[
3+2=5
\]
decimal places. Therefore,
\[
0.075\cdot 2.56=\frac{19200}{100000}=0.192.
\]

ANSWER 5: B

Problem 6:  
The speech must last between one-half hour and three-quarters of an hour, which is between
\[
30\text{ minutes}
\]
and
\[
45\text{ minutes}.
\]
At 150 words per minute, the acceptable word counts are between
\[
150\cdot 30=4500
\]
and
\[
150\cdot 45=6750.
\]
Among the choices, only 5650 lies between 4500 and 6750.

ANSWER 6: E

Problem 7:  
Each day he went 2 miles.

Monday: speed 5 mph, time
\[
\frac{2}{5}\text{ hour}=24\text{ minutes}.
\]
Wednesday: speed 3 mph, time
\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]
Friday: speed 4 mph, time
\[
\frac{2}{4}\text{ hour}=30\text{ minutes}.
\]
Total time:
\[
24+40+30=94\text{ minutes}.
\]
If he had walked all 6 miles at 4 mph, the time would be
\[
\frac{6}{4}\text{ hour}=1.5\text{ hours}=90\text{ minutes}.
\]
The difference is
\[
94-90=4\text{ minutes}.
\]

ANSWER 7: D

Problem 8:  
Let four consecutive odd integers be
\[
2k+1,\;2k+3,\;2k+5,\;2k+7.
\]
Their sum is
\[
(2k+1)+(2k+3)+(2k+5)+(2k+7)=8k+16=8(k+2).
\]
So any sum of four consecutive odd integers must be divisible by 8.

Check the choices:
- 16 is divisible by 8.
- 40 is divisible by 8.
- 72 is divisible by 8.
- 100 is not divisible by 8.
- 200 is divisible by 8.

Therefore, 100 cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

Problem 9:  
Blake’s average was 78 on four tests, so his total score was
\[
4\cdot 78=312.
\]
Jenny’s scores differed from Blake’s by
\[
+10,\;-10,\;+20,\;+20.
\]
The total difference is
\[
10-10+20+20=40.
\]
So Jenny’s total score was 40 points higher than Blake’s. Since there are 4 tests, her average was higher by
\[
\frac{40}{4}=10.
\]

ANSWER 9: A

Problem 10:  
Compute:
\[
(0.2)^3=0.008
\]
and
\[
(0.02)^2=0.0004.
\]
Thus
\[
\frac{(0.2)^3}{(0.02)^2}=\frac{0.008}{0.0004}=20.
\]

ANSWER 10: E

Problem 1:
Each factor simplifies: (1 − 1/2) = 1/2, (1 − 1/3) = 2/3, …, (1 − 1/10) = 9/10.

Product = (1/2)(2/3)(3/4)⋯(9/10).

Everything telescopes: the numerator of each fraction cancels the denominator of the previous one, leaving 1/10.

(Choice B, 1/9, would come from miscounting the last denominator; the product is definitely less than 1/2, ruling out C, D, E.)

ANSWER 1: A

Problem 2:
Factor 2016: 2016 = 2 · 1008 = 2·2·504 = 2·2·2·252 = 2^5 · 63 = 2^5 · 3^2 · 7.

Distinct primes: 2, 3, 7. Sum = 2 + 3 + 7 = 12.

(9 = 2+7 misses 3; 16 = 2·5+... etc.; 49 and 63 come from adding non‑prime factors or all factors with multiplicity.)

ANSWER 2: B

Problem 3:
Start: 1998 → 200 people. Tripling every 25 years:
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16200

The population reaches about 6000 shortly after 2073, so the best estimate among the choices is 2075.

(2050 gives ~1800, 2100 gives ~16000 — both far off.)

ANSWER 3: B

Problem 4:
We need 74% of the number of people to be a whole number: 0.74n = 37n/50 must be an integer.

Since 37/50 is in lowest terms (37 is prime), n must be a multiple of 50. The smallest is n = 50 (then 37 people said yes).

Check the smaller options: 74% of 10 = 7.4, of 20 = 14.8, of 25 = 18.5 — none are whole numbers.

ANSWER 4: D

Problem 5:
Ignoring decimals: 75 × 256 = 19200, matching the display.

Count decimal places: 0.075 has 3, 2.56 has 2, total 5 decimal places.

So 0.075 × 2.56 = 19200 with the decimal point moved 5 places left = 0.19200 = 0.192.

Sanity check: 0.075 ≈ 0.08 and 2.56 ≈ 2.5, product ≈ 0.2. So 0.192 is right (not 0.0192 or 1.92).

ANSWER 5: B

Problem 6:
Time allowed: 30 minutes to 45 minutes.

At 150 words per minute:
- Minimum words = 150 × 30 = 4500
- Maximum words = 150 × 45 = 6750

We need a count between 4500 and 6750. Only 5650 qualifies.

(2250 would take 15 min, 3000 → 20 min, 4200 → 28 min, 4350 → 29 min — all too short.)

ANSWER 6: E

Problem 7:
Time = distance ÷ speed, with 2 miles each day.

- Monday at 5 mph: 2/5 h = 24 min
- Wednesday at 3 mph: 2/3 h = 40 min
- Friday at 4 mph: 2/4 h = 30 min

Total actual = 24 + 40 + 30 = 94 minutes.

If always 4 mph: 30 min each day → 3 × 30 = 90 minutes.

Difference: 94 − 90 = 4 minutes less.

ANSWER 7: D

Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6 with a odd.

Sum = 4a + 12 = 4(a + 3). Since a is odd, a + 3 is even, say a + 3 = 2k, so the sum = 8k — a multiple of 8.

Check the choices for divisibility by 8:
- 16 = 8·2 ✓ (1+3+5+7)
- 40 = 8·5 ✓ (7+9+11+13)
- 72 = 8·9 ✓ (15+17+19+21)
- 100 ÷ 8 = 12.5 ✗
- 200 = 8·25 ✓ (47+49+51+53)

So 100 cannot be written that way.

ANSWER 8: D

Problem 9:
Jenny's total differs from Blake's total by:
+10 − 10 + 20 + 20 = +40 points.

Average difference = 40 ÷ 4 = 10 points.

(40 is the total difference, not the average — that's the trap in choice E. Blake's 78 average is not even needed, though Jenny averages 88, which is under 100, so it's consistent.)

ANSWER 9: A

Problem 10:
(0.2)³ = 0.008 and (0.02)² = 0.0004.

0.008 ÷ 0.0004 = 80/4 = 20.

(Alternatively: (2×10⁻¹)³ / (2×10⁻²)² = 8×10⁻³ / (4×10⁻⁴) = 2 × 10¹ = 20.)

ANSWER 10: E

Problem 1:

We need to compute

\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots \left(1-\frac1{10}\right).
\]

Rewrite each factor:

\[
1-\frac12=\frac12,\qquad
1-\frac13=\frac23,\qquad
1-\frac14=\frac34,
\]

and so on. Thus the product is

\[
\frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdots\frac9{10}.
\]

Nearly everything cancels:

\[
\frac{\cancel1}{2}\cdot\frac{\cancel2}{3}\cdot\frac{\cancel3}{4}\cdots
\frac{\cancel9}{10}
=\frac1{10}.
\]

Choice B, \(\frac19\), might seem plausible if one thinks there are 9 factors, but the cancellation leaves the first numerator \(1\) and final denominator \(10\).

\[
\boxed{\frac1{10}}
\]

ANSWER 1: A

Problem 2:

We need the sum of the distinct prime divisors of \(2016\).

Factor \(2016\):

\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot252=2^4\cdot126=2^5\cdot63.
\]

Since

\[
63=3^2\cdot7,
\]

we have

\[
2016=2^5\cdot3^2\cdot7.
\]

The distinct prime divisors are

\[
2,\ 3,\ 7.
\]

Their sum is

\[
2+3+7=12.
\]

Do not add repeated prime factors such as all five 2s and both 3s; the question asks for distinct primes only.

\[
\boxed{12}
\]

ANSWER 2: B

Problem 3:

The population starts at \(200\) in 1998 and triples every 25 years.

We want it to be about \(6000\). Compare:

\[
\frac{6000}{200}=30.
\]

So the population must grow by a factor of about \(30\).

Powers of \(3\) are:

\[
3^1=3,\qquad 3^2=9,\qquad 3^3=27,\qquad 3^4=81.
\]

A factor of \(27\) is close to \(30\), so we need about 3 periods of 25 years:

\[
3\cdot25=75\text{ years}.
\]

Therefore the estimated year is

\[
1998+75=2073.
\]

The nearest choice is \(2075\).

Indeed, after 75 years the estimated population is

\[
200\cdot 27=5400,
\]

which is reasonably close to \(6000\). Waiting 100 years would give \(200\cdot81=16200\), much too large.

\[
\boxed{2075}
\]

ANSWER 3: B

Problem 4:

Exactly \(74\%\) of the people answered yes.

Convert \(74\%\) to a fraction:

\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]

For the number of yes answers to be a whole number, the total number surveyed must be a multiple of \(50\).

The smallest possible total is therefore

\[
50.
\]

For example,

\[
74\%\text{ of }50=\frac{74}{100}\cdot 50=37,
\]

which is an integer number of people.

Smaller choices such as 10, 20, and 25 do not work, since \(74\%\) of them is not a whole number.

\[
\boxed{50}
\]

ANSWER 4: D

Problem 5:

Without decimal points, Betty entered

\[
75\times256=19200.
\]

Now place the decimal points correctly.

The number \(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places. Therefore the product must have

\[
3+2=5
\]

decimal places.

So move the decimal point in \(19200\) five places left:

\[
19200\longrightarrow 0.19200=0.192.
\]

Thus

\[
0.075\times2.56=0.192.
\]

Choice A, \(0.0192\), has one too many decimal places, while \(1.92\) and larger answers are too large because \(0.075\) is less than \(0.1\).

\[
\boxed{0.192}
\]

ANSWER 5: B

Problem 6:

Julie’s speech must last between one-half hour and three-quarters of an hour.

Convert these times to minutes:

\[
\frac12\text{ hour}=30\text{ minutes},
\qquad
\frac34\text{ hour}=45\text{ minutes}.
\]

At \(150\) words per minute, the minimum appropriate word count is

\[
30\cdot150=4500.
\]

The maximum appropriate word count is

\[
45\cdot150=6750.
\]

So her speech should contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) and \(3000\) are too few.
- \(4200\) and \(4350\) are still below \(4500\).
- \(5650\) is between \(4500\) and \(6750\).

Therefore the appropriate word count is

\[
\boxed{5650}.
\]

ANSWER 6: E

Problem 7:

Grandfather traveled 2 miles on each of 3 days.

Find his actual treadmill time.

On Monday, he jogged at 5 mph:

\[
\frac{2}{5}\text{ hour}=0.4\text{ hour}=24\text{ minutes}.
\]

On Wednesday, he walked at 3 mph:

\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]

On Friday, he walked at 4 mph:

\[
\frac24\text{ hour}=\frac12\text{ hour}=30\text{ minutes}.
\]

So his actual total time was

\[
24+40+30=94\text{ minutes}.
\]

If he had walked at 4 mph on all three days, each 2-mile trip would take 30 minutes:

\[
3\cdot30=90\text{ minutes}.
\]

Thus he would have spent

\[
94-90=4
\]

fewer minutes on the treadmill.

The 5-mph jogging day saves time, but the 3-mph walking day costs more time; together the actual time is 4 minutes longer than always walking at 4 mph.

\[
\boxed{4}
\]

ANSWER 7: D

Problem 8:

Let the four consecutive odd integers be

\[
n,\ n+2,\ n+4,\ n+6,
\]

where \(n\) is odd.

Their sum is

\[
n+(n+2)+(n+4)+(n+6)=4n+12.
\]

Factor:

\[
4n+12=4(n+3).
\]

Since \(n\) is odd, \(n+3\) is even. Therefore \(4(n+3)\) must be divisible by \(8\).

So a number can be written as the sum of four consecutive odd integers only if it is divisible by \(8\).

Check the choices:

\[
16\div8=2,
\quad
40\div8=5,
\quad
72\div8=9,
\quad
200\div8=25,
\]

so all of these are divisible by 8.

But

\[
100\div8=12.5,
\]

so \(100\) is not divisible by 8. Therefore it cannot be the sum of four consecutive odd integers.

For example, \(16=1+3+5+7\), showing that the condition is possible for the other type of choices.

\[
\boxed{100}
\]

ANSWER 8: D

Problem 9:

Blake’s average is not actually needed, because we can compare Jenny’s scores directly to Blake’s.

Jenny’s score changes relative to Blake’s are:

- first test: \(+10\),
- second test: \(-10\),
- third test: \(+20\),
- fourth test: \(+20\).

Her total score exceeds Blake’s by

\[
10-10+20+20=40.
\]

Since there are four tests, her average exceeds Blake’s by

\[
\frac{40}{4}=10.
\]

The \(40\) is the difference in total points, not the difference in average points.

\[
\boxed{10}
\]

ANSWER 9: A

Problem 10:

Compute

\[
\frac{(0.2)^3}{(0.02)^2}.
\]

First,

\[
(0.2)^3=0.2\cdot0.2\cdot0.2=0.008.
\]

Next,

\[
(0.02)^2=0.02\cdot0.02=0.0004.
\]

Therefore,

\[
\frac{0.008}{0.0004}=20.
\]

Another way is to use powers of 10:

\[
0.2=2\cdot10^{-1},\qquad 0.02=2\cdot10^{-2}.
\]

Then

\[
\fra

Problem 1:
We are asked for the value of $(1-1/2)(1-1/3)\cdots (1-1/10)$.

Write each factor as a single fraction:
$1-1/2 = 1/2$, $1-1/3 = 2/3$, $1-1/4 = 3/4$, ..., $1-1/10 = 9/10$.

So the product is:
$$\frac12 \cdot \frac23 \cdot \frac34 \cdots \frac89 \cdot \frac9{10}.$$

This telescopes: the numerator 2 cancels denominator 2, numerator 3 cancels denominator 3, etc. All cancel except the first denominator 2? Let's see:
$$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10} = \frac{1}{10}.$$
Formally numerator $=1\cdot2\cdots9$, denominator $=2\cdot3\cdots10$, ratio $=1/10$.

Choices B $1/9$, C $1/2$, D $10/11$, E $11/2$ do not result from cancellation.
ANSWER 1: A

Problem 2:
We need the sum of the distinct prime divisors of 2016.

Factor 2016. It is even:
$2016 = 2\times1008 = 2\times2\times504 = 2^2\times504$.
$504 = 2\times252 = 2^2\times252$? Continue systematically:
$2016 /2 =1008$, $/2=504$, $/2=252$, $/2=126$, $/2=63$.
So $2016 = 2^5 \times 63$.
$63 = 7\times9 = 3^2\times7$.

Hence $2016 = 2^5 \cdot 3^2 \cdot 7^1$.
The distinct primes occurring are $2,3,7$.

Their sum is $2+3+7 = 12$.

This rules out A 9, C 16, D 49, E 63 which would come from missing a prime or including powers.
ANSWER 2: B

Problem 3:
In 1998 population is 200, triples every 25 years. When is it about 6000?

Ratio wanted: $6000/200 = 30$. So we need about 30 times growth.

Each 25 years multiply by 3:
After 25 yr: $200\times3=600$.
After 50 yr: $600\times3=1800$.
After 75 yr: $1800\times3=5400$.

5400 is about 6000. So about 3 triplings, i.e. about $3\times25=75$ years after 1998.

$1998+75 = 2073$, closest to 2075 among choices.

2050 is only ~2 triplings, 2100+ is too many.
ANSWER 3: B

Problem 4:
Some $n$ people asked, exactly $74\%$ said yes. Find smallest possible $n$.

Let $k$ be number saying yes. Then $k/n = 74\% = 74/100 = 37/50$.

So $k = 37n/50$ must be an integer. Since 37 and 50 are coprime (gcd=1), $n$ must be a multiple of 50 for $37n/50$ to be integer.

Smallest positive multiple is $n=50$, giving $k=37$.

$n=10,20,25$ give $7.4,14.8,18.5$ yes-votes, impossible. $n=100$ works but is not fewest.
ANSWER 4: D

Problem 5:
True product is $0.075 \times 2.56$. Entering $75 \times 256$ gave $19200$. What is correct with decimals?

$0.075$ has 3 decimal places: $0.075 = 75/1000$.
$2.56$ has 2 decimal places: $2.56 = 256/100$.

So product $= 75\times256 / (1000\times100) = 19200 / 100000$.

$19200/100000 = 0.192$.

In general move decimal 3+2=5 places left. .0192 would be 6 places, 1.92 would be 4 places, etc.
ANSWER 5: B

Problem 6:
Speech must be $1/2$ hour to $3/4$ hour at 150 words per minute. Which word count fits?

Convert hours to minutes: $1/2$ hr $=30$ min, $3/4$ hr $=45$ min.

At 150 wpm:
minimum words $=30\times150 = 4500$.
maximum words $=45\times150 = 6750$.

We need word count between 4500 and 6750 inclusive.

A 2250 is 15 min, B 3000 is 20 min, C 4200 is 28 min, D 4350 is 29 min — all too short. Only E 5650 satisfies $4500 \le 5650 \le 6750$.
ANSWER 6: E

Problem 7:
2 miles each day for 3 days. Speeds: Monday 5 mph, Wednesday 3 mph, Friday 4 mph. Compare total time to time if all 3 days at 4 mph. How many minutes saved?

Time = distance / speed.

Actual:
Monday: $2/5$ hr $=0.4$ hr $=0.4\times60=24$ min.
Wednesday: $2/3$ hr $= (2/3)\times60 =40$ min.
Friday: $2/4=1/2$ hr $=30$ min.
Total $=24+40+30=94$ min.

If always 4 mph: each day $30$ min, total $3\times30=90$ min.

Saving $=94-90=4$ minutes.
ANSWER 7: D

Problem 8:
Which integer cannot be sum of four consecutive odd integers?

Let the four be $n, n+2, n+4, n+6$ with $n$ odd.
Sum $=4n+12 =4(n+3)$.

Since $n$ is odd, $n+3$ is even, say $n+3=2m$. Then sum $=4\cdot2m=8m$, a multiple of 8. So any such sum must be divisible by 8.

Test:
16: $16/8=2$ possible; indeed $1+3+5+7=16$.
40: $40/8=5$ possible; indeed $7+9+11+13=40$.
72: $72/8=9$ possible; e.g. $15+17+19+21=72$.
200: $200/8=25$ possible; e.g. $47+49+51+53=200$.
100: $100/8=12.5$ not integer, so impossible.

Thus 100 cannot be written that way.
ANSWER 8: D

Problem 9:
Blake averaged 78 over four tests. Jenny vs Blake: +10, -10, +20, +20. By how much does Jenny's average exceed Blake's?

Only differences matter. Total points difference over 4 tests:
$+10-10+20+20 = +40$.

Average is total/4, so Jenny's average is $40/4=10$ points higher.

Blake's 78 is irrelevant to the difference. Choices 15,20,25,40 confuse total with average.
ANSWER 9: A

Problem 10:
Compute $(.2)^3/(.02)^2$.

$(.2)^3 = 0.2\times0.2\times0.2 = 0.04\times0.2 = 0.008$.
$(.02)^2 = 0.02\times0.02 = 0.0004$.

Ratio $=0.008/0.0004 = 8/0.4 = 80/4 =20$.

Alternatively with powers of 10: $(2\times10^{-1})^3=8\times10^{-3}$, $(2\times10^{-2})^2=4\times10^{-4}$, ratio $=2\times10^{1}=20$.

So not .2, 2, 10, or 15.
ANSWER 10: E

Problem 1:
We are asked to evaluate the product of the 9 factors:
$$\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{10}\right)$$

First, simplify each factor:
$$1 - \frac{1}{2} = \frac{1}{2}$$
$$1 - \frac{1}{3} = \frac{2}{3}$$
$$1 - \frac{1}{4} = \frac{3}{4}$$
$$\vdots$$
$$1 - \frac{1}{10} = \frac{9}{10}$$

Now, multiply the fractions together:
$$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{9}{10}$$

Notice that this is a telescoping product where the denominator of each fraction cancels with the numerator of the following fraction:
$$\frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \cdots \times \frac{\cancel{9}}{10} = \frac{1}{10}$$

ANSWER 1: A

---

Problem 2:
We are asked to find the sum of the distinct prime integer divisors of $2016$.

First, find the prime factorization of $2016$:
- $2016 = 2 \times 1008$
- $1008 = 2 \times 504$
- $504 = 2 \times 252$
- $252 = 2 \times 126$
- $126 = 2 \times 63$
- $63 = 9 \times 7 = 3^2 \times 7$

Thus, the prime factorization of $2016$ is:
$$2016 = 2^5 \times 3^2 \times 7$$

The distinct prime divisors are $2$, $3$, and $7$. 
Their sum is:
$$2 + 3 + 7 = 12$$

ANSWER 2: B

---

Problem 3:
We are given that the population of Nisos Isles is $200$ in the year $1998$, and it triples every $25$ years. We want to find the year in which the population reaches approximately $6000$.

Let $n$ be the number of $25$-year periods that pass after $1998$. The population can be represented as:
$$P = 200 \times 3^n$$

We want $P \approx 6000$:
$$200 \times 3^n \approx 6000 \implies 3^n \approx \frac{6000}{200} = 30$$

Let's test powers of $3$:
- For $n = 1$: $3^1 = 3 \implies P = 600$
- For $n = 2$: $3^2 = 9 \implies P = 1800$
- For $n = 3$: $3^3 = 27 \implies P = 5400$
- For $n = 4$: $3^4 = 81 \implies P = 16200$

$5400$ is very close to $6000$, so about $n = 3$ periods of $25$ years are needed:
$$3 \times 25 = 75\text{ years}$$

Adding $75$ years to $1998$:
$$1998 + 75 = 2073 \approx 2075$$

ANSWER 3: B

---

Problem 4:
We are given that exactly $74\%$ of the people surveyed answered "yes." We need to find the minimum number of people who could have been surveyed.

Let $N$ be the total number of people surveyed. The number of people who answered "yes" is:
$$74\% \times N = \frac{74}{100} N = \frac{37}{50} N$$

Since the number of people who answered "yes" must be a whole number, $\frac{37}{50} N$ must be an integer. 
Since $37$ and $50$ share no common factors other than $1$ (they are coprime), $N$ must be a multiple of $50$. 

The smallest positive integer multiple of $50$ is $50$. 
With $N = 50$, exactly $\frac{37}{50} \times 50 = 37$ people answered "yes."

ANSWER 4: D

---

Problem 5:
Betty is multiplying $0.075$ by $2.56$.
She calculated $75 \times 256 = 19200$.

Count the total number of digits to the right of the decimal point in the original numbers:
- $0.075$ has $3$ decimal places.
- $2.56$ has $2$ decimal places.

The product must therefore have $3 + 2 = 5$ decimal places:
$$19200 \times 10^{-5} = 0.19200 = 0.192$$

ANSWER 5: B

---

Problem 6:
Julie's speech must last between one-half hour ($30$ minutes) and three-quarters of an hour ($45$ minutes). Her speaking rate is $150$ words per minute.

Calculate the minimum and maximum acceptable word counts:
- Minimum words: $30\text{ minutes} \times 150\text{ words/minute} = 4500\text{ words}$
- Maximum words: $45\text{ minutes} \times 150\text{ words/minute} = 6750\text{ words}$

An appropriate word count must lie in the range $[4500, 6750]$. 
Looking at the choices:
- A. $2250$ (too low)
- B. $3000$ (too low)
- C. $4200$ (too low)
- D. $4350$ (too low)
- E. $5650$ is between $4500$ and $6750$ ($\frac{5650}{150} \approx 37.7$ minutes).

ANSWER 6: E

---

Problem 7:
Ted's grandfather covered $2$ miles on each of the $3$ days. 
First, find the time he actually spent each day using $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
- **Monday:** $\frac{2\text{ miles}}{5\text{ mph}} = \frac{2}{5}\text{ hour} = \frac{2}{5} \times 60 = 24\text{ minutes}$.
- **Wednesday:** $\frac{2\text{ miles}}{3\text{ mph}} = \frac{2}{3}\text{ hour} = \frac{2}{3} \times 60 = 40\text{ minutes}$.
- **Friday:** $\frac{2\text{ miles}}{4\text{ mph}} = \frac{2}{4}\text{ hour} = \frac{1}{2} \times 60 = 30\text{ minutes}$.

Total actual time spent:
$$24 + 40 + 30 = 94\text{ minutes}$$

If he had walked at $4\text{ mph}$ every day, the time spent each day would be $\frac{2}{4}\text{ hour} = 30\text{ minutes}$, so for all $3$ days:
$$3 \times 30 = 90\text{ minutes}$$

He would have spent:
$$94 - 90 = 4\text{ minutes less}$$

ANSWER 7: D

---

Problem 8:
Let four consecutive odd integers be represented as:
$$n, \quad n+2, \quad n+4, \quad n+6$$
where $n$ is an odd integer.

Their sum is:
$$S = n + (n+2) + (n+4) + (n+6) = 4n + 12 = 4(n+3)$$

Since $n$ is an odd integer, $n+3$ must be an even integer, which can be written as $2k$ for some integer $k$.
Substituting this in gives:
$$S = 4(2k) = 8k$$

Therefore, the sum of four consecutive odd integers must be a multiple of $8$.
Let's check the given options:
- A. $16 = 8 \times 2$ ($1 + 3 + 5 + 7 = 16$)
- B. $40 = 8 \times 5$ ($7 + 9 + 11 + 13 = 40$)
- C. $72 = 8 \times 9$ ($15 + 17 + 19 + 21 = 72$)
- D. $100$: not divisible by $8$ ($100 / 8 = 12.5$)
- E. $200 = 8 \times 25$ ($47 + 49 + 51 + 53 = 200$)

Hence, $100$ cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

---

Problem 9:
Let Blake's scores on the four tests be $B_1, B_2, B_3, B_4$.
Jenny's scores are:
- First test: $B_1 + 10$
- Second test: $B_2 - 10$
- Third test: $B_3 + 20$
- Fourth test: $B_4 + 20$

The sum of Jenny's scores minus the sum of Blake's scores is:
$$(B_1 + 10 + B_2 - 10 + B_3 + 20 + B_4 + 20) - (B_1 + B_2 + B_3 + B_4) = 10 - 10 + 20 + 20 = 40\text{ points}$$

Since there are $4$ tests, the difference between Jenny's average and Blake's aver

# Problem 1:
We need the product of the 9 factors (1 − 1/n) for n = 2, 3, 4, …, 10.

Rewrite each factor: 1 − 1/n = (n−1)/n. So the product becomes

(1/2)(2/3)(3/4)(4/5)(5/6)(6/7)(7/8)(8/9)(9/10)

This **telescopes**: each numerator cancels the denominator before it (2 cancels 2, 3 cancels 3, …, 9 cancels 9), leaving only the first numerator and the last denominator:

Product = 1/10

Ruling out traps: every factor is less than 1, so the product must be *smaller* than any single factor — this eliminates D (10/11) and E (11/2) immediately. Choice C (1/2) would be the answer if only the first factor were used, and B (1/9) would result from miscounting the factors (stopping at 1 − 1/9). Since n runs from 2 to 10 inclusive, there are exactly 9 factors and the answer is 1/10.

ANSWER 1: A

---

# Problem 2:
Factor 2016 into primes by repeated division:

2016 = 2 × 1008 = 2² × 504 = 2³ × 252 = 2⁴ × 126 = 2⁵ × 63, and 63 = 3² × 7.

So 2016 = 2⁵ × 3² × 7.

The **distinct** prime divisors are just 2, 3, and 7 (the exponents don't matter — we count each prime once).

Sum = 2 + 3 + 7 = **12**

Ruling out traps: A (9) would come from forgetting the prime 3; E (63) is the odd part 3² × 7 of 2016, not a sum of primes; adding primes *with repetition* gives 2+2+2+2+2+3+3+7 = 23, which isn't even a choice. The correct distinct-prime sum is 12.

ANSWER 2: B

---

# Problem 3:
Starting at 200 people in 1998, the population triples every 25 years:

- 1998: 200
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16 200

We want about 6000. Check how many triplings are needed: 6000/200 = 30, and 3³ = 27 ≈ 30 (while 3⁴ = 81 is way too big). So about **three triplings ≈ 75 years** after 1998, i.e., around 2073.

Comparing to the choices: at 2073 the population is 5400 (off by only 600), while at 2098 it is 16 200 (off by over 10 000). The closest estimate offered is **2075**.

Ruling out traps: 2050 and 2100 bracket the answer but are farther from the true crossing point than 2075; 2125 and 2150 correspond to populations of tens of thousands.

ANSWER 3: B

---

# Problem 4:
If N people were asked and exactly 74% said yes, then 0.74N must be a whole number of people:

0.74N = (74/100)N = (37/50)N

Since 37 is prime and shares no factor with 50, the fraction 37/50 is fully reduced. For (37/50)N to be an integer, **50 must divide N**. The smallest such N is 50.

Check: with 50 people, 0.74 × 50 = 37 people said yes — exactly 74%. ✓

Ruling out the smaller choices: 10 people → 7.4 yes-votes; 20 → 14.8; 25 → 18.5 — none of these is a whole number of people, so each is impossible. Hence 50 is the fewest.

ANSWER 4: D

---

# Problem 5:
First, confirm the calculator display: entering 075 × 256 (no decimal points) gives 75 × 256 = 19 200. ✓ That matches.

Now place the decimal correctly. The rule for multiplying decimals: the product has as many decimal places as the two factors combined.

- 0.075 has **3** decimal places
- 2.56 has **2** decimal places
- Product has 3 + 2 = **5** decimal places

Start from 19 200 and move the decimal point 5 places to the left:

19 200 → 0.19200 = **0.192**

Ruling out traps: 192 (E) and 19.2 (D) come from shifting the point only 2 or 3 places; 0.0192 (A) from shifting 6 places. A quick sanity check: 0.075 × 2.56 should be a bit less than 0.1 × 2.56 = 0.256, and 0.192 fits perfectly.

ANSWER 5: B

---

# Problem 6:
Convert the time limits to minutes, then to words:

- Minimum time: 1/2 hour = 30 minutes → 150 × 30 = **4500 words**
- Maximum time: 3/4 hour = 45 minutes → 150 × 45 = **6750 words**

So the speech needs between 4500 and 6750 words.

Check each choice: 2250 words = 15 min (too short); 3000 = 20 min (too short); 4200 = 28 min (too short); 4350 = 29 min (just under a half hour — too short); 5650 = 5650/150 ≈ 37.7 minutes, which lies nicely between 30 and 45 minutes. ✓

Only 5650 works.

ANSWER 6: E

---

# Problem 7:
Use time = distance ÷ speed, and convert to minutes (1 hour = 60 min):

- Monday: 2 miles at 5 mph → 2/5 hour = 24 minutes
- Wednesday: 2 miles at 3 mph → 2/3 hour = 40 minutes
- Friday: 2 miles at 4 mph → 2/4 = 1/2 hour = 30 minutes

Actual total: 24 + 40 + 30 = **94 minutes**.

If he had always gone 4 mph: total distance 6 miles at 4 mph → 6/4 = 1.5 hours = 90 minutes.

Difference: 94 − 90 = **4 minutes less**.

Ruling out traps: note Monday would actually take *longer* at 4 mph (30 vs. 24 min), so you can't just assume each day saves time — Wednesday's big savings (40 → 30) must be combined with Monday's loss (24 → 30). Net saving = 10 − 6 = 4 minutes, matching the totals.

ANSWER 7: D

---

# Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6, where a is odd. Their sum is

a + (a+2) + (a+4) + (a+6) = 4a + 12 = 4(a + 3).

Since a is odd, a + 3 is **even**, so the sum equals 4 × (an even number), i.e., every such sum is **divisible by 8**.

Test divisibility by 8 for each choice:

- 16 ÷ 8 = 2 ✓ (indeed 1 + 3 + 5 + 7 = 16)
- 40 ÷ 8 = 5 ✓ (7 + 9 + 11 + 13 = 40)
- 72 ÷ 8 = 9 ✓ (15 + 17 + 19 + 21 = 72)
- 100 ÷ 8 = 12.5 ✗ — **not divisible by 8**
- 200 ÷ 8 = 25 ✓ (47 + 49 + 51 + 53 = 200)

So 100 cannot be written as such a sum. (Check directly: 100 = 4(a+3) forces a + 3 = 25, so a = 22 — but 22 is even, a contradiction.)

ANSWER 8: D

---

# Problem 9:
Blake's total over four tests: 4 × 78 = 312 points.

Jenny's scores compared to Blake's, test by test: +10, −10, +20, +20.

Total difference: +10 − 10 + 20 + 20 = **+40 points**.

Jenny's total: 312 + 40 = 352, so her average is 352 ÷ 4 = 88.

Jenny's average exceeds Blake's by 88 − 78 = **10 points**.

(Equivalently: the *average* difference is the total difference divided by the number of tests: 40 ÷ 4 = 10.)

Ruling out traps: E (40) is the total point difference — it forgets to divide by the 4 tests; C (20) might come from averaging only the +20's and ignoring the +10/−10 pair; B and D have no valid basis.

ANSWER 9: A

---

# Problem 10:
Compute the numerator and denominator separately:

- (0.2)³ = 

Problem 1:  
The product is  
\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots\left(1-\frac1{10}\right).
\]
Rewrite each factor:
\[
\frac12\cdot \frac23\cdot \frac34\cdot \frac45\cdots \frac9{10}.
\]
Everything cancels except the first numerator and the last denominator:
\[
\frac1{10}.
\]
So the answer is A.

ANSWER 1: A

Problem 2:  
We need the sum of the distinct prime factors of 2016. Factor:
\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot 252=2^4\cdot 126=2^5\cdot 63.
\]
Also,
\[
63=3^2\cdot 7.
\]
Thus
\[
2016=2^5\cdot 3^2\cdot 7.
\]
The distinct prime divisors are 2, 3, and 7. Their sum is
\[
2+3+7=12.
\]

ANSWER 2: B

Problem 3:  
The population triples every 25 years. Starting with 200 in 1998, after \(t\) years the population is approximately
\[
200\cdot 3^{t/25}.
\]
We want this to be about 6000:
\[
200\cdot 3^{t/25}=6000.
\]
Divide by 200:
\[
3^{t/25}=30.
\]
Now,
\[
3^3=27
\]
and
\[
3^4=81.
\]
So \(t/25\) is a little more than 3, meaning \(t\) is a little more than 75 years.  
\[
1998+75=2073,
\]
so the closest choice is 2075.

ANSWER 3: B

Problem 4:  
Exactly 74% answered “yes.” Write 74% as a fraction in lowest terms:
\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]
For the number of “yes” answers to be an integer, the total number of people asked must be a multiple of 50. The smallest such number among the choices is 50.  
Check: \(\frac{37}{50}=74\%\).

ANSWER 4: D

Problem 5:  
Betty forgot the decimal points, so she effectively multiplied
\[
75\cdot 256=19200.
\]
Now restore the decimal places.  
\(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places, so the correct product must have
\[
3+2=5
\]
decimal places. Therefore,
\[
0.075\cdot 2.56=\frac{19200}{100000}=0.192.
\]

ANSWER 5: B

Problem 6:  
The speech must last between one-half hour and three-quarters of an hour, which is between
\[
30\text{ minutes}
\]
and
\[
45\text{ minutes}.
\]
At 150 words per minute, the acceptable word counts are between
\[
150\cdot 30=4500
\]
and
\[
150\cdot 45=6750.
\]
Among the choices, only 5650 lies between 4500 and 6750.

ANSWER 6: E

Problem 7:  
Each day he went 2 miles.

Monday: speed 5 mph, time
\[
\frac{2}{5}\text{ hour}=24\text{ minutes}.
\]
Wednesday: speed 3 mph, time
\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]
Friday: speed 4 mph, time
\[
\frac{2}{4}\text{ hour}=30\text{ minutes}.
\]
Total time:
\[
24+40+30=94\text{ minutes}.
\]
If he had walked all 6 miles at 4 mph, the time would be
\[
\frac{6}{4}\text{ hour}=1.5\text{ hours}=90\text{ minutes}.
\]
The difference is
\[
94-90=4\text{ minutes}.
\]

ANSWER 7: D

Problem 8:  
Let four consecutive odd integers be
\[
2k+1,\;2k+3,\;2k+5,\;2k+7.
\]
Their sum is
\[
(2k+1)+(2k+3)+(2k+5)+(2k+7)=8k+16=8(k+2).
\]
So any sum of four consecutive odd integers must be divisible by 8.

Check the choices:
- 16 is divisible by 8.
- 40 is divisible by 8.
- 72 is divisible by 8.
- 100 is not divisible by 8.
- 200 is divisible by 8.

Therefore, 100 cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

Problem 9:  
Blake’s average was 78 on four tests, so his total score was
\[
4\cdot 78=312.
\]
Jenny’s scores differed from Blake’s by
\[
+10,\;-10,\;+20,\;+20.
\]
The total difference is
\[
10-10+20+20=40.
\]
So Jenny’s total score was 40 points higher than Blake’s. Since there are 4 tests, her average was higher by
\[
\frac{40}{4}=10.
\]

ANSWER 9: A

Problem 10:  
Compute:
\[
(0.2)^3=0.008
\]
and
\[
(0.02)^2=0.0004.
\]
Thus
\[
\frac{(0.2)^3}{(0.02)^2}=\frac{0.008}{0.0004}=20.
\]

ANSWER 10: E

Problem 1:
Each factor simplifies: (1 − 1/2) = 1/2, (1 − 1/3) = 2/3, …, (1 − 1/10) = 9/10.

Product = (1/2)(2/3)(3/4)⋯(9/10).

Everything telescopes: the numerator of each fraction cancels the denominator of the previous one, leaving 1/10.

(Choice B, 1/9, would come from miscounting the last denominator; the product is definitely less than 1/2, ruling out C, D, E.)

ANSWER 1: A

Problem 2:
Factor 2016: 2016 = 2 · 1008 = 2·2·504 = 2·2·2·252 = 2^5 · 63 = 2^5 · 3^2 · 7.

Distinct primes: 2, 3, 7. Sum = 2 + 3 + 7 = 12.

(9 = 2+7 misses 3; 16 = 2·5+... etc.; 49 and 63 come from adding non‑prime factors or all factors with multiplicity.)

ANSWER 2: B

Problem 3:
Start: 1998 → 200 people. Tripling every 25 years:
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16200

The population reaches about 6000 shortly after 2073, so the best estimate among the choices is 2075.

(2050 gives ~1800, 2100 gives ~16000 — both far off.)

ANSWER 3: B

Problem 4:
We need 74% of the number of people to be a whole number: 0.74n = 37n/50 must be an integer.

Since 37/50 is in lowest terms (37 is prime), n must be a multiple of 50. The smallest is n = 50 (then 37 people said yes).

Check the smaller options: 74% of 10 = 7.4, of 20 = 14.8, of 25 = 18.5 — none are whole numbers.

ANSWER 4: D

Problem 5:
Ignoring decimals: 75 × 256 = 19200, matching the display.

Count decimal places: 0.075 has 3, 2.56 has 2, total 5 decimal places.

So 0.075 × 2.56 = 19200 with the decimal point moved 5 places left = 0.19200 = 0.192.

Sanity check: 0.075 ≈ 0.08 and 2.56 ≈ 2.5, product ≈ 0.2. So 0.192 is right (not 0.0192 or 1.92).

ANSWER 5: B

Problem 6:
Time allowed: 30 minutes to 45 minutes.

At 150 words per minute:
- Minimum words = 150 × 30 = 4500
- Maximum words = 150 × 45 = 6750

We need a count between 4500 and 6750. Only 5650 qualifies.

(2250 would take 15 min, 3000 → 20 min, 4200 → 28 min, 4350 → 29 min — all too short.)

ANSWER 6: E

Problem 7:
Time = distance ÷ speed, with 2 miles each day.

- Monday at 5 mph: 2/5 h = 24 min
- Wednesday at 3 mph: 2/3 h = 40 min
- Friday at 4 mph: 2/4 h = 30 min

Total actual = 24 + 40 + 30 = 94 minutes.

If always 4 mph: 30 min each day → 3 × 30 = 90 minutes.

Difference: 94 − 90 = 4 minutes less.

ANSWER 7: D

Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6 with a odd.

Sum = 4a + 12 = 4(a + 3). Since a is odd, a + 3 is even, say a + 3 = 2k, so the sum = 8k — a multiple of 8.

Check the choices for divisibility by 8:
- 16 = 8·2 ✓ (1+3+5+7)
- 40 = 8·5 ✓ (7+9+11+13)
- 72 = 8·9 ✓ (15+17+19+21)
- 100 ÷ 8 = 12.5 ✗
- 200 = 8·25 ✓ (47+49+51+53)

So 100 cannot be written that way.

ANSWER 8: D

Problem 9:
Jenny's total differs from Blake's total by:
+10 − 10 + 20 + 20 = +40 points.

Average difference = 40 ÷ 4 = 10 points.

(40 is the total difference, not the average — that's the trap in choice E. Blake's 78 average is not even needed, though Jenny averages 88, which is under 100, so it's consistent.)

ANSWER 9: A

Problem 10:
(0.2)³ = 0.008 and (0.02)² = 0.0004.

0.008 ÷ 0.0004 = 80/4 = 20.

(Alternatively: (2×10⁻¹)³ / (2×10⁻²)² = 8×10⁻³ / (4×10⁻⁴) = 2 × 10¹ = 20.)

ANSWER 10: E

Problem 1:

We need to compute

\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots \left(1-\frac1{10}\right).
\]

Rewrite each factor:

\[
1-\frac12=\frac12,\qquad
1-\frac13=\frac23,\qquad
1-\frac14=\frac34,
\]

and so on. Thus the product is

\[
\frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdots\frac9{10}.
\]

Nearly everything cancels:

\[
\frac{\cancel1}{2}\cdot\frac{\cancel2}{3}\cdot\frac{\cancel3}{4}\cdots
\frac{\cancel9}{10}
=\frac1{10}.
\]

Choice B, \(\frac19\), might seem plausible if one thinks there are 9 factors, but the cancellation leaves the first numerator \(1\) and final denominator \(10\).

\[
\boxed{\frac1{10}}
\]

ANSWER 1: A

Problem 2:

We need the sum of the distinct prime divisors of \(2016\).

Factor \(2016\):

\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot252=2^4\cdot126=2^5\cdot63.
\]

Since

\[
63=3^2\cdot7,
\]

we have

\[
2016=2^5\cdot3^2\cdot7.
\]

The distinct prime divisors are

\[
2,\ 3,\ 7.
\]

Their sum is

\[
2+3+7=12.
\]

Do not add repeated prime factors such as all five 2s and both 3s; the question asks for distinct primes only.

\[
\boxed{12}
\]

ANSWER 2: B

Problem 3:

The population starts at \(200\) in 1998 and triples every 25 years.

We want it to be about \(6000\). Compare:

\[
\frac{6000}{200}=30.
\]

So the population must grow by a factor of about \(30\).

Powers of \(3\) are:

\[
3^1=3,\qquad 3^2=9,\qquad 3^3=27,\qquad 3^4=81.
\]

A factor of \(27\) is close to \(30\), so we need about 3 periods of 25 years:

\[
3\cdot25=75\text{ years}.
\]

Therefore the estimated year is

\[
1998+75=2073.
\]

The nearest choice is \(2075\).

Indeed, after 75 years the estimated population is

\[
200\cdot 27=5400,
\]

which is reasonably close to \(6000\). Waiting 100 years would give \(200\cdot81=16200\), much too large.

\[
\boxed{2075}
\]

ANSWER 3: B

Problem 4:

Exactly \(74\%\) of the people answered yes.

Convert \(74\%\) to a fraction:

\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]

For the number of yes answers to be a whole number, the total number surveyed must be a multiple of \(50\).

The smallest possible total is therefore

\[
50.
\]

For example,

\[
74\%\text{ of }50=\frac{74}{100}\cdot 50=37,
\]

which is an integer number of people.

Smaller choices such as 10, 20, and 25 do not work, since \(74\%\) of them is not a whole number.

\[
\boxed{50}
\]

ANSWER 4: D

Problem 5:

Without decimal points, Betty entered

\[
75\times256=19200.
\]

Now place the decimal points correctly.

The number \(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places. Therefore the product must have

\[
3+2=5
\]

decimal places.

So move the decimal point in \(19200\) five places left:

\[
19200\longrightarrow 0.19200=0.192.
\]

Thus

\[
0.075\times2.56=0.192.
\]

Choice A, \(0.0192\), has one too many decimal places, while \(1.92\) and larger answers are too large because \(0.075\) is less than \(0.1\).

\[
\boxed{0.192}
\]

ANSWER 5: B

Problem 6:

Julie’s speech must last between one-half hour and three-quarters of an hour.

Convert these times to minutes:

\[
\frac12\text{ hour}=30\text{ minutes},
\qquad
\frac34\text{ hour}=45\text{ minutes}.
\]

At \(150\) words per minute, the minimum appropriate word count is

\[
30\cdot150=4500.
\]

The maximum appropriate word count is

\[
45\cdot150=6750.
\]

So her speech should contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) and \(3000\) are too few.
- \(4200\) and \(4350\) are still below \(4500\).
- \(5650\) is between \(4500\) and \(6750\).

Therefore the appropriate word count is

\[
\boxed{5650}.
\]

ANSWER 6: E

Problem 7:

Grandfather traveled 2 miles on each of 3 days.

Find his actual treadmill time.

On Monday, he jogged at 5 mph:

\[
\frac{2}{5}\text{ hour}=0.4\text{ hour}=24\text{ minutes}.
\]

On Wednesday, he walked at 3 mph:

\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]

On Friday, he walked at 4 mph:

\[
\frac24\text{ hour}=\frac12\text{ hour}=30\text{ minutes}.
\]

So his actual total time was

\[
24+40+30=94\text{ minutes}.
\]

If he had walked at 4 mph on all three days, each 2-mile trip would take 30 minutes:

\[
3\cdot30=90\text{ minutes}.
\]

Thus he would have spent

\[
94-90=4
\]

fewer minutes on the treadmill.

The 5-mph jogging day saves time, but the 3-mph walking day costs more time; together the actual time is 4 minutes longer than always walking at 4 mph.

\[
\boxed{4}
\]

ANSWER 7: D

Problem 8:

Let the four consecutive odd integers be

\[
n,\ n+2,\ n+4,\ n+6,
\]

where \(n\) is odd.

Their sum is

\[
n+(n+2)+(n+4)+(n+6)=4n+12.
\]

Factor:

\[
4n+12=4(n+3).
\]

Since \(n\) is odd, \(n+3\) is even. Therefore \(4(n+3)\) must be divisible by \(8\).

So a number can be written as the sum of four consecutive odd integers only if it is divisible by \(8\).

Check the choices:

\[
16\div8=2,
\quad
40\div8=5,
\quad
72\div8=9,
\quad
200\div8=25,
\]

so all of these are divisible by 8.

But

\[
100\div8=12.5,
\]

so \(100\) is not divisible by 8. Therefore it cannot be the sum of four consecutive odd integers.

For example, \(16=1+3+5+7\), showing that the condition is possible for the other type of choices.

\[
\boxed{100}
\]

ANSWER 8: D

Problem 9:

Blake’s average is not actually needed, because we can compare Jenny’s scores directly to Blake’s.

Jenny’s score changes relative to Blake’s are:

- first test: \(+10\),
- second test: \(-10\),
- third test: \(+20\),
- fourth test: \(+20\).

Her total score exceeds Blake’s by

\[
10-10+20+20=40.
\]

Since there are four tests, her average exceeds Blake’s by

\[
\frac{40}{4}=10.
\]

The \(40\) is the difference in total points, not the difference in average points.

\[
\boxed{10}
\]

ANSWER 9: A

Problem 10:

Compute

\[
\frac{(0.2)^3}{(0.02)^2}.
\]

First,

\[
(0.2)^3=0.2\cdot0.2\cdot0.2=0.008.
\]

Next,

\[
(0.02)^2=0.02\cdot0.02=0.0004.
\]

Therefore,

\[
\frac{0.008}{0.0004}=20.
\]

Another way is to use powers of 10:

\[
0.2=2\cdot10^{-1},\qquad 0.02=2\cdot10^{-2}.
\]

Then

\[
\fra

Problem 1:
We are asked for the value of $(1-1/2)(1-1/3)\cdots (1-1/10)$.

Write each factor as a single fraction:
$1-1/2 = 1/2$, $1-1/3 = 2/3$, $1-1/4 = 3/4$, ..., $1-1/10 = 9/10$.

So the product is:
$$\frac12 \cdot \frac23 \cdot \frac34 \cdots \frac89 \cdot \frac9{10}.$$

This telescopes: the numerator 2 cancels denominator 2, numerator 3 cancels denominator 3, etc. All cancel except the first denominator 2? Let's see:
$$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10} = \frac{1}{10}.$$
Formally numerator $=1\cdot2\cdots9$, denominator $=2\cdot3\cdots10$, ratio $=1/10$.

Choices B $1/9$, C $1/2$, D $10/11$, E $11/2$ do not result from cancellation.
ANSWER 1: A

Problem 2:
We need the sum of the distinct prime divisors of 2016.

Factor 2016. It is even:
$2016 = 2\times1008 = 2\times2\times504 = 2^2\times504$.
$504 = 2\times252 = 2^2\times252$? Continue systematically:
$2016 /2 =1008$, $/2=504$, $/2=252$, $/2=126$, $/2=63$.
So $2016 = 2^5 \times 63$.
$63 = 7\times9 = 3^2\times7$.

Hence $2016 = 2^5 \cdot 3^2 \cdot 7^1$.
The distinct primes occurring are $2,3,7$.

Their sum is $2+3+7 = 12$.

This rules out A 9, C 16, D 49, E 63 which would come from missing a prime or including powers.
ANSWER 2: B

Problem 3:
In 1998 population is 200, triples every 25 years. When is it about 6000?

Ratio wanted: $6000/200 = 30$. So we need about 30 times growth.

Each 25 years multiply by 3:
After 25 yr: $200\times3=600$.
After 50 yr: $600\times3=1800$.
After 75 yr: $1800\times3=5400$.

5400 is about 6000. So about 3 triplings, i.e. about $3\times25=75$ years after 1998.

$1998+75 = 2073$, closest to 2075 among choices.

2050 is only ~2 triplings, 2100+ is too many.
ANSWER 3: B

Problem 4:
Some $n$ people asked, exactly $74\%$ said yes. Find smallest possible $n$.

Let $k$ be number saying yes. Then $k/n = 74\% = 74/100 = 37/50$.

So $k = 37n/50$ must be an integer. Since 37 and 50 are coprime (gcd=1), $n$ must be a multiple of 50 for $37n/50$ to be integer.

Smallest positive multiple is $n=50$, giving $k=37$.

$n=10,20,25$ give $7.4,14.8,18.5$ yes-votes, impossible. $n=100$ works but is not fewest.
ANSWER 4: D

Problem 5:
True product is $0.075 \times 2.56$. Entering $75 \times 256$ gave $19200$. What is correct with decimals?

$0.075$ has 3 decimal places: $0.075 = 75/1000$.
$2.56$ has 2 decimal places: $2.56 = 256/100$.

So product $= 75\times256 / (1000\times100) = 19200 / 100000$.

$19200/100000 = 0.192$.

In general move decimal 3+2=5 places left. .0192 would be 6 places, 1.92 would be 4 places, etc.
ANSWER 5: B

Problem 6:
Speech must be $1/2$ hour to $3/4$ hour at 150 words per minute. Which word count fits?

Convert hours to minutes: $1/2$ hr $=30$ min, $3/4$ hr $=45$ min.

At 150 wpm:
minimum words $=30\times150 = 4500$.
maximum words $=45\times150 = 6750$.

We need word count between 4500 and 6750 inclusive.

A 2250 is 15 min, B 3000 is 20 min, C 4200 is 28 min, D 4350 is 29 min — all too short. Only E 5650 satisfies $4500 \le 5650 \le 6750$.
ANSWER 6: E

Problem 7:
2 miles each day for 3 days. Speeds: Monday 5 mph, Wednesday 3 mph, Friday 4 mph. Compare total time to time if all 3 days at 4 mph. How many minutes saved?

Time = distance / speed.

Actual:
Monday: $2/5$ hr $=0.4$ hr $=0.4\times60=24$ min.
Wednesday: $2/3$ hr $= (2/3)\times60 =40$ min.
Friday: $2/4=1/2$ hr $=30$ min.
Total $=24+40+30=94$ min.

If always 4 mph: each day $30$ min, total $3\times30=90$ min.

Saving $=94-90=4$ minutes.
ANSWER 7: D

Problem 8:
Which integer cannot be sum of four consecutive odd integers?

Let the four be $n, n+2, n+4, n+6$ with $n$ odd.
Sum $=4n+12 =4(n+3)$.

Since $n$ is odd, $n+3$ is even, say $n+3=2m$. Then sum $=4\cdot2m=8m$, a multiple of 8. So any such sum must be divisible by 8.

Test:
16: $16/8=2$ possible; indeed $1+3+5+7=16$.
40: $40/8=5$ possible; indeed $7+9+11+13=40$.
72: $72/8=9$ possible; e.g. $15+17+19+21=72$.
200: $200/8=25$ possible; e.g. $47+49+51+53=200$.
100: $100/8=12.5$ not integer, so impossible.

Thus 100 cannot be written that way.
ANSWER 8: D

Problem 9:
Blake averaged 78 over four tests. Jenny vs Blake: +10, -10, +20, +20. By how much does Jenny's average exceed Blake's?

Only differences matter. Total points difference over 4 tests:
$+10-10+20+20 = +40$.

Average is total/4, so Jenny's average is $40/4=10$ points higher.

Blake's 78 is irrelevant to the difference. Choices 15,20,25,40 confuse total with average.
ANSWER 9: A

Problem 10:
Compute $(.2)^3/(.02)^2$.

$(.2)^3 = 0.2\times0.2\times0.2 = 0.04\times0.2 = 0.008$.
$(.02)^2 = 0.02\times0.02 = 0.0004$.

Ratio $=0.008/0.0004 = 8/0.4 = 80/4 =20$.

Alternatively with powers of 10: $(2\times10^{-1})^3=8\times10^{-3}$, $(2\times10^{-2})^2=4\times10^{-4}$, ratio $=2\times10^{1}=20$.

So not .2, 2, 10, or 15.
ANSWER 10: E

Problem 1:
We are asked to evaluate the product of the 9 factors:
$$\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{10}\right)$$

First, simplify each factor:
$$1 - \frac{1}{2} = \frac{1}{2}$$
$$1 - \frac{1}{3} = \frac{2}{3}$$
$$1 - \frac{1}{4} = \frac{3}{4}$$
$$\vdots$$
$$1 - \frac{1}{10} = \frac{9}{10}$$

Now, multiply the fractions together:
$$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{9}{10}$$

Notice that this is a telescoping product where the denominator of each fraction cancels with the numerator of the following fraction:
$$\frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \cdots \times \frac{\cancel{9}}{10} = \frac{1}{10}$$

ANSWER 1: A

---

Problem 2:
We are asked to find the sum of the distinct prime integer divisors of $2016$.

First, find the prime factorization of $2016$:
- $2016 = 2 \times 1008$
- $1008 = 2 \times 504$
- $504 = 2 \times 252$
- $252 = 2 \times 126$
- $126 = 2 \times 63$
- $63 = 9 \times 7 = 3^2 \times 7$

Thus, the prime factorization of $2016$ is:
$$2016 = 2^5 \times 3^2 \times 7$$

The distinct prime divisors are $2$, $3$, and $7$. 
Their sum is:
$$2 + 3 + 7 = 12$$

ANSWER 2: B

---

Problem 3:
We are given that the population of Nisos Isles is $200$ in the year $1998$, and it triples every $25$ years. We want to find the year in which the population reaches approximately $6000$.

Let $n$ be the number of $25$-year periods that pass after $1998$. The population can be represented as:
$$P = 200 \times 3^n$$

We want $P \approx 6000$:
$$200 \times 3^n \approx 6000 \implies 3^n \approx \frac{6000}{200} = 30$$

Let's test powers of $3$:
- For $n = 1$: $3^1 = 3 \implies P = 600$
- For $n = 2$: $3^2 = 9 \implies P = 1800$
- For $n = 3$: $3^3 = 27 \implies P = 5400$
- For $n = 4$: $3^4 = 81 \implies P = 16200$

$5400$ is very close to $6000$, so about $n = 3$ periods of $25$ years are needed:
$$3 \times 25 = 75\text{ years}$$

Adding $75$ years to $1998$:
$$1998 + 75 = 2073 \approx 2075$$

ANSWER 3: B

---

Problem 4:
We are given that exactly $74\%$ of the people surveyed answered "yes." We need to find the minimum number of people who could have been surveyed.

Let $N$ be the total number of people surveyed. The number of people who answered "yes" is:
$$74\% \times N = \frac{74}{100} N = \frac{37}{50} N$$

Since the number of people who answered "yes" must be a whole number, $\frac{37}{50} N$ must be an integer. 
Since $37$ and $50$ share no common factors other than $1$ (they are coprime), $N$ must be a multiple of $50$. 

The smallest positive integer multiple of $50$ is $50$. 
With $N = 50$, exactly $\frac{37}{50} \times 50 = 37$ people answered "yes."

ANSWER 4: D

---

Problem 5:
Betty is multiplying $0.075$ by $2.56$.
She calculated $75 \times 256 = 19200$.

Count the total number of digits to the right of the decimal point in the original numbers:
- $0.075$ has $3$ decimal places.
- $2.56$ has $2$ decimal places.

The product must therefore have $3 + 2 = 5$ decimal places:
$$19200 \times 10^{-5} = 0.19200 = 0.192$$

ANSWER 5: B

---

Problem 6:
Julie's speech must last between one-half hour ($30$ minutes) and three-quarters of an hour ($45$ minutes). Her speaking rate is $150$ words per minute.

Calculate the minimum and maximum acceptable word counts:
- Minimum words: $30\text{ minutes} \times 150\text{ words/minute} = 4500\text{ words}$
- Maximum words: $45\text{ minutes} \times 150\text{ words/minute} = 6750\text{ words}$

An appropriate word count must lie in the range $[4500, 6750]$. 
Looking at the choices:
- A. $2250$ (too low)
- B. $3000$ (too low)
- C. $4200$ (too low)
- D. $4350$ (too low)
- E. $5650$ is between $4500$ and $6750$ ($\frac{5650}{150} \approx 37.7$ minutes).

ANSWER 6: E

---

Problem 7:
Ted's grandfather covered $2$ miles on each of the $3$ days. 
First, find the time he actually spent each day using $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
- **Monday:** $\frac{2\text{ miles}}{5\text{ mph}} = \frac{2}{5}\text{ hour} = \frac{2}{5} \times 60 = 24\text{ minutes}$.
- **Wednesday:** $\frac{2\text{ miles}}{3\text{ mph}} = \frac{2}{3}\text{ hour} = \frac{2}{3} \times 60 = 40\text{ minutes}$.
- **Friday:** $\frac{2\text{ miles}}{4\text{ mph}} = \frac{2}{4}\text{ hour} = \frac{1}{2} \times 60 = 30\text{ minutes}$.

Total actual time spent:
$$24 + 40 + 30 = 94\text{ minutes}$$

If he had walked at $4\text{ mph}$ every day, the time spent each day would be $\frac{2}{4}\text{ hour} = 30\text{ minutes}$, so for all $3$ days:
$$3 \times 30 = 90\text{ minutes}$$

He would have spent:
$$94 - 90 = 4\text{ minutes less}$$

ANSWER 7: D

---

Problem 8:
Let four consecutive odd integers be represented as:
$$n, \quad n+2, \quad n+4, \quad n+6$$
where $n$ is an odd integer.

Their sum is:
$$S = n + (n+2) + (n+4) + (n+6) = 4n + 12 = 4(n+3)$$

Since $n$ is an odd integer, $n+3$ must be an even integer, which can be written as $2k$ for some integer $k$.
Substituting this in gives:
$$S = 4(2k) = 8k$$

Therefore, the sum of four consecutive odd integers must be a multiple of $8$.
Let's check the given options:
- A. $16 = 8 \times 2$ ($1 + 3 + 5 + 7 = 16$)
- B. $40 = 8 \times 5$ ($7 + 9 + 11 + 13 = 40$)
- C. $72 = 8 \times 9$ ($15 + 17 + 19 + 21 = 72$)
- D. $100$: not divisible by $8$ ($100 / 8 = 12.5$)
- E. $200 = 8 \times 25$ ($47 + 49 + 51 + 53 = 200$)

Hence, $100$ cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

---

Problem 9:
Let Blake's scores on the four tests be $B_1, B_2, B_3, B_4$.
Jenny's scores are:
- First test: $B_1 + 10$
- Second test: $B_2 - 10$
- Third test: $B_3 + 20$
- Fourth test: $B_4 + 20$

The sum of Jenny's scores minus the sum of Blake's scores is:
$$(B_1 + 10 + B_2 - 10 + B_3 + 20 + B_4 + 20) - (B_1 + B_2 + B_3 + B_4) = 10 - 10 + 20 + 20 = 40\text{ points}$$

Since there are $4$ tests, the difference between Jenny's average and Blake's aver

# Problem 1:
We need the product of the 9 factors (1 − 1/n) for n = 2, 3, 4, …, 10.

Rewrite each factor: 1 − 1/n = (n−1)/n. So the product becomes

(1/2)(2/3)(3/4)(4/5)(5/6)(6/7)(7/8)(8/9)(9/10)

This **telescopes**: each numerator cancels the denominator before it (2 cancels 2, 3 cancels 3, …, 9 cancels 9), leaving only the first numerator and the last denominator:

Product = 1/10

Ruling out traps: every factor is less than 1, so the product must be *smaller* than any single factor — this eliminates D (10/11) and E (11/2) immediately. Choice C (1/2) would be the answer if only the first factor were used, and B (1/9) would result from miscounting the factors (stopping at 1 − 1/9). Since n runs from 2 to 10 inclusive, there are exactly 9 factors and the answer is 1/10.

ANSWER 1: A

---

# Problem 2:
Factor 2016 into primes by repeated division:

2016 = 2 × 1008 = 2² × 504 = 2³ × 252 = 2⁴ × 126 = 2⁵ × 63, and 63 = 3² × 7.

So 2016 = 2⁵ × 3² × 7.

The **distinct** prime divisors are just 2, 3, and 7 (the exponents don't matter — we count each prime once).

Sum = 2 + 3 + 7 = **12**

Ruling out traps: A (9) would come from forgetting the prime 3; E (63) is the odd part 3² × 7 of 2016, not a sum of primes; adding primes *with repetition* gives 2+2+2+2+2+3+3+7 = 23, which isn't even a choice. The correct distinct-prime sum is 12.

ANSWER 2: B

---

# Problem 3:
Starting at 200 people in 1998, the population triples every 25 years:

- 1998: 200
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16 200

We want about 6000. Check how many triplings are needed: 6000/200 = 30, and 3³ = 27 ≈ 30 (while 3⁴ = 81 is way too big). So about **three triplings ≈ 75 years** after 1998, i.e., around 2073.

Comparing to the choices: at 2073 the population is 5400 (off by only 600), while at 2098 it is 16 200 (off by over 10 000). The closest estimate offered is **2075**.

Ruling out traps: 2050 and 2100 bracket the answer but are farther from the true crossing point than 2075; 2125 and 2150 correspond to populations of tens of thousands.

ANSWER 3: B

---

# Problem 4:
If N people were asked and exactly 74% said yes, then 0.74N must be a whole number of people:

0.74N = (74/100)N = (37/50)N

Since 37 is prime and shares no factor with 50, the fraction 37/50 is fully reduced. For (37/50)N to be an integer, **50 must divide N**. The smallest such N is 50.

Check: with 50 people, 0.74 × 50 = 37 people said yes — exactly 74%. ✓

Ruling out the smaller choices: 10 people → 7.4 yes-votes; 20 → 14.8; 25 → 18.5 — none of these is a whole number of people, so each is impossible. Hence 50 is the fewest.

ANSWER 4: D

---

# Problem 5:
First, confirm the calculator display: entering 075 × 256 (no decimal points) gives 75 × 256 = 19 200. ✓ That matches.

Now place the decimal correctly. The rule for multiplying decimals: the product has as many decimal places as the two factors combined.

- 0.075 has **3** decimal places
- 2.56 has **2** decimal places
- Product has 3 + 2 = **5** decimal places

Start from 19 200 and move the decimal point 5 places to the left:

19 200 → 0.19200 = **0.192**

Ruling out traps: 192 (E) and 19.2 (D) come from shifting the point only 2 or 3 places; 0.0192 (A) from shifting 6 places. A quick sanity check: 0.075 × 2.56 should be a bit less than 0.1 × 2.56 = 0.256, and 0.192 fits perfectly.

ANSWER 5: B

---

# Problem 6:
Convert the time limits to minutes, then to words:

- Minimum time: 1/2 hour = 30 minutes → 150 × 30 = **4500 words**
- Maximum time: 3/4 hour = 45 minutes → 150 × 45 = **6750 words**

So the speech needs between 4500 and 6750 words.

Check each choice: 2250 words = 15 min (too short); 3000 = 20 min (too short); 4200 = 28 min (too short); 4350 = 29 min (just under a half hour — too short); 5650 = 5650/150 ≈ 37.7 minutes, which lies nicely between 30 and 45 minutes. ✓

Only 5650 works.

ANSWER 6: E

---

# Problem 7:
Use time = distance ÷ speed, and convert to minutes (1 hour = 60 min):

- Monday: 2 miles at 5 mph → 2/5 hour = 24 minutes
- Wednesday: 2 miles at 3 mph → 2/3 hour = 40 minutes
- Friday: 2 miles at 4 mph → 2/4 = 1/2 hour = 30 minutes

Actual total: 24 + 40 + 30 = **94 minutes**.

If he had always gone 4 mph: total distance 6 miles at 4 mph → 6/4 = 1.5 hours = 90 minutes.

Difference: 94 − 90 = **4 minutes less**.

Ruling out traps: note Monday would actually take *longer* at 4 mph (30 vs. 24 min), so you can't just assume each day saves time — Wednesday's big savings (40 → 30) must be combined with Monday's loss (24 → 30). Net saving = 10 − 6 = 4 minutes, matching the totals.

ANSWER 7: D

---

# Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6, where a is odd. Their sum is

a + (a+2) + (a+4) + (a+6) = 4a + 12 = 4(a + 3).

Since a is odd, a + 3 is **even**, so the sum equals 4 × (an even number), i.e., every such sum is **divisible by 8**.

Test divisibility by 8 for each choice:

- 16 ÷ 8 = 2 ✓ (indeed 1 + 3 + 5 + 7 = 16)
- 40 ÷ 8 = 5 ✓ (7 + 9 + 11 + 13 = 40)
- 72 ÷ 8 = 9 ✓ (15 + 17 + 19 + 21 = 72)
- 100 ÷ 8 = 12.5 ✗ — **not divisible by 8**
- 200 ÷ 8 = 25 ✓ (47 + 49 + 51 + 53 = 200)

So 100 cannot be written as such a sum. (Check directly: 100 = 4(a+3) forces a + 3 = 25, so a = 22 — but 22 is even, a contradiction.)

ANSWER 8: D

---

# Problem 9:
Blake's total over four tests: 4 × 78 = 312 points.

Jenny's scores compared to Blake's, test by test: +10, −10, +20, +20.

Total difference: +10 − 10 + 20 + 20 = **+40 points**.

Jenny's total: 312 + 40 = 352, so her average is 352 ÷ 4 = 88.

Jenny's average exceeds Blake's by 88 − 78 = **10 points**.

(Equivalently: the *average* difference is the total difference divided by the number of tests: 40 ÷ 4 = 10.)

Ruling out traps: E (40) is the total point difference — it forgets to divide by the 4 tests; C (20) might come from averaging only the +20's and ignoring the +10/−10 pair; B and D have no valid basis.

ANSWER 9: A

---

# Problem 10:
Compute the numerator and denominator separately:

- (0.2)³ = 

Problem 1:  
The product is  
\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots\left(1-\frac1{10}\right).
\]
Rewrite each factor:
\[
\frac12\cdot \frac23\cdot \frac34\cdot \frac45\cdots \frac9{10}.
\]
Everything cancels except the first numerator and the last denominator:
\[
\frac1{10}.
\]
So the answer is A.

ANSWER 1: A

Problem 2:  
We need the sum of the distinct prime factors of 2016. Factor:
\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot 252=2^4\cdot 126=2^5\cdot 63.
\]
Also,
\[
63=3^2\cdot 7.
\]
Thus
\[
2016=2^5\cdot 3^2\cdot 7.
\]
The distinct prime divisors are 2, 3, and 7. Their sum is
\[
2+3+7=12.
\]

ANSWER 2: B

Problem 3:  
The population triples every 25 years. Starting with 200 in 1998, after \(t\) years the population is approximately
\[
200\cdot 3^{t/25}.
\]
We want this to be about 6000:
\[
200\cdot 3^{t/25}=6000.
\]
Divide by 200:
\[
3^{t/25}=30.
\]
Now,
\[
3^3=27
\]
and
\[
3^4=81.
\]
So \(t/25\) is a little more than 3, meaning \(t\) is a little more than 75 years.  
\[
1998+75=2073,
\]
so the closest choice is 2075.

ANSWER 3: B

Problem 4:  
Exactly 74% answered “yes.” Write 74% as a fraction in lowest terms:
\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]
For the number of “yes” answers to be an integer, the total number of people asked must be a multiple of 50. The smallest such number among the choices is 50.  
Check: \(\frac{37}{50}=74\%\).

ANSWER 4: D

Problem 5:  
Betty forgot the decimal points, so she effectively multiplied
\[
75\cdot 256=19200.
\]
Now restore the decimal places.  
\(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places, so the correct product must have
\[
3+2=5
\]
decimal places. Therefore,
\[
0.075\cdot 2.56=\frac{19200}{100000}=0.192.
\]

ANSWER 5: B

Problem 6:  
The speech must last between one-half hour and three-quarters of an hour, which is between
\[
30\text{ minutes}
\]
and
\[
45\text{ minutes}.
\]
At 150 words per minute, the acceptable word counts are between
\[
150\cdot 30=4500
\]
and
\[
150\cdot 45=6750.
\]
Among the choices, only 5650 lies between 4500 and 6750.

ANSWER 6: E

Problem 7:  
Each day he went 2 miles.

Monday: speed 5 mph, time
\[
\frac{2}{5}\text{ hour}=24\text{ minutes}.
\]
Wednesday: speed 3 mph, time
\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]
Friday: speed 4 mph, time
\[
\frac{2}{4}\text{ hour}=30\text{ minutes}.
\]
Total time:
\[
24+40+30=94\text{ minutes}.
\]
If he had walked all 6 miles at 4 mph, the time would be
\[
\frac{6}{4}\text{ hour}=1.5\text{ hours}=90\text{ minutes}.
\]
The difference is
\[
94-90=4\text{ minutes}.
\]

ANSWER 7: D

Problem 8:  
Let four consecutive odd integers be
\[
2k+1,\;2k+3,\;2k+5,\;2k+7.
\]
Their sum is
\[
(2k+1)+(2k+3)+(2k+5)+(2k+7)=8k+16=8(k+2).
\]
So any sum of four consecutive odd integers must be divisible by 8.

Check the choices:
- 16 is divisible by 8.
- 40 is divisible by 8.
- 72 is divisible by 8.
- 100 is not divisible by 8.
- 200 is divisible by 8.

Therefore, 100 cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

Problem 9:  
Blake’s average was 78 on four tests, so his total score was
\[
4\cdot 78=312.
\]
Jenny’s scores differed from Blake’s by
\[
+10,\;-10,\;+20,\;+20.
\]
The total difference is
\[
10-10+20+20=40.
\]
So Jenny’s total score was 40 points higher than Blake’s. Since there are 4 tests, her average was higher by
\[
\frac{40}{4}=10.
\]

ANSWER 9: A

Problem 10:  
Compute:
\[
(0.2)^3=0.008
\]
and
\[
(0.02)^2=0.0004.
\]
Thus
\[
\frac{(0.2)^3}{(0.02)^2}=\frac{0.008}{0.0004}=20.
\]

ANSWER 10: E

Problem 1:
Each factor simplifies: (1 − 1/2) = 1/2, (1 − 1/3) = 2/3, …, (1 − 1/10) = 9/10.

Product = (1/2)(2/3)(3/4)⋯(9/10).

Everything telescopes: the numerator of each fraction cancels the denominator of the previous one, leaving 1/10.

(Choice B, 1/9, would come from miscounting the last denominator; the product is definitely less than 1/2, ruling out C, D, E.)

ANSWER 1: A

Problem 2:
Factor 2016: 2016 = 2 · 1008 = 2·2·504 = 2·2·2·252 = 2^5 · 63 = 2^5 · 3^2 · 7.

Distinct primes: 2, 3, 7. Sum = 2 + 3 + 7 = 12.

(9 = 2+7 misses 3; 16 = 2·5+... etc.; 49 and 63 come from adding non‑prime factors or all factors with multiplicity.)

ANSWER 2: B

Problem 3:
Start: 1998 → 200 people. Tripling every 25 years:
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16200

The population reaches about 6000 shortly after 2073, so the best estimate among the choices is 2075.

(2050 gives ~1800, 2100 gives ~16000 — both far off.)

ANSWER 3: B

Problem 4:
We need 74% of the number of people to be a whole number: 0.74n = 37n/50 must be an integer.

Since 37/50 is in lowest terms (37 is prime), n must be a multiple of 50. The smallest is n = 50 (then 37 people said yes).

Check the smaller options: 74% of 10 = 7.4, of 20 = 14.8, of 25 = 18.5 — none are whole numbers.

ANSWER 4: D

Problem 5:
Ignoring decimals: 75 × 256 = 19200, matching the display.

Count decimal places: 0.075 has 3, 2.56 has 2, total 5 decimal places.

So 0.075 × 2.56 = 19200 with the decimal point moved 5 places left = 0.19200 = 0.192.

Sanity check: 0.075 ≈ 0.08 and 2.56 ≈ 2.5, product ≈ 0.2. So 0.192 is right (not 0.0192 or 1.92).

ANSWER 5: B

Problem 6:
Time allowed: 30 minutes to 45 minutes.

At 150 words per minute:
- Minimum words = 150 × 30 = 4500
- Maximum words = 150 × 45 = 6750

We need a count between 4500 and 6750. Only 5650 qualifies.

(2250 would take 15 min, 3000 → 20 min, 4200 → 28 min, 4350 → 29 min — all too short.)

ANSWER 6: E

Problem 7:
Time = distance ÷ speed, with 2 miles each day.

- Monday at 5 mph: 2/5 h = 24 min
- Wednesday at 3 mph: 2/3 h = 40 min
- Friday at 4 mph: 2/4 h = 30 min

Total actual = 24 + 40 + 30 = 94 minutes.

If always 4 mph: 30 min each day → 3 × 30 = 90 minutes.

Difference: 94 − 90 = 4 minutes less.

ANSWER 7: D

Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6 with a odd.

Sum = 4a + 12 = 4(a + 3). Since a is odd, a + 3 is even, say a + 3 = 2k, so the sum = 8k — a multiple of 8.

Check the choices for divisibility by 8:
- 16 = 8·2 ✓ (1+3+5+7)
- 40 = 8·5 ✓ (7+9+11+13)
- 72 = 8·9 ✓ (15+17+19+21)
- 100 ÷ 8 = 12.5 ✗
- 200 = 8·25 ✓ (47+49+51+53)

So 100 cannot be written that way.

ANSWER 8: D

Problem 9:
Jenny's total differs from Blake's total by:
+10 − 10 + 20 + 20 = +40 points.

Average difference = 40 ÷ 4 = 10 points.

(40 is the total difference, not the average — that's the trap in choice E. Blake's 78 average is not even needed, though Jenny averages 88, which is under 100, so it's consistent.)

ANSWER 9: A

Problem 10:
(0.2)³ = 0.008 and (0.02)² = 0.0004.

0.008 ÷ 0.0004 = 80/4 = 20.

(Alternatively: (2×10⁻¹)³ / (2×10⁻²)² = 8×10⁻³ / (4×10⁻⁴) = 2 × 10¹ = 20.)

ANSWER 10: E

Problem 1:

We need to compute

\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots \left(1-\frac1{10}\right).
\]

Rewrite each factor:

\[
1-\frac12=\frac12,\qquad
1-\frac13=\frac23,\qquad
1-\frac14=\frac34,
\]

and so on. Thus the product is

\[
\frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdots\frac9{10}.
\]

Nearly everything cancels:

\[
\frac{\cancel1}{2}\cdot\frac{\cancel2}{3}\cdot\frac{\cancel3}{4}\cdots
\frac{\cancel9}{10}
=\frac1{10}.
\]

Choice B, \(\frac19\), might seem plausible if one thinks there are 9 factors, but the cancellation leaves the first numerator \(1\) and final denominator \(10\).

\[
\boxed{\frac1{10}}
\]

ANSWER 1: A

Problem 2:

We need the sum of the distinct prime divisors of \(2016\).

Factor \(2016\):

\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot252=2^4\cdot126=2^5\cdot63.
\]

Since

\[
63=3^2\cdot7,
\]

we have

\[
2016=2^5\cdot3^2\cdot7.
\]

The distinct prime divisors are

\[
2,\ 3,\ 7.
\]

Their sum is

\[
2+3+7=12.
\]

Do not add repeated prime factors such as all five 2s and both 3s; the question asks for distinct primes only.

\[
\boxed{12}
\]

ANSWER 2: B

Problem 3:

The population starts at \(200\) in 1998 and triples every 25 years.

We want it to be about \(6000\). Compare:

\[
\frac{6000}{200}=30.
\]

So the population must grow by a factor of about \(30\).

Powers of \(3\) are:

\[
3^1=3,\qquad 3^2=9,\qquad 3^3=27,\qquad 3^4=81.
\]

A factor of \(27\) is close to \(30\), so we need about 3 periods of 25 years:

\[
3\cdot25=75\text{ years}.
\]

Therefore the estimated year is

\[
1998+75=2073.
\]

The nearest choice is \(2075\).

Indeed, after 75 years the estimated population is

\[
200\cdot 27=5400,
\]

which is reasonably close to \(6000\). Waiting 100 years would give \(200\cdot81=16200\), much too large.

\[
\boxed{2075}
\]

ANSWER 3: B

Problem 4:

Exactly \(74\%\) of the people answered yes.

Convert \(74\%\) to a fraction:

\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]

For the number of yes answers to be a whole number, the total number surveyed must be a multiple of \(50\).

The smallest possible total is therefore

\[
50.
\]

For example,

\[
74\%\text{ of }50=\frac{74}{100}\cdot 50=37,
\]

which is an integer number of people.

Smaller choices such as 10, 20, and 25 do not work, since \(74\%\) of them is not a whole number.

\[
\boxed{50}
\]

ANSWER 4: D

Problem 5:

Without decimal points, Betty entered

\[
75\times256=19200.
\]

Now place the decimal points correctly.

The number \(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places. Therefore the product must have

\[
3+2=5
\]

decimal places.

So move the decimal point in \(19200\) five places left:

\[
19200\longrightarrow 0.19200=0.192.
\]

Thus

\[
0.075\times2.56=0.192.
\]

Choice A, \(0.0192\), has one too many decimal places, while \(1.92\) and larger answers are too large because \(0.075\) is less than \(0.1\).

\[
\boxed{0.192}
\]

ANSWER 5: B

Problem 6:

Julie’s speech must last between one-half hour and three-quarters of an hour.

Convert these times to minutes:

\[
\frac12\text{ hour}=30\text{ minutes},
\qquad
\frac34\text{ hour}=45\text{ minutes}.
\]

At \(150\) words per minute, the minimum appropriate word count is

\[
30\cdot150=4500.
\]

The maximum appropriate word count is

\[
45\cdot150=6750.
\]

So her speech should contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) and \(3000\) are too few.
- \(4200\) and \(4350\) are still below \(4500\).
- \(5650\) is between \(4500\) and \(6750\).

Therefore the appropriate word count is

\[
\boxed{5650}.
\]

ANSWER 6: E

Problem 7:

Grandfather traveled 2 miles on each of 3 days.

Find his actual treadmill time.

On Monday, he jogged at 5 mph:

\[
\frac{2}{5}\text{ hour}=0.4\text{ hour}=24\text{ minutes}.
\]

On Wednesday, he walked at 3 mph:

\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]

On Friday, he walked at 4 mph:

\[
\frac24\text{ hour}=\frac12\text{ hour}=30\text{ minutes}.
\]

So his actual total time was

\[
24+40+30=94\text{ minutes}.
\]

If he had walked at 4 mph on all three days, each 2-mile trip would take 30 minutes:

\[
3\cdot30=90\text{ minutes}.
\]

Thus he would have spent

\[
94-90=4
\]

fewer minutes on the treadmill.

The 5-mph jogging day saves time, but the 3-mph walking day costs more time; together the actual time is 4 minutes longer than always walking at 4 mph.

\[
\boxed{4}
\]

ANSWER 7: D

Problem 8:

Let the four consecutive odd integers be

\[
n,\ n+2,\ n+4,\ n+6,
\]

where \(n\) is odd.

Their sum is

\[
n+(n+2)+(n+4)+(n+6)=4n+12.
\]

Factor:

\[
4n+12=4(n+3).
\]

Since \(n\) is odd, \(n+3\) is even. Therefore \(4(n+3)\) must be divisible by \(8\).

So a number can be written as the sum of four consecutive odd integers only if it is divisible by \(8\).

Check the choices:

\[
16\div8=2,
\quad
40\div8=5,
\quad
72\div8=9,
\quad
200\div8=25,
\]

so all of these are divisible by 8.

But

\[
100\div8=12.5,
\]

so \(100\) is not divisible by 8. Therefore it cannot be the sum of four consecutive odd integers.

For example, \(16=1+3+5+7\), showing that the condition is possible for the other type of choices.

\[
\boxed{100}
\]

ANSWER 8: D

Problem 9:

Blake’s average is not actually needed, because we can compare Jenny’s scores directly to Blake’s.

Jenny’s score changes relative to Blake’s are:

- first test: \(+10\),
- second test: \(-10\),
- third test: \(+20\),
- fourth test: \(+20\).

Her total score exceeds Blake’s by

\[
10-10+20+20=40.
\]

Since there are four tests, her average exceeds Blake’s by

\[
\frac{40}{4}=10.
\]

The \(40\) is the difference in total points, not the difference in average points.

\[
\boxed{10}
\]

ANSWER 9: A

Problem 10:

Compute

\[
\frac{(0.2)^3}{(0.02)^2}.
\]

First,

\[
(0.2)^3=0.2\cdot0.2\cdot0.2=0.008.
\]

Next,

\[
(0.02)^2=0.02\cdot0.02=0.0004.
\]

Therefore,

\[
\frac{0.008}{0.0004}=20.
\]

Another way is to use powers of 10:

\[
0.2=2\cdot10^{-1},\qquad 0.02=2\cdot10^{-2}.
\]

Then

\[
\fra

Problem 1:
We are asked for the value of $(1-1/2)(1-1/3)\cdots (1-1/10)$.

Write each factor as a single fraction:
$1-1/2 = 1/2$, $1-1/3 = 2/3$, $1-1/4 = 3/4$, ..., $1-1/10 = 9/10$.

So the product is:
$$\frac12 \cdot \frac23 \cdot \frac34 \cdots \frac89 \cdot \frac9{10}.$$

This telescopes: the numerator 2 cancels denominator 2, numerator 3 cancels denominator 3, etc. All cancel except the first denominator 2? Let's see:
$$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10} = \frac{1}{10}.$$
Formally numerator $=1\cdot2\cdots9$, denominator $=2\cdot3\cdots10$, ratio $=1/10$.

Choices B $1/9$, C $1/2$, D $10/11$, E $11/2$ do not result from cancellation.
ANSWER 1: A

Problem 2:
We need the sum of the distinct prime divisors of 2016.

Factor 2016. It is even:
$2016 = 2\times1008 = 2\times2\times504 = 2^2\times504$.
$504 = 2\times252 = 2^2\times252$? Continue systematically:
$2016 /2 =1008$, $/2=504$, $/2=252$, $/2=126$, $/2=63$.
So $2016 = 2^5 \times 63$.
$63 = 7\times9 = 3^2\times7$.

Hence $2016 = 2^5 \cdot 3^2 \cdot 7^1$.
The distinct primes occurring are $2,3,7$.

Their sum is $2+3+7 = 12$.

This rules out A 9, C 16, D 49, E 63 which would come from missing a prime or including powers.
ANSWER 2: B

Problem 3:
In 1998 population is 200, triples every 25 years. When is it about 6000?

Ratio wanted: $6000/200 = 30$. So we need about 30 times growth.

Each 25 years multiply by 3:
After 25 yr: $200\times3=600$.
After 50 yr: $600\times3=1800$.
After 75 yr: $1800\times3=5400$.

5400 is about 6000. So about 3 triplings, i.e. about $3\times25=75$ years after 1998.

$1998+75 = 2073$, closest to 2075 among choices.

2050 is only ~2 triplings, 2100+ is too many.
ANSWER 3: B

Problem 4:
Some $n$ people asked, exactly $74\%$ said yes. Find smallest possible $n$.

Let $k$ be number saying yes. Then $k/n = 74\% = 74/100 = 37/50$.

So $k = 37n/50$ must be an integer. Since 37 and 50 are coprime (gcd=1), $n$ must be a multiple of 50 for $37n/50$ to be integer.

Smallest positive multiple is $n=50$, giving $k=37$.

$n=10,20,25$ give $7.4,14.8,18.5$ yes-votes, impossible. $n=100$ works but is not fewest.
ANSWER 4: D

Problem 5:
True product is $0.075 \times 2.56$. Entering $75 \times 256$ gave $19200$. What is correct with decimals?

$0.075$ has 3 decimal places: $0.075 = 75/1000$.
$2.56$ has 2 decimal places: $2.56 = 256/100$.

So product $= 75\times256 / (1000\times100) = 19200 / 100000$.

$19200/100000 = 0.192$.

In general move decimal 3+2=5 places left. .0192 would be 6 places, 1.92 would be 4 places, etc.
ANSWER 5: B

Problem 6:
Speech must be $1/2$ hour to $3/4$ hour at 150 words per minute. Which word count fits?

Convert hours to minutes: $1/2$ hr $=30$ min, $3/4$ hr $=45$ min.

At 150 wpm:
minimum words $=30\times150 = 4500$.
maximum words $=45\times150 = 6750$.

We need word count between 4500 and 6750 inclusive.

A 2250 is 15 min, B 3000 is 20 min, C 4200 is 28 min, D 4350 is 29 min — all too short. Only E 5650 satisfies $4500 \le 5650 \le 6750$.
ANSWER 6: E

Problem 7:
2 miles each day for 3 days. Speeds: Monday 5 mph, Wednesday 3 mph, Friday 4 mph. Compare total time to time if all 3 days at 4 mph. How many minutes saved?

Time = distance / speed.

Actual:
Monday: $2/5$ hr $=0.4$ hr $=0.4\times60=24$ min.
Wednesday: $2/3$ hr $= (2/3)\times60 =40$ min.
Friday: $2/4=1/2$ hr $=30$ min.
Total $=24+40+30=94$ min.

If always 4 mph: each day $30$ min, total $3\times30=90$ min.

Saving $=94-90=4$ minutes.
ANSWER 7: D

Problem 8:
Which integer cannot be sum of four consecutive odd integers?

Let the four be $n, n+2, n+4, n+6$ with $n$ odd.
Sum $=4n+12 =4(n+3)$.

Since $n$ is odd, $n+3$ is even, say $n+3=2m$. Then sum $=4\cdot2m=8m$, a multiple of 8. So any such sum must be divisible by 8.

Test:
16: $16/8=2$ possible; indeed $1+3+5+7=16$.
40: $40/8=5$ possible; indeed $7+9+11+13=40$.
72: $72/8=9$ possible; e.g. $15+17+19+21=72$.
200: $200/8=25$ possible; e.g. $47+49+51+53=200$.
100: $100/8=12.5$ not integer, so impossible.

Thus 100 cannot be written that way.
ANSWER 8: D

Problem 9:
Blake averaged 78 over four tests. Jenny vs Blake: +10, -10, +20, +20. By how much does Jenny's average exceed Blake's?

Only differences matter. Total points difference over 4 tests:
$+10-10+20+20 = +40$.

Average is total/4, so Jenny's average is $40/4=10$ points higher.

Blake's 78 is irrelevant to the difference. Choices 15,20,25,40 confuse total with average.
ANSWER 9: A

Problem 10:
Compute $(.2)^3/(.02)^2$.

$(.2)^3 = 0.2\times0.2\times0.2 = 0.04\times0.2 = 0.008$.
$(.02)^2 = 0.02\times0.02 = 0.0004$.

Ratio $=0.008/0.0004 = 8/0.4 = 80/4 =20$.

Alternatively with powers of 10: $(2\times10^{-1})^3=8\times10^{-3}$, $(2\times10^{-2})^2=4\times10^{-4}$, ratio $=2\times10^{1}=20$.

So not .2, 2, 10, or 15.
ANSWER 10: E

Problem 1:
We are asked to evaluate the product of the 9 factors:
$$\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{10}\right)$$

First, simplify each factor:
$$1 - \frac{1}{2} = \frac{1}{2}$$
$$1 - \frac{1}{3} = \frac{2}{3}$$
$$1 - \frac{1}{4} = \frac{3}{4}$$
$$\vdots$$
$$1 - \frac{1}{10} = \frac{9}{10}$$

Now, multiply the fractions together:
$$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{9}{10}$$

Notice that this is a telescoping product where the denominator of each fraction cancels with the numerator of the following fraction:
$$\frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \cdots \times \frac{\cancel{9}}{10} = \frac{1}{10}$$

ANSWER 1: A

---

Problem 2:
We are asked to find the sum of the distinct prime integer divisors of $2016$.

First, find the prime factorization of $2016$:
- $2016 = 2 \times 1008$
- $1008 = 2 \times 504$
- $504 = 2 \times 252$
- $252 = 2 \times 126$
- $126 = 2 \times 63$
- $63 = 9 \times 7 = 3^2 \times 7$

Thus, the prime factorization of $2016$ is:
$$2016 = 2^5 \times 3^2 \times 7$$

The distinct prime divisors are $2$, $3$, and $7$. 
Their sum is:
$$2 + 3 + 7 = 12$$

ANSWER 2: B

---

Problem 3:
We are given that the population of Nisos Isles is $200$ in the year $1998$, and it triples every $25$ years. We want to find the year in which the population reaches approximately $6000$.

Let $n$ be the number of $25$-year periods that pass after $1998$. The population can be represented as:
$$P = 200 \times 3^n$$

We want $P \approx 6000$:
$$200 \times 3^n \approx 6000 \implies 3^n \approx \frac{6000}{200} = 30$$

Let's test powers of $3$:
- For $n = 1$: $3^1 = 3 \implies P = 600$
- For $n = 2$: $3^2 = 9 \implies P = 1800$
- For $n = 3$: $3^3 = 27 \implies P = 5400$
- For $n = 4$: $3^4 = 81 \implies P = 16200$

$5400$ is very close to $6000$, so about $n = 3$ periods of $25$ years are needed:
$$3 \times 25 = 75\text{ years}$$

Adding $75$ years to $1998$:
$$1998 + 75 = 2073 \approx 2075$$

ANSWER 3: B

---

Problem 4:
We are given that exactly $74\%$ of the people surveyed answered "yes." We need to find the minimum number of people who could have been surveyed.

Let $N$ be the total number of people surveyed. The number of people who answered "yes" is:
$$74\% \times N = \frac{74}{100} N = \frac{37}{50} N$$

Since the number of people who answered "yes" must be a whole number, $\frac{37}{50} N$ must be an integer. 
Since $37$ and $50$ share no common factors other than $1$ (they are coprime), $N$ must be a multiple of $50$. 

The smallest positive integer multiple of $50$ is $50$. 
With $N = 50$, exactly $\frac{37}{50} \times 50 = 37$ people answered "yes."

ANSWER 4: D

---

Problem 5:
Betty is multiplying $0.075$ by $2.56$.
She calculated $75 \times 256 = 19200$.

Count the total number of digits to the right of the decimal point in the original numbers:
- $0.075$ has $3$ decimal places.
- $2.56$ has $2$ decimal places.

The product must therefore have $3 + 2 = 5$ decimal places:
$$19200 \times 10^{-5} = 0.19200 = 0.192$$

ANSWER 5: B

---

Problem 6:
Julie's speech must last between one-half hour ($30$ minutes) and three-quarters of an hour ($45$ minutes). Her speaking rate is $150$ words per minute.

Calculate the minimum and maximum acceptable word counts:
- Minimum words: $30\text{ minutes} \times 150\text{ words/minute} = 4500\text{ words}$
- Maximum words: $45\text{ minutes} \times 150\text{ words/minute} = 6750\text{ words}$

An appropriate word count must lie in the range $[4500, 6750]$. 
Looking at the choices:
- A. $2250$ (too low)
- B. $3000$ (too low)
- C. $4200$ (too low)
- D. $4350$ (too low)
- E. $5650$ is between $4500$ and $6750$ ($\frac{5650}{150} \approx 37.7$ minutes).

ANSWER 6: E

---

Problem 7:
Ted's grandfather covered $2$ miles on each of the $3$ days. 
First, find the time he actually spent each day using $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
- **Monday:** $\frac{2\text{ miles}}{5\text{ mph}} = \frac{2}{5}\text{ hour} = \frac{2}{5} \times 60 = 24\text{ minutes}$.
- **Wednesday:** $\frac{2\text{ miles}}{3\text{ mph}} = \frac{2}{3}\text{ hour} = \frac{2}{3} \times 60 = 40\text{ minutes}$.
- **Friday:** $\frac{2\text{ miles}}{4\text{ mph}} = \frac{2}{4}\text{ hour} = \frac{1}{2} \times 60 = 30\text{ minutes}$.

Total actual time spent:
$$24 + 40 + 30 = 94\text{ minutes}$$

If he had walked at $4\text{ mph}$ every day, the time spent each day would be $\frac{2}{4}\text{ hour} = 30\text{ minutes}$, so for all $3$ days:
$$3 \times 30 = 90\text{ minutes}$$

He would have spent:
$$94 - 90 = 4\text{ minutes less}$$

ANSWER 7: D

---

Problem 8:
Let four consecutive odd integers be represented as:
$$n, \quad n+2, \quad n+4, \quad n+6$$
where $n$ is an odd integer.

Their sum is:
$$S = n + (n+2) + (n+4) + (n+6) = 4n + 12 = 4(n+3)$$

Since $n$ is an odd integer, $n+3$ must be an even integer, which can be written as $2k$ for some integer $k$.
Substituting this in gives:
$$S = 4(2k) = 8k$$

Therefore, the sum of four consecutive odd integers must be a multiple of $8$.
Let's check the given options:
- A. $16 = 8 \times 2$ ($1 + 3 + 5 + 7 = 16$)
- B. $40 = 8 \times 5$ ($7 + 9 + 11 + 13 = 40$)
- C. $72 = 8 \times 9$ ($15 + 17 + 19 + 21 = 72$)
- D. $100$: not divisible by $8$ ($100 / 8 = 12.5$)
- E. $200 = 8 \times 25$ ($47 + 49 + 51 + 53 = 200$)

Hence, $100$ cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

---

Problem 9:
Let Blake's scores on the four tests be $B_1, B_2, B_3, B_4$.
Jenny's scores are:
- First test: $B_1 + 10$
- Second test: $B_2 - 10$
- Third test: $B_3 + 20$
- Fourth test: $B_4 + 20$

The sum of Jenny's scores minus the sum of Blake's scores is:
$$(B_1 + 10 + B_2 - 10 + B_3 + 20 + B_4 + 20) - (B_1 + B_2 + B_3 + B_4) = 10 - 10 + 20 + 20 = 40\text{ points}$$

Since there are $4$ tests, the difference between Jenny's average and Blake's aver

# Problem 1:
We need the product of the 9 factors (1 − 1/n) for n = 2, 3, 4, …, 10.

Rewrite each factor: 1 − 1/n = (n−1)/n. So the product becomes

(1/2)(2/3)(3/4)(4/5)(5/6)(6/7)(7/8)(8/9)(9/10)

This **telescopes**: each numerator cancels the denominator before it (2 cancels 2, 3 cancels 3, …, 9 cancels 9), leaving only the first numerator and the last denominator:

Product = 1/10

Ruling out traps: every factor is less than 1, so the product must be *smaller* than any single factor — this eliminates D (10/11) and E (11/2) immediately. Choice C (1/2) would be the answer if only the first factor were used, and B (1/9) would result from miscounting the factors (stopping at 1 − 1/9). Since n runs from 2 to 10 inclusive, there are exactly 9 factors and the answer is 1/10.

ANSWER 1: A

---

# Problem 2:
Factor 2016 into primes by repeated division:

2016 = 2 × 1008 = 2² × 504 = 2³ × 252 = 2⁴ × 126 = 2⁵ × 63, and 63 = 3² × 7.

So 2016 = 2⁵ × 3² × 7.

The **distinct** prime divisors are just 2, 3, and 7 (the exponents don't matter — we count each prime once).

Sum = 2 + 3 + 7 = **12**

Ruling out traps: A (9) would come from forgetting the prime 3; E (63) is the odd part 3² × 7 of 2016, not a sum of primes; adding primes *with repetition* gives 2+2+2+2+2+3+3+7 = 23, which isn't even a choice. The correct distinct-prime sum is 12.

ANSWER 2: B

---

# Problem 3:
Starting at 200 people in 1998, the population triples every 25 years:

- 1998: 200
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16 200

We want about 6000. Check how many triplings are needed: 6000/200 = 30, and 3³ = 27 ≈ 30 (while 3⁴ = 81 is way too big). So about **three triplings ≈ 75 years** after 1998, i.e., around 2073.

Comparing to the choices: at 2073 the population is 5400 (off by only 600), while at 2098 it is 16 200 (off by over 10 000). The closest estimate offered is **2075**.

Ruling out traps: 2050 and 2100 bracket the answer but are farther from the true crossing point than 2075; 2125 and 2150 correspond to populations of tens of thousands.

ANSWER 3: B

---

# Problem 4:
If N people were asked and exactly 74% said yes, then 0.74N must be a whole number of people:

0.74N = (74/100)N = (37/50)N

Since 37 is prime and shares no factor with 50, the fraction 37/50 is fully reduced. For (37/50)N to be an integer, **50 must divide N**. The smallest such N is 50.

Check: with 50 people, 0.74 × 50 = 37 people said yes — exactly 74%. ✓

Ruling out the smaller choices: 10 people → 7.4 yes-votes; 20 → 14.8; 25 → 18.5 — none of these is a whole number of people, so each is impossible. Hence 50 is the fewest.

ANSWER 4: D

---

# Problem 5:
First, confirm the calculator display: entering 075 × 256 (no decimal points) gives 75 × 256 = 19 200. ✓ That matches.

Now place the decimal correctly. The rule for multiplying decimals: the product has as many decimal places as the two factors combined.

- 0.075 has **3** decimal places
- 2.56 has **2** decimal places
- Product has 3 + 2 = **5** decimal places

Start from 19 200 and move the decimal point 5 places to the left:

19 200 → 0.19200 = **0.192**

Ruling out traps: 192 (E) and 19.2 (D) come from shifting the point only 2 or 3 places; 0.0192 (A) from shifting 6 places. A quick sanity check: 0.075 × 2.56 should be a bit less than 0.1 × 2.56 = 0.256, and 0.192 fits perfectly.

ANSWER 5: B

---

# Problem 6:
Convert the time limits to minutes, then to words:

- Minimum time: 1/2 hour = 30 minutes → 150 × 30 = **4500 words**
- Maximum time: 3/4 hour = 45 minutes → 150 × 45 = **6750 words**

So the speech needs between 4500 and 6750 words.

Check each choice: 2250 words = 15 min (too short); 3000 = 20 min (too short); 4200 = 28 min (too short); 4350 = 29 min (just under a half hour — too short); 5650 = 5650/150 ≈ 37.7 minutes, which lies nicely between 30 and 45 minutes. ✓

Only 5650 works.

ANSWER 6: E

---

# Problem 7:
Use time = distance ÷ speed, and convert to minutes (1 hour = 60 min):

- Monday: 2 miles at 5 mph → 2/5 hour = 24 minutes
- Wednesday: 2 miles at 3 mph → 2/3 hour = 40 minutes
- Friday: 2 miles at 4 mph → 2/4 = 1/2 hour = 30 minutes

Actual total: 24 + 40 + 30 = **94 minutes**.

If he had always gone 4 mph: total distance 6 miles at 4 mph → 6/4 = 1.5 hours = 90 minutes.

Difference: 94 − 90 = **4 minutes less**.

Ruling out traps: note Monday would actually take *longer* at 4 mph (30 vs. 24 min), so you can't just assume each day saves time — Wednesday's big savings (40 → 30) must be combined with Monday's loss (24 → 30). Net saving = 10 − 6 = 4 minutes, matching the totals.

ANSWER 7: D

---

# Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6, where a is odd. Their sum is

a + (a+2) + (a+4) + (a+6) = 4a + 12 = 4(a + 3).

Since a is odd, a + 3 is **even**, so the sum equals 4 × (an even number), i.e., every such sum is **divisible by 8**.

Test divisibility by 8 for each choice:

- 16 ÷ 8 = 2 ✓ (indeed 1 + 3 + 5 + 7 = 16)
- 40 ÷ 8 = 5 ✓ (7 + 9 + 11 + 13 = 40)
- 72 ÷ 8 = 9 ✓ (15 + 17 + 19 + 21 = 72)
- 100 ÷ 8 = 12.5 ✗ — **not divisible by 8**
- 200 ÷ 8 = 25 ✓ (47 + 49 + 51 + 53 = 200)

So 100 cannot be written as such a sum. (Check directly: 100 = 4(a+3) forces a + 3 = 25, so a = 22 — but 22 is even, a contradiction.)

ANSWER 8: D

---

# Problem 9:
Blake's total over four tests: 4 × 78 = 312 points.

Jenny's scores compared to Blake's, test by test: +10, −10, +20, +20.

Total difference: +10 − 10 + 20 + 20 = **+40 points**.

Jenny's total: 312 + 40 = 352, so her average is 352 ÷ 4 = 88.

Jenny's average exceeds Blake's by 88 − 78 = **10 points**.

(Equivalently: the *average* difference is the total difference divided by the number of tests: 40 ÷ 4 = 10.)

Ruling out traps: E (40) is the total point difference — it forgets to divide by the 4 tests; C (20) might come from averaging only the +20's and ignoring the +10/−10 pair; B and D have no valid basis.

ANSWER 9: A

---

# Problem 10:
Compute the numerator and denominator separately:

- (0.2)³ = 

Problem 1:  
The product is  
\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots\left(1-\frac1{10}\right).
\]
Rewrite each factor:
\[
\frac12\cdot \frac23\cdot \frac34\cdot \frac45\cdots \frac9{10}.
\]
Everything cancels except the first numerator and the last denominator:
\[
\frac1{10}.
\]
So the answer is A.

ANSWER 1: A

Problem 2:  
We need the sum of the distinct prime factors of 2016. Factor:
\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot 252=2^4\cdot 126=2^5\cdot 63.
\]
Also,
\[
63=3^2\cdot 7.
\]
Thus
\[
2016=2^5\cdot 3^2\cdot 7.
\]
The distinct prime divisors are 2, 3, and 7. Their sum is
\[
2+3+7=12.
\]

ANSWER 2: B

Problem 3:  
The population triples every 25 years. Starting with 200 in 1998, after \(t\) years the population is approximately
\[
200\cdot 3^{t/25}.
\]
We want this to be about 6000:
\[
200\cdot 3^{t/25}=6000.
\]
Divide by 200:
\[
3^{t/25}=30.
\]
Now,
\[
3^3=27
\]
and
\[
3^4=81.
\]
So \(t/25\) is a little more than 3, meaning \(t\) is a little more than 75 years.  
\[
1998+75=2073,
\]
so the closest choice is 2075.

ANSWER 3: B

Problem 4:  
Exactly 74% answered “yes.” Write 74% as a fraction in lowest terms:
\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]
For the number of “yes” answers to be an integer, the total number of people asked must be a multiple of 50. The smallest such number among the choices is 50.  
Check: \(\frac{37}{50}=74\%\).

ANSWER 4: D

Problem 5:  
Betty forgot the decimal points, so she effectively multiplied
\[
75\cdot 256=19200.
\]
Now restore the decimal places.  
\(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places, so the correct product must have
\[
3+2=5
\]
decimal places. Therefore,
\[
0.075\cdot 2.56=\frac{19200}{100000}=0.192.
\]

ANSWER 5: B

Problem 6:  
The speech must last between one-half hour and three-quarters of an hour, which is between
\[
30\text{ minutes}
\]
and
\[
45\text{ minutes}.
\]
At 150 words per minute, the acceptable word counts are between
\[
150\cdot 30=4500
\]
and
\[
150\cdot 45=6750.
\]
Among the choices, only 5650 lies between 4500 and 6750.

ANSWER 6: E

Problem 7:  
Each day he went 2 miles.

Monday: speed 5 mph, time
\[
\frac{2}{5}\text{ hour}=24\text{ minutes}.
\]
Wednesday: speed 3 mph, time
\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]
Friday: speed 4 mph, time
\[
\frac{2}{4}\text{ hour}=30\text{ minutes}.
\]
Total time:
\[
24+40+30=94\text{ minutes}.
\]
If he had walked all 6 miles at 4 mph, the time would be
\[
\frac{6}{4}\text{ hour}=1.5\text{ hours}=90\text{ minutes}.
\]
The difference is
\[
94-90=4\text{ minutes}.
\]

ANSWER 7: D

Problem 8:  
Let four consecutive odd integers be
\[
2k+1,\;2k+3,\;2k+5,\;2k+7.
\]
Their sum is
\[
(2k+1)+(2k+3)+(2k+5)+(2k+7)=8k+16=8(k+2).
\]
So any sum of four consecutive odd integers must be divisible by 8.

Check the choices:
- 16 is divisible by 8.
- 40 is divisible by 8.
- 72 is divisible by 8.
- 100 is not divisible by 8.
- 200 is divisible by 8.

Therefore, 100 cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

Problem 9:  
Blake’s average was 78 on four tests, so his total score was
\[
4\cdot 78=312.
\]
Jenny’s scores differed from Blake’s by
\[
+10,\;-10,\;+20,\;+20.
\]
The total difference is
\[
10-10+20+20=40.
\]
So Jenny’s total score was 40 points higher than Blake’s. Since there are 4 tests, her average was higher by
\[
\frac{40}{4}=10.
\]

ANSWER 9: A

Problem 10:  
Compute:
\[
(0.2)^3=0.008
\]
and
\[
(0.02)^2=0.0004.
\]
Thus
\[
\frac{(0.2)^3}{(0.02)^2}=\frac{0.008}{0.0004}=20.
\]

ANSWER 10: E

Problem 1:
Each factor simplifies: (1 − 1/2) = 1/2, (1 − 1/3) = 2/3, …, (1 − 1/10) = 9/10.

Product = (1/2)(2/3)(3/4)⋯(9/10).

Everything telescopes: the numerator of each fraction cancels the denominator of the previous one, leaving 1/10.

(Choice B, 1/9, would come from miscounting the last denominator; the product is definitely less than 1/2, ruling out C, D, E.)

ANSWER 1: A

Problem 2:
Factor 2016: 2016 = 2 · 1008 = 2·2·504 = 2·2·2·252 = 2^5 · 63 = 2^5 · 3^2 · 7.

Distinct primes: 2, 3, 7. Sum = 2 + 3 + 7 = 12.

(9 = 2+7 misses 3; 16 = 2·5+... etc.; 49 and 63 come from adding non‑prime factors or all factors with multiplicity.)

ANSWER 2: B

Problem 3:
Start: 1998 → 200 people. Tripling every 25 years:
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16200

The population reaches about 6000 shortly after 2073, so the best estimate among the choices is 2075.

(2050 gives ~1800, 2100 gives ~16000 — both far off.)

ANSWER 3: B

Problem 4:
We need 74% of the number of people to be a whole number: 0.74n = 37n/50 must be an integer.

Since 37/50 is in lowest terms (37 is prime), n must be a multiple of 50. The smallest is n = 50 (then 37 people said yes).

Check the smaller options: 74% of 10 = 7.4, of 20 = 14.8, of 25 = 18.5 — none are whole numbers.

ANSWER 4: D

Problem 5:
Ignoring decimals: 75 × 256 = 19200, matching the display.

Count decimal places: 0.075 has 3, 2.56 has 2, total 5 decimal places.

So 0.075 × 2.56 = 19200 with the decimal point moved 5 places left = 0.19200 = 0.192.

Sanity check: 0.075 ≈ 0.08 and 2.56 ≈ 2.5, product ≈ 0.2. So 0.192 is right (not 0.0192 or 1.92).

ANSWER 5: B

Problem 6:
Time allowed: 30 minutes to 45 minutes.

At 150 words per minute:
- Minimum words = 150 × 30 = 4500
- Maximum words = 150 × 45 = 6750

We need a count between 4500 and 6750. Only 5650 qualifies.

(2250 would take 15 min, 3000 → 20 min, 4200 → 28 min, 4350 → 29 min — all too short.)

ANSWER 6: E

Problem 7:
Time = distance ÷ speed, with 2 miles each day.

- Monday at 5 mph: 2/5 h = 24 min
- Wednesday at 3 mph: 2/3 h = 40 min
- Friday at 4 mph: 2/4 h = 30 min

Total actual = 24 + 40 + 30 = 94 minutes.

If always 4 mph: 30 min each day → 3 × 30 = 90 minutes.

Difference: 94 − 90 = 4 minutes less.

ANSWER 7: D

Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6 with a odd.

Sum = 4a + 12 = 4(a + 3). Since a is odd, a + 3 is even, say a + 3 = 2k, so the sum = 8k — a multiple of 8.

Check the choices for divisibility by 8:
- 16 = 8·2 ✓ (1+3+5+7)
- 40 = 8·5 ✓ (7+9+11+13)
- 72 = 8·9 ✓ (15+17+19+21)
- 100 ÷ 8 = 12.5 ✗
- 200 = 8·25 ✓ (47+49+51+53)

So 100 cannot be written that way.

ANSWER 8: D

Problem 9:
Jenny's total differs from Blake's total by:
+10 − 10 + 20 + 20 = +40 points.

Average difference = 40 ÷ 4 = 10 points.

(40 is the total difference, not the average — that's the trap in choice E. Blake's 78 average is not even needed, though Jenny averages 88, which is under 100, so it's consistent.)

ANSWER 9: A

Problem 10:
(0.2)³ = 0.008 and (0.02)² = 0.0004.

0.008 ÷ 0.0004 = 80/4 = 20.

(Alternatively: (2×10⁻¹)³ / (2×10⁻²)² = 8×10⁻³ / (4×10⁻⁴) = 2 × 10¹ = 20.)

ANSWER 10: E

Problem 1:

We need to compute

\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots \left(1-\frac1{10}\right).
\]

Rewrite each factor:

\[
1-\frac12=\frac12,\qquad
1-\frac13=\frac23,\qquad
1-\frac14=\frac34,
\]

and so on. Thus the product is

\[
\frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdots\frac9{10}.
\]

Nearly everything cancels:

\[
\frac{\cancel1}{2}\cdot\frac{\cancel2}{3}\cdot\frac{\cancel3}{4}\cdots
\frac{\cancel9}{10}
=\frac1{10}.
\]

Choice B, \(\frac19\), might seem plausible if one thinks there are 9 factors, but the cancellation leaves the first numerator \(1\) and final denominator \(10\).

\[
\boxed{\frac1{10}}
\]

ANSWER 1: A

Problem 2:

We need the sum of the distinct prime divisors of \(2016\).

Factor \(2016\):

\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot252=2^4\cdot126=2^5\cdot63.
\]

Since

\[
63=3^2\cdot7,
\]

we have

\[
2016=2^5\cdot3^2\cdot7.
\]

The distinct prime divisors are

\[
2,\ 3,\ 7.
\]

Their sum is

\[
2+3+7=12.
\]

Do not add repeated prime factors such as all five 2s and both 3s; the question asks for distinct primes only.

\[
\boxed{12}
\]

ANSWER 2: B

Problem 3:

The population starts at \(200\) in 1998 and triples every 25 years.

We want it to be about \(6000\). Compare:

\[
\frac{6000}{200}=30.
\]

So the population must grow by a factor of about \(30\).

Powers of \(3\) are:

\[
3^1=3,\qquad 3^2=9,\qquad 3^3=27,\qquad 3^4=81.
\]

A factor of \(27\) is close to \(30\), so we need about 3 periods of 25 years:

\[
3\cdot25=75\text{ years}.
\]

Therefore the estimated year is

\[
1998+75=2073.
\]

The nearest choice is \(2075\).

Indeed, after 75 years the estimated population is

\[
200\cdot 27=5400,
\]

which is reasonably close to \(6000\). Waiting 100 years would give \(200\cdot81=16200\), much too large.

\[
\boxed{2075}
\]

ANSWER 3: B

Problem 4:

Exactly \(74\%\) of the people answered yes.

Convert \(74\%\) to a fraction:

\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]

For the number of yes answers to be a whole number, the total number surveyed must be a multiple of \(50\).

The smallest possible total is therefore

\[
50.
\]

For example,

\[
74\%\text{ of }50=\frac{74}{100}\cdot 50=37,
\]

which is an integer number of people.

Smaller choices such as 10, 20, and 25 do not work, since \(74\%\) of them is not a whole number.

\[
\boxed{50}
\]

ANSWER 4: D

Problem 5:

Without decimal points, Betty entered

\[
75\times256=19200.
\]

Now place the decimal points correctly.

The number \(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places. Therefore the product must have

\[
3+2=5
\]

decimal places.

So move the decimal point in \(19200\) five places left:

\[
19200\longrightarrow 0.19200=0.192.
\]

Thus

\[
0.075\times2.56=0.192.
\]

Choice A, \(0.0192\), has one too many decimal places, while \(1.92\) and larger answers are too large because \(0.075\) is less than \(0.1\).

\[
\boxed{0.192}
\]

ANSWER 5: B

Problem 6:

Julie’s speech must last between one-half hour and three-quarters of an hour.

Convert these times to minutes:

\[
\frac12\text{ hour}=30\text{ minutes},
\qquad
\frac34\text{ hour}=45\text{ minutes}.
\]

At \(150\) words per minute, the minimum appropriate word count is

\[
30\cdot150=4500.
\]

The maximum appropriate word count is

\[
45\cdot150=6750.
\]

So her speech should contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) and \(3000\) are too few.
- \(4200\) and \(4350\) are still below \(4500\).
- \(5650\) is between \(4500\) and \(6750\).

Therefore the appropriate word count is

\[
\boxed{5650}.
\]

ANSWER 6: E

Problem 7:

Grandfather traveled 2 miles on each of 3 days.

Find his actual treadmill time.

On Monday, he jogged at 5 mph:

\[
\frac{2}{5}\text{ hour}=0.4\text{ hour}=24\text{ minutes}.
\]

On Wednesday, he walked at 3 mph:

\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]

On Friday, he walked at 4 mph:

\[
\frac24\text{ hour}=\frac12\text{ hour}=30\text{ minutes}.
\]

So his actual total time was

\[
24+40+30=94\text{ minutes}.
\]

If he had walked at 4 mph on all three days, each 2-mile trip would take 30 minutes:

\[
3\cdot30=90\text{ minutes}.
\]

Thus he would have spent

\[
94-90=4
\]

fewer minutes on the treadmill.

The 5-mph jogging day saves time, but the 3-mph walking day costs more time; together the actual time is 4 minutes longer than always walking at 4 mph.

\[
\boxed{4}
\]

ANSWER 7: D

Problem 8:

Let the four consecutive odd integers be

\[
n,\ n+2,\ n+4,\ n+6,
\]

where \(n\) is odd.

Their sum is

\[
n+(n+2)+(n+4)+(n+6)=4n+12.
\]

Factor:

\[
4n+12=4(n+3).
\]

Since \(n\) is odd, \(n+3\) is even. Therefore \(4(n+3)\) must be divisible by \(8\).

So a number can be written as the sum of four consecutive odd integers only if it is divisible by \(8\).

Check the choices:

\[
16\div8=2,
\quad
40\div8=5,
\quad
72\div8=9,
\quad
200\div8=25,
\]

so all of these are divisible by 8.

But

\[
100\div8=12.5,
\]

so \(100\) is not divisible by 8. Therefore it cannot be the sum of four consecutive odd integers.

For example, \(16=1+3+5+7\), showing that the condition is possible for the other type of choices.

\[
\boxed{100}
\]

ANSWER 8: D

Problem 9:

Blake’s average is not actually needed, because we can compare Jenny’s scores directly to Blake’s.

Jenny’s score changes relative to Blake’s are:

- first test: \(+10\),
- second test: \(-10\),
- third test: \(+20\),
- fourth test: \(+20\).

Her total score exceeds Blake’s by

\[
10-10+20+20=40.
\]

Since there are four tests, her average exceeds Blake’s by

\[
\frac{40}{4}=10.
\]

The \(40\) is the difference in total points, not the difference in average points.

\[
\boxed{10}
\]

ANSWER 9: A

Problem 10:

Compute

\[
\frac{(0.2)^3}{(0.02)^2}.
\]

First,

\[
(0.2)^3=0.2\cdot0.2\cdot0.2=0.008.
\]

Next,

\[
(0.02)^2=0.02\cdot0.02=0.0004.
\]

Therefore,

\[
\frac{0.008}{0.0004}=20.
\]

Another way is to use powers of 10:

\[
0.2=2\cdot10^{-1},\qquad 0.02=2\cdot10^{-2}.
\]

Then

\[
\fra

Problem 1:
We are asked for the value of $(1-1/2)(1-1/3)\cdots (1-1/10)$.

Write each factor as a single fraction:
$1-1/2 = 1/2$, $1-1/3 = 2/3$, $1-1/4 = 3/4$, ..., $1-1/10 = 9/10$.

So the product is:
$$\frac12 \cdot \frac23 \cdot \frac34 \cdots \frac89 \cdot \frac9{10}.$$

This telescopes: the numerator 2 cancels denominator 2, numerator 3 cancels denominator 3, etc. All cancel except the first denominator 2? Let's see:
$$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10} = \frac{1}{10}.$$
Formally numerator $=1\cdot2\cdots9$, denominator $=2\cdot3\cdots10$, ratio $=1/10$.

Choices B $1/9$, C $1/2$, D $10/11$, E $11/2$ do not result from cancellation.
ANSWER 1: A

Problem 2:
We need the sum of the distinct prime divisors of 2016.

Factor 2016. It is even:
$2016 = 2\times1008 = 2\times2\times504 = 2^2\times504$.
$504 = 2\times252 = 2^2\times252$? Continue systematically:
$2016 /2 =1008$, $/2=504$, $/2=252$, $/2=126$, $/2=63$.
So $2016 = 2^5 \times 63$.
$63 = 7\times9 = 3^2\times7$.

Hence $2016 = 2^5 \cdot 3^2 \cdot 7^1$.
The distinct primes occurring are $2,3,7$.

Their sum is $2+3+7 = 12$.

This rules out A 9, C 16, D 49, E 63 which would come from missing a prime or including powers.
ANSWER 2: B

Problem 3:
In 1998 population is 200, triples every 25 years. When is it about 6000?

Ratio wanted: $6000/200 = 30$. So we need about 30 times growth.

Each 25 years multiply by 3:
After 25 yr: $200\times3=600$.
After 50 yr: $600\times3=1800$.
After 75 yr: $1800\times3=5400$.

5400 is about 6000. So about 3 triplings, i.e. about $3\times25=75$ years after 1998.

$1998+75 = 2073$, closest to 2075 among choices.

2050 is only ~2 triplings, 2100+ is too many.
ANSWER 3: B

Problem 4:
Some $n$ people asked, exactly $74\%$ said yes. Find smallest possible $n$.

Let $k$ be number saying yes. Then $k/n = 74\% = 74/100 = 37/50$.

So $k = 37n/50$ must be an integer. Since 37 and 50 are coprime (gcd=1), $n$ must be a multiple of 50 for $37n/50$ to be integer.

Smallest positive multiple is $n=50$, giving $k=37$.

$n=10,20,25$ give $7.4,14.8,18.5$ yes-votes, impossible. $n=100$ works but is not fewest.
ANSWER 4: D

Problem 5:
True product is $0.075 \times 2.56$. Entering $75 \times 256$ gave $19200$. What is correct with decimals?

$0.075$ has 3 decimal places: $0.075 = 75/1000$.
$2.56$ has 2 decimal places: $2.56 = 256/100$.

So product $= 75\times256 / (1000\times100) = 19200 / 100000$.

$19200/100000 = 0.192$.

In general move decimal 3+2=5 places left. .0192 would be 6 places, 1.92 would be 4 places, etc.
ANSWER 5: B

Problem 6:
Speech must be $1/2$ hour to $3/4$ hour at 150 words per minute. Which word count fits?

Convert hours to minutes: $1/2$ hr $=30$ min, $3/4$ hr $=45$ min.

At 150 wpm:
minimum words $=30\times150 = 4500$.
maximum words $=45\times150 = 6750$.

We need word count between 4500 and 6750 inclusive.

A 2250 is 15 min, B 3000 is 20 min, C 4200 is 28 min, D 4350 is 29 min — all too short. Only E 5650 satisfies $4500 \le 5650 \le 6750$.
ANSWER 6: E

Problem 7:
2 miles each day for 3 days. Speeds: Monday 5 mph, Wednesday 3 mph, Friday 4 mph. Compare total time to time if all 3 days at 4 mph. How many minutes saved?

Time = distance / speed.

Actual:
Monday: $2/5$ hr $=0.4$ hr $=0.4\times60=24$ min.
Wednesday: $2/3$ hr $= (2/3)\times60 =40$ min.
Friday: $2/4=1/2$ hr $=30$ min.
Total $=24+40+30=94$ min.

If always 4 mph: each day $30$ min, total $3\times30=90$ min.

Saving $=94-90=4$ minutes.
ANSWER 7: D

Problem 8:
Which integer cannot be sum of four consecutive odd integers?

Let the four be $n, n+2, n+4, n+6$ with $n$ odd.
Sum $=4n+12 =4(n+3)$.

Since $n$ is odd, $n+3$ is even, say $n+3=2m$. Then sum $=4\cdot2m=8m$, a multiple of 8. So any such sum must be divisible by 8.

Test:
16: $16/8=2$ possible; indeed $1+3+5+7=16$.
40: $40/8=5$ possible; indeed $7+9+11+13=40$.
72: $72/8=9$ possible; e.g. $15+17+19+21=72$.
200: $200/8=25$ possible; e.g. $47+49+51+53=200$.
100: $100/8=12.5$ not integer, so impossible.

Thus 100 cannot be written that way.
ANSWER 8: D

Problem 9:
Blake averaged 78 over four tests. Jenny vs Blake: +10, -10, +20, +20. By how much does Jenny's average exceed Blake's?

Only differences matter. Total points difference over 4 tests:
$+10-10+20+20 = +40$.

Average is total/4, so Jenny's average is $40/4=10$ points higher.

Blake's 78 is irrelevant to the difference. Choices 15,20,25,40 confuse total with average.
ANSWER 9: A

Problem 10:
Compute $(.2)^3/(.02)^2$.

$(.2)^3 = 0.2\times0.2\times0.2 = 0.04\times0.2 = 0.008$.
$(.02)^2 = 0.02\times0.02 = 0.0004$.

Ratio $=0.008/0.0004 = 8/0.4 = 80/4 =20$.

Alternatively with powers of 10: $(2\times10^{-1})^3=8\times10^{-3}$, $(2\times10^{-2})^2=4\times10^{-4}$, ratio $=2\times10^{1}=20$.

So not .2, 2, 10, or 15.
ANSWER 10: E

Problem 1:
We are asked to evaluate the product of the 9 factors:
$$\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{10}\right)$$

First, simplify each factor:
$$1 - \frac{1}{2} = \frac{1}{2}$$
$$1 - \frac{1}{3} = \frac{2}{3}$$
$$1 - \frac{1}{4} = \frac{3}{4}$$
$$\vdots$$
$$1 - \frac{1}{10} = \frac{9}{10}$$

Now, multiply the fractions together:
$$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{9}{10}$$

Notice that this is a telescoping product where the denominator of each fraction cancels with the numerator of the following fraction:
$$\frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \cdots \times \frac{\cancel{9}}{10} = \frac{1}{10}$$

ANSWER 1: A

---

Problem 2:
We are asked to find the sum of the distinct prime integer divisors of $2016$.

First, find the prime factorization of $2016$:
- $2016 = 2 \times 1008$
- $1008 = 2 \times 504$
- $504 = 2 \times 252$
- $252 = 2 \times 126$
- $126 = 2 \times 63$
- $63 = 9 \times 7 = 3^2 \times 7$

Thus, the prime factorization of $2016$ is:
$$2016 = 2^5 \times 3^2 \times 7$$

The distinct prime divisors are $2$, $3$, and $7$. 
Their sum is:
$$2 + 3 + 7 = 12$$

ANSWER 2: B

---

Problem 3:
We are given that the population of Nisos Isles is $200$ in the year $1998$, and it triples every $25$ years. We want to find the year in which the population reaches approximately $6000$.

Let $n$ be the number of $25$-year periods that pass after $1998$. The population can be represented as:
$$P = 200 \times 3^n$$

We want $P \approx 6000$:
$$200 \times 3^n \approx 6000 \implies 3^n \approx \frac{6000}{200} = 30$$

Let's test powers of $3$:
- For $n = 1$: $3^1 = 3 \implies P = 600$
- For $n = 2$: $3^2 = 9 \implies P = 1800$
- For $n = 3$: $3^3 = 27 \implies P = 5400$
- For $n = 4$: $3^4 = 81 \implies P = 16200$

$5400$ is very close to $6000$, so about $n = 3$ periods of $25$ years are needed:
$$3 \times 25 = 75\text{ years}$$

Adding $75$ years to $1998$:
$$1998 + 75 = 2073 \approx 2075$$

ANSWER 3: B

---

Problem 4:
We are given that exactly $74\%$ of the people surveyed answered "yes." We need to find the minimum number of people who could have been surveyed.

Let $N$ be the total number of people surveyed. The number of people who answered "yes" is:
$$74\% \times N = \frac{74}{100} N = \frac{37}{50} N$$

Since the number of people who answered "yes" must be a whole number, $\frac{37}{50} N$ must be an integer. 
Since $37$ and $50$ share no common factors other than $1$ (they are coprime), $N$ must be a multiple of $50$. 

The smallest positive integer multiple of $50$ is $50$. 
With $N = 50$, exactly $\frac{37}{50} \times 50 = 37$ people answered "yes."

ANSWER 4: D

---

Problem 5:
Betty is multiplying $0.075$ by $2.56$.
She calculated $75 \times 256 = 19200$.

Count the total number of digits to the right of the decimal point in the original numbers:
- $0.075$ has $3$ decimal places.
- $2.56$ has $2$ decimal places.

The product must therefore have $3 + 2 = 5$ decimal places:
$$19200 \times 10^{-5} = 0.19200 = 0.192$$

ANSWER 5: B

---

Problem 6:
Julie's speech must last between one-half hour ($30$ minutes) and three-quarters of an hour ($45$ minutes). Her speaking rate is $150$ words per minute.

Calculate the minimum and maximum acceptable word counts:
- Minimum words: $30\text{ minutes} \times 150\text{ words/minute} = 4500\text{ words}$
- Maximum words: $45\text{ minutes} \times 150\text{ words/minute} = 6750\text{ words}$

An appropriate word count must lie in the range $[4500, 6750]$. 
Looking at the choices:
- A. $2250$ (too low)
- B. $3000$ (too low)
- C. $4200$ (too low)
- D. $4350$ (too low)
- E. $5650$ is between $4500$ and $6750$ ($\frac{5650}{150} \approx 37.7$ minutes).

ANSWER 6: E

---

Problem 7:
Ted's grandfather covered $2$ miles on each of the $3$ days. 
First, find the time he actually spent each day using $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
- **Monday:** $\frac{2\text{ miles}}{5\text{ mph}} = \frac{2}{5}\text{ hour} = \frac{2}{5} \times 60 = 24\text{ minutes}$.
- **Wednesday:** $\frac{2\text{ miles}}{3\text{ mph}} = \frac{2}{3}\text{ hour} = \frac{2}{3} \times 60 = 40\text{ minutes}$.
- **Friday:** $\frac{2\text{ miles}}{4\text{ mph}} = \frac{2}{4}\text{ hour} = \frac{1}{2} \times 60 = 30\text{ minutes}$.

Total actual time spent:
$$24 + 40 + 30 = 94\text{ minutes}$$

If he had walked at $4\text{ mph}$ every day, the time spent each day would be $\frac{2}{4}\text{ hour} = 30\text{ minutes}$, so for all $3$ days:
$$3 \times 30 = 90\text{ minutes}$$

He would have spent:
$$94 - 90 = 4\text{ minutes less}$$

ANSWER 7: D

---

Problem 8:
Let four consecutive odd integers be represented as:
$$n, \quad n+2, \quad n+4, \quad n+6$$
where $n$ is an odd integer.

Their sum is:
$$S = n + (n+2) + (n+4) + (n+6) = 4n + 12 = 4(n+3)$$

Since $n$ is an odd integer, $n+3$ must be an even integer, which can be written as $2k$ for some integer $k$.
Substituting this in gives:
$$S = 4(2k) = 8k$$

Therefore, the sum of four consecutive odd integers must be a multiple of $8$.
Let's check the given options:
- A. $16 = 8 \times 2$ ($1 + 3 + 5 + 7 = 16$)
- B. $40 = 8 \times 5$ ($7 + 9 + 11 + 13 = 40$)
- C. $72 = 8 \times 9$ ($15 + 17 + 19 + 21 = 72$)
- D. $100$: not divisible by $8$ ($100 / 8 = 12.5$)
- E. $200 = 8 \times 25$ ($47 + 49 + 51 + 53 = 200$)

Hence, $100$ cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

---

Problem 9:
Let Blake's scores on the four tests be $B_1, B_2, B_3, B_4$.
Jenny's scores are:
- First test: $B_1 + 10$
- Second test: $B_2 - 10$
- Third test: $B_3 + 20$
- Fourth test: $B_4 + 20$

The sum of Jenny's scores minus the sum of Blake's scores is:
$$(B_1 + 10 + B_2 - 10 + B_3 + 20 + B_4 + 20) - (B_1 + B_2 + B_3 + B_4) = 10 - 10 + 20 + 20 = 40\text{ points}$$

Since there are $4$ tests, the difference between Jenny's average and Blake's aver

# Problem 1:
We need the product of the 9 factors (1 − 1/n) for n = 2, 3, 4, …, 10.

Rewrite each factor: 1 − 1/n = (n−1)/n. So the product becomes

(1/2)(2/3)(3/4)(4/5)(5/6)(6/7)(7/8)(8/9)(9/10)

This **telescopes**: each numerator cancels the denominator before it (2 cancels 2, 3 cancels 3, …, 9 cancels 9), leaving only the first numerator and the last denominator:

Product = 1/10

Ruling out traps: every factor is less than 1, so the product must be *smaller* than any single factor — this eliminates D (10/11) and E (11/2) immediately. Choice C (1/2) would be the answer if only the first factor were used, and B (1/9) would result from miscounting the factors (stopping at 1 − 1/9). Since n runs from 2 to 10 inclusive, there are exactly 9 factors and the answer is 1/10.

ANSWER 1: A

---

# Problem 2:
Factor 2016 into primes by repeated division:

2016 = 2 × 1008 = 2² × 504 = 2³ × 252 = 2⁴ × 126 = 2⁵ × 63, and 63 = 3² × 7.

So 2016 = 2⁵ × 3² × 7.

The **distinct** prime divisors are just 2, 3, and 7 (the exponents don't matter — we count each prime once).

Sum = 2 + 3 + 7 = **12**

Ruling out traps: A (9) would come from forgetting the prime 3; E (63) is the odd part 3² × 7 of 2016, not a sum of primes; adding primes *with repetition* gives 2+2+2+2+2+3+3+7 = 23, which isn't even a choice. The correct distinct-prime sum is 12.

ANSWER 2: B

---

# Problem 3:
Starting at 200 people in 1998, the population triples every 25 years:

- 1998: 200
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16 200

We want about 6000. Check how many triplings are needed: 6000/200 = 30, and 3³ = 27 ≈ 30 (while 3⁴ = 81 is way too big). So about **three triplings ≈ 75 years** after 1998, i.e., around 2073.

Comparing to the choices: at 2073 the population is 5400 (off by only 600), while at 2098 it is 16 200 (off by over 10 000). The closest estimate offered is **2075**.

Ruling out traps: 2050 and 2100 bracket the answer but are farther from the true crossing point than 2075; 2125 and 2150 correspond to populations of tens of thousands.

ANSWER 3: B

---

# Problem 4:
If N people were asked and exactly 74% said yes, then 0.74N must be a whole number of people:

0.74N = (74/100)N = (37/50)N

Since 37 is prime and shares no factor with 50, the fraction 37/50 is fully reduced. For (37/50)N to be an integer, **50 must divide N**. The smallest such N is 50.

Check: with 50 people, 0.74 × 50 = 37 people said yes — exactly 74%. ✓

Ruling out the smaller choices: 10 people → 7.4 yes-votes; 20 → 14.8; 25 → 18.5 — none of these is a whole number of people, so each is impossible. Hence 50 is the fewest.

ANSWER 4: D

---

# Problem 5:
First, confirm the calculator display: entering 075 × 256 (no decimal points) gives 75 × 256 = 19 200. ✓ That matches.

Now place the decimal correctly. The rule for multiplying decimals: the product has as many decimal places as the two factors combined.

- 0.075 has **3** decimal places
- 2.56 has **2** decimal places
- Product has 3 + 2 = **5** decimal places

Start from 19 200 and move the decimal point 5 places to the left:

19 200 → 0.19200 = **0.192**

Ruling out traps: 192 (E) and 19.2 (D) come from shifting the point only 2 or 3 places; 0.0192 (A) from shifting 6 places. A quick sanity check: 0.075 × 2.56 should be a bit less than 0.1 × 2.56 = 0.256, and 0.192 fits perfectly.

ANSWER 5: B

---

# Problem 6:
Convert the time limits to minutes, then to words:

- Minimum time: 1/2 hour = 30 minutes → 150 × 30 = **4500 words**
- Maximum time: 3/4 hour = 45 minutes → 150 × 45 = **6750 words**

So the speech needs between 4500 and 6750 words.

Check each choice: 2250 words = 15 min (too short); 3000 = 20 min (too short); 4200 = 28 min (too short); 4350 = 29 min (just under a half hour — too short); 5650 = 5650/150 ≈ 37.7 minutes, which lies nicely between 30 and 45 minutes. ✓

Only 5650 works.

ANSWER 6: E

---

# Problem 7:
Use time = distance ÷ speed, and convert to minutes (1 hour = 60 min):

- Monday: 2 miles at 5 mph → 2/5 hour = 24 minutes
- Wednesday: 2 miles at 3 mph → 2/3 hour = 40 minutes
- Friday: 2 miles at 4 mph → 2/4 = 1/2 hour = 30 minutes

Actual total: 24 + 40 + 30 = **94 minutes**.

If he had always gone 4 mph: total distance 6 miles at 4 mph → 6/4 = 1.5 hours = 90 minutes.

Difference: 94 − 90 = **4 minutes less**.

Ruling out traps: note Monday would actually take *longer* at 4 mph (30 vs. 24 min), so you can't just assume each day saves time — Wednesday's big savings (40 → 30) must be combined with Monday's loss (24 → 30). Net saving = 10 − 6 = 4 minutes, matching the totals.

ANSWER 7: D

---

# Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6, where a is odd. Their sum is

a + (a+2) + (a+4) + (a+6) = 4a + 12 = 4(a + 3).

Since a is odd, a + 3 is **even**, so the sum equals 4 × (an even number), i.e., every such sum is **divisible by 8**.

Test divisibility by 8 for each choice:

- 16 ÷ 8 = 2 ✓ (indeed 1 + 3 + 5 + 7 = 16)
- 40 ÷ 8 = 5 ✓ (7 + 9 + 11 + 13 = 40)
- 72 ÷ 8 = 9 ✓ (15 + 17 + 19 + 21 = 72)
- 100 ÷ 8 = 12.5 ✗ — **not divisible by 8**
- 200 ÷ 8 = 25 ✓ (47 + 49 + 51 + 53 = 200)

So 100 cannot be written as such a sum. (Check directly: 100 = 4(a+3) forces a + 3 = 25, so a = 22 — but 22 is even, a contradiction.)

ANSWER 8: D

---

# Problem 9:
Blake's total over four tests: 4 × 78 = 312 points.

Jenny's scores compared to Blake's, test by test: +10, −10, +20, +20.

Total difference: +10 − 10 + 20 + 20 = **+40 points**.

Jenny's total: 312 + 40 = 352, so her average is 352 ÷ 4 = 88.

Jenny's average exceeds Blake's by 88 − 78 = **10 points**.

(Equivalently: the *average* difference is the total difference divided by the number of tests: 40 ÷ 4 = 10.)

Ruling out traps: E (40) is the total point difference — it forgets to divide by the 4 tests; C (20) might come from averaging only the +20's and ignoring the +10/−10 pair; B and D have no valid basis.

ANSWER 9: A

---

# Problem 10:
Compute the numerator and denominator separately:

- (0.2)³ = 

Problem 1:  
The product is  
\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots\left(1-\frac1{10}\right).
\]
Rewrite each factor:
\[
\frac12\cdot \frac23\cdot \frac34\cdot \frac45\cdots \frac9{10}.
\]
Everything cancels except the first numerator and the last denominator:
\[
\frac1{10}.
\]
So the answer is A.

ANSWER 1: A

Problem 2:  
We need the sum of the distinct prime factors of 2016. Factor:
\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot 252=2^4\cdot 126=2^5\cdot 63.
\]
Also,
\[
63=3^2\cdot 7.
\]
Thus
\[
2016=2^5\cdot 3^2\cdot 7.
\]
The distinct prime divisors are 2, 3, and 7. Their sum is
\[
2+3+7=12.
\]

ANSWER 2: B

Problem 3:  
The population triples every 25 years. Starting with 200 in 1998, after \(t\) years the population is approximately
\[
200\cdot 3^{t/25}.
\]
We want this to be about 6000:
\[
200\cdot 3^{t/25}=6000.
\]
Divide by 200:
\[
3^{t/25}=30.
\]
Now,
\[
3^3=27
\]
and
\[
3^4=81.
\]
So \(t/25\) is a little more than 3, meaning \(t\) is a little more than 75 years.  
\[
1998+75=2073,
\]
so the closest choice is 2075.

ANSWER 3: B

Problem 4:  
Exactly 74% answered “yes.” Write 74% as a fraction in lowest terms:
\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]
For the number of “yes” answers to be an integer, the total number of people asked must be a multiple of 50. The smallest such number among the choices is 50.  
Check: \(\frac{37}{50}=74\%\).

ANSWER 4: D

Problem 5:  
Betty forgot the decimal points, so she effectively multiplied
\[
75\cdot 256=19200.
\]
Now restore the decimal places.  
\(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places, so the correct product must have
\[
3+2=5
\]
decimal places. Therefore,
\[
0.075\cdot 2.56=\frac{19200}{100000}=0.192.
\]

ANSWER 5: B

Problem 6:  
The speech must last between one-half hour and three-quarters of an hour, which is between
\[
30\text{ minutes}
\]
and
\[
45\text{ minutes}.
\]
At 150 words per minute, the acceptable word counts are between
\[
150\cdot 30=4500
\]
and
\[
150\cdot 45=6750.
\]
Among the choices, only 5650 lies between 4500 and 6750.

ANSWER 6: E

Problem 7:  
Each day he went 2 miles.

Monday: speed 5 mph, time
\[
\frac{2}{5}\text{ hour}=24\text{ minutes}.
\]
Wednesday: speed 3 mph, time
\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]
Friday: speed 4 mph, time
\[
\frac{2}{4}\text{ hour}=30\text{ minutes}.
\]
Total time:
\[
24+40+30=94\text{ minutes}.
\]
If he had walked all 6 miles at 4 mph, the time would be
\[
\frac{6}{4}\text{ hour}=1.5\text{ hours}=90\text{ minutes}.
\]
The difference is
\[
94-90=4\text{ minutes}.
\]

ANSWER 7: D

Problem 8:  
Let four consecutive odd integers be
\[
2k+1,\;2k+3,\;2k+5,\;2k+7.
\]
Their sum is
\[
(2k+1)+(2k+3)+(2k+5)+(2k+7)=8k+16=8(k+2).
\]
So any sum of four consecutive odd integers must be divisible by 8.

Check the choices:
- 16 is divisible by 8.
- 40 is divisible by 8.
- 72 is divisible by 8.
- 100 is not divisible by 8.
- 200 is divisible by 8.

Therefore, 100 cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

Problem 9:  
Blake’s average was 78 on four tests, so his total score was
\[
4\cdot 78=312.
\]
Jenny’s scores differed from Blake’s by
\[
+10,\;-10,\;+20,\;+20.
\]
The total difference is
\[
10-10+20+20=40.
\]
So Jenny’s total score was 40 points higher than Blake’s. Since there are 4 tests, her average was higher by
\[
\frac{40}{4}=10.
\]

ANSWER 9: A

Problem 10:  
Compute:
\[
(0.2)^3=0.008
\]
and
\[
(0.02)^2=0.0004.
\]
Thus
\[
\frac{(0.2)^3}{(0.02)^2}=\frac{0.008}{0.0004}=20.
\]

ANSWER 10: E

Problem 1:
Each factor simplifies: (1 − 1/2) = 1/2, (1 − 1/3) = 2/3, …, (1 − 1/10) = 9/10.

Product = (1/2)(2/3)(3/4)⋯(9/10).

Everything telescopes: the numerator of each fraction cancels the denominator of the previous one, leaving 1/10.

(Choice B, 1/9, would come from miscounting the last denominator; the product is definitely less than 1/2, ruling out C, D, E.)

ANSWER 1: A

Problem 2:
Factor 2016: 2016 = 2 · 1008 = 2·2·504 = 2·2·2·252 = 2^5 · 63 = 2^5 · 3^2 · 7.

Distinct primes: 2, 3, 7. Sum = 2 + 3 + 7 = 12.

(9 = 2+7 misses 3; 16 = 2·5+... etc.; 49 and 63 come from adding non‑prime factors or all factors with multiplicity.)

ANSWER 2: B

Problem 3:
Start: 1998 → 200 people. Tripling every 25 years:
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16200

The population reaches about 6000 shortly after 2073, so the best estimate among the choices is 2075.

(2050 gives ~1800, 2100 gives ~16000 — both far off.)

ANSWER 3: B

Problem 4:
We need 74% of the number of people to be a whole number: 0.74n = 37n/50 must be an integer.

Since 37/50 is in lowest terms (37 is prime), n must be a multiple of 50. The smallest is n = 50 (then 37 people said yes).

Check the smaller options: 74% of 10 = 7.4, of 20 = 14.8, of 25 = 18.5 — none are whole numbers.

ANSWER 4: D

Problem 5:
Ignoring decimals: 75 × 256 = 19200, matching the display.

Count decimal places: 0.075 has 3, 2.56 has 2, total 5 decimal places.

So 0.075 × 2.56 = 19200 with the decimal point moved 5 places left = 0.19200 = 0.192.

Sanity check: 0.075 ≈ 0.08 and 2.56 ≈ 2.5, product ≈ 0.2. So 0.192 is right (not 0.0192 or 1.92).

ANSWER 5: B

Problem 6:
Time allowed: 30 minutes to 45 minutes.

At 150 words per minute:
- Minimum words = 150 × 30 = 4500
- Maximum words = 150 × 45 = 6750

We need a count between 4500 and 6750. Only 5650 qualifies.

(2250 would take 15 min, 3000 → 20 min, 4200 → 28 min, 4350 → 29 min — all too short.)

ANSWER 6: E

Problem 7:
Time = distance ÷ speed, with 2 miles each day.

- Monday at 5 mph: 2/5 h = 24 min
- Wednesday at 3 mph: 2/3 h = 40 min
- Friday at 4 mph: 2/4 h = 30 min

Total actual = 24 + 40 + 30 = 94 minutes.

If always 4 mph: 30 min each day → 3 × 30 = 90 minutes.

Difference: 94 − 90 = 4 minutes less.

ANSWER 7: D

Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6 with a odd.

Sum = 4a + 12 = 4(a + 3). Since a is odd, a + 3 is even, say a + 3 = 2k, so the sum = 8k — a multiple of 8.

Check the choices for divisibility by 8:
- 16 = 8·2 ✓ (1+3+5+7)
- 40 = 8·5 ✓ (7+9+11+13)
- 72 = 8·9 ✓ (15+17+19+21)
- 100 ÷ 8 = 12.5 ✗
- 200 = 8·25 ✓ (47+49+51+53)

So 100 cannot be written that way.

ANSWER 8: D

Problem 9:
Jenny's total differs from Blake's total by:
+10 − 10 + 20 + 20 = +40 points.

Average difference = 40 ÷ 4 = 10 points.

(40 is the total difference, not the average — that's the trap in choice E. Blake's 78 average is not even needed, though Jenny averages 88, which is under 100, so it's consistent.)

ANSWER 9: A

Problem 10:
(0.2)³ = 0.008 and (0.02)² = 0.0004.

0.008 ÷ 0.0004 = 80/4 = 20.

(Alternatively: (2×10⁻¹)³ / (2×10⁻²)² = 8×10⁻³ / (4×10⁻⁴) = 2 × 10¹ = 20.)

ANSWER 10: E

Problem 1:

We need to compute

\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots \left(1-\frac1{10}\right).
\]

Rewrite each factor:

\[
1-\frac12=\frac12,\qquad
1-\frac13=\frac23,\qquad
1-\frac14=\frac34,
\]

and so on. Thus the product is

\[
\frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdots\frac9{10}.
\]

Nearly everything cancels:

\[
\frac{\cancel1}{2}\cdot\frac{\cancel2}{3}\cdot\frac{\cancel3}{4}\cdots
\frac{\cancel9}{10}
=\frac1{10}.
\]

Choice B, \(\frac19\), might seem plausible if one thinks there are 9 factors, but the cancellation leaves the first numerator \(1\) and final denominator \(10\).

\[
\boxed{\frac1{10}}
\]

ANSWER 1: A

Problem 2:

We need the sum of the distinct prime divisors of \(2016\).

Factor \(2016\):

\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot252=2^4\cdot126=2^5\cdot63.
\]

Since

\[
63=3^2\cdot7,
\]

we have

\[
2016=2^5\cdot3^2\cdot7.
\]

The distinct prime divisors are

\[
2,\ 3,\ 7.
\]

Their sum is

\[
2+3+7=12.
\]

Do not add repeated prime factors such as all five 2s and both 3s; the question asks for distinct primes only.

\[
\boxed{12}
\]

ANSWER 2: B

Problem 3:

The population starts at \(200\) in 1998 and triples every 25 years.

We want it to be about \(6000\). Compare:

\[
\frac{6000}{200}=30.
\]

So the population must grow by a factor of about \(30\).

Powers of \(3\) are:

\[
3^1=3,\qquad 3^2=9,\qquad 3^3=27,\qquad 3^4=81.
\]

A factor of \(27\) is close to \(30\), so we need about 3 periods of 25 years:

\[
3\cdot25=75\text{ years}.
\]

Therefore the estimated year is

\[
1998+75=2073.
\]

The nearest choice is \(2075\).

Indeed, after 75 years the estimated population is

\[
200\cdot 27=5400,
\]

which is reasonably close to \(6000\). Waiting 100 years would give \(200\cdot81=16200\), much too large.

\[
\boxed{2075}
\]

ANSWER 3: B

Problem 4:

Exactly \(74\%\) of the people answered yes.

Convert \(74\%\) to a fraction:

\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]

For the number of yes answers to be a whole number, the total number surveyed must be a multiple of \(50\).

The smallest possible total is therefore

\[
50.
\]

For example,

\[
74\%\text{ of }50=\frac{74}{100}\cdot 50=37,
\]

which is an integer number of people.

Smaller choices such as 10, 20, and 25 do not work, since \(74\%\) of them is not a whole number.

\[
\boxed{50}
\]

ANSWER 4: D

Problem 5:

Without decimal points, Betty entered

\[
75\times256=19200.
\]

Now place the decimal points correctly.

The number \(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places. Therefore the product must have

\[
3+2=5
\]

decimal places.

So move the decimal point in \(19200\) five places left:

\[
19200\longrightarrow 0.19200=0.192.
\]

Thus

\[
0.075\times2.56=0.192.
\]

Choice A, \(0.0192\), has one too many decimal places, while \(1.92\) and larger answers are too large because \(0.075\) is less than \(0.1\).

\[
\boxed{0.192}
\]

ANSWER 5: B

Problem 6:

Julie’s speech must last between one-half hour and three-quarters of an hour.

Convert these times to minutes:

\[
\frac12\text{ hour}=30\text{ minutes},
\qquad
\frac34\text{ hour}=45\text{ minutes}.
\]

At \(150\) words per minute, the minimum appropriate word count is

\[
30\cdot150=4500.
\]

The maximum appropriate word count is

\[
45\cdot150=6750.
\]

So her speech should contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) and \(3000\) are too few.
- \(4200\) and \(4350\) are still below \(4500\).
- \(5650\) is between \(4500\) and \(6750\).

Therefore the appropriate word count is

\[
\boxed{5650}.
\]

ANSWER 6: E

Problem 7:

Grandfather traveled 2 miles on each of 3 days.

Find his actual treadmill time.

On Monday, he jogged at 5 mph:

\[
\frac{2}{5}\text{ hour}=0.4\text{ hour}=24\text{ minutes}.
\]

On Wednesday, he walked at 3 mph:

\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]

On Friday, he walked at 4 mph:

\[
\frac24\text{ hour}=\frac12\text{ hour}=30\text{ minutes}.
\]

So his actual total time was

\[
24+40+30=94\text{ minutes}.
\]

If he had walked at 4 mph on all three days, each 2-mile trip would take 30 minutes:

\[
3\cdot30=90\text{ minutes}.
\]

Thus he would have spent

\[
94-90=4
\]

fewer minutes on the treadmill.

The 5-mph jogging day saves time, but the 3-mph walking day costs more time; together the actual time is 4 minutes longer than always walking at 4 mph.

\[
\boxed{4}
\]

ANSWER 7: D

Problem 8:

Let the four consecutive odd integers be

\[
n,\ n+2,\ n+4,\ n+6,
\]

where \(n\) is odd.

Their sum is

\[
n+(n+2)+(n+4)+(n+6)=4n+12.
\]

Factor:

\[
4n+12=4(n+3).
\]

Since \(n\) is odd, \(n+3\) is even. Therefore \(4(n+3)\) must be divisible by \(8\).

So a number can be written as the sum of four consecutive odd integers only if it is divisible by \(8\).

Check the choices:

\[
16\div8=2,
\quad
40\div8=5,
\quad
72\div8=9,
\quad
200\div8=25,
\]

so all of these are divisible by 8.

But

\[
100\div8=12.5,
\]

so \(100\) is not divisible by 8. Therefore it cannot be the sum of four consecutive odd integers.

For example, \(16=1+3+5+7\), showing that the condition is possible for the other type of choices.

\[
\boxed{100}
\]

ANSWER 8: D

Problem 9:

Blake’s average is not actually needed, because we can compare Jenny’s scores directly to Blake’s.

Jenny’s score changes relative to Blake’s are:

- first test: \(+10\),
- second test: \(-10\),
- third test: \(+20\),
- fourth test: \(+20\).

Her total score exceeds Blake’s by

\[
10-10+20+20=40.
\]

Since there are four tests, her average exceeds Blake’s by

\[
\frac{40}{4}=10.
\]

The \(40\) is the difference in total points, not the difference in average points.

\[
\boxed{10}
\]

ANSWER 9: A

Problem 10:

Compute

\[
\frac{(0.2)^3}{(0.02)^2}.
\]

First,

\[
(0.2)^3=0.2\cdot0.2\cdot0.2=0.008.
\]

Next,

\[
(0.02)^2=0.02\cdot0.02=0.0004.
\]

Therefore,

\[
\frac{0.008}{0.0004}=20.
\]

Another way is to use powers of 10:

\[
0.2=2\cdot10^{-1},\qquad 0.02=2\cdot10^{-2}.
\]

Then

\[
\fra

Problem 1:
We are asked for the value of $(1-1/2)(1-1/3)\cdots (1-1/10)$.

Write each factor as a single fraction:
$1-1/2 = 1/2$, $1-1/3 = 2/3$, $1-1/4 = 3/4$, ..., $1-1/10 = 9/10$.

So the product is:
$$\frac12 \cdot \frac23 \cdot \frac34 \cdots \frac89 \cdot \frac9{10}.$$

This telescopes: the numerator 2 cancels denominator 2, numerator 3 cancels denominator 3, etc. All cancel except the first denominator 2? Let's see:
$$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10} = \frac{1}{10}.$$
Formally numerator $=1\cdot2\cdots9$, denominator $=2\cdot3\cdots10$, ratio $=1/10$.

Choices B $1/9$, C $1/2$, D $10/11$, E $11/2$ do not result from cancellation.
ANSWER 1: A

Problem 2:
We need the sum of the distinct prime divisors of 2016.

Factor 2016. It is even:
$2016 = 2\times1008 = 2\times2\times504 = 2^2\times504$.
$504 = 2\times252 = 2^2\times252$? Continue systematically:
$2016 /2 =1008$, $/2=504$, $/2=252$, $/2=126$, $/2=63$.
So $2016 = 2^5 \times 63$.
$63 = 7\times9 = 3^2\times7$.

Hence $2016 = 2^5 \cdot 3^2 \cdot 7^1$.
The distinct primes occurring are $2,3,7$.

Their sum is $2+3+7 = 12$.

This rules out A 9, C 16, D 49, E 63 which would come from missing a prime or including powers.
ANSWER 2: B

Problem 3:
In 1998 population is 200, triples every 25 years. When is it about 6000?

Ratio wanted: $6000/200 = 30$. So we need about 30 times growth.

Each 25 years multiply by 3:
After 25 yr: $200\times3=600$.
After 50 yr: $600\times3=1800$.
After 75 yr: $1800\times3=5400$.

5400 is about 6000. So about 3 triplings, i.e. about $3\times25=75$ years after 1998.

$1998+75 = 2073$, closest to 2075 among choices.

2050 is only ~2 triplings, 2100+ is too many.
ANSWER 3: B

Problem 4:
Some $n$ people asked, exactly $74\%$ said yes. Find smallest possible $n$.

Let $k$ be number saying yes. Then $k/n = 74\% = 74/100 = 37/50$.

So $k = 37n/50$ must be an integer. Since 37 and 50 are coprime (gcd=1), $n$ must be a multiple of 50 for $37n/50$ to be integer.

Smallest positive multiple is $n=50$, giving $k=37$.

$n=10,20,25$ give $7.4,14.8,18.5$ yes-votes, impossible. $n=100$ works but is not fewest.
ANSWER 4: D

Problem 5:
True product is $0.075 \times 2.56$. Entering $75 \times 256$ gave $19200$. What is correct with decimals?

$0.075$ has 3 decimal places: $0.075 = 75/1000$.
$2.56$ has 2 decimal places: $2.56 = 256/100$.

So product $= 75\times256 / (1000\times100) = 19200 / 100000$.

$19200/100000 = 0.192$.

In general move decimal 3+2=5 places left. .0192 would be 6 places, 1.92 would be 4 places, etc.
ANSWER 5: B

Problem 6:
Speech must be $1/2$ hour to $3/4$ hour at 150 words per minute. Which word count fits?

Convert hours to minutes: $1/2$ hr $=30$ min, $3/4$ hr $=45$ min.

At 150 wpm:
minimum words $=30\times150 = 4500$.
maximum words $=45\times150 = 6750$.

We need word count between 4500 and 6750 inclusive.

A 2250 is 15 min, B 3000 is 20 min, C 4200 is 28 min, D 4350 is 29 min — all too short. Only E 5650 satisfies $4500 \le 5650 \le 6750$.
ANSWER 6: E

Problem 7:
2 miles each day for 3 days. Speeds: Monday 5 mph, Wednesday 3 mph, Friday 4 mph. Compare total time to time if all 3 days at 4 mph. How many minutes saved?

Time = distance / speed.

Actual:
Monday: $2/5$ hr $=0.4$ hr $=0.4\times60=24$ min.
Wednesday: $2/3$ hr $= (2/3)\times60 =40$ min.
Friday: $2/4=1/2$ hr $=30$ min.
Total $=24+40+30=94$ min.

If always 4 mph: each day $30$ min, total $3\times30=90$ min.

Saving $=94-90=4$ minutes.
ANSWER 7: D

Problem 8:
Which integer cannot be sum of four consecutive odd integers?

Let the four be $n, n+2, n+4, n+6$ with $n$ odd.
Sum $=4n+12 =4(n+3)$.

Since $n$ is odd, $n+3$ is even, say $n+3=2m$. Then sum $=4\cdot2m=8m$, a multiple of 8. So any such sum must be divisible by 8.

Test:
16: $16/8=2$ possible; indeed $1+3+5+7=16$.
40: $40/8=5$ possible; indeed $7+9+11+13=40$.
72: $72/8=9$ possible; e.g. $15+17+19+21=72$.
200: $200/8=25$ possible; e.g. $47+49+51+53=200$.
100: $100/8=12.5$ not integer, so impossible.

Thus 100 cannot be written that way.
ANSWER 8: D

Problem 9:
Blake averaged 78 over four tests. Jenny vs Blake: +10, -10, +20, +20. By how much does Jenny's average exceed Blake's?

Only differences matter. Total points difference over 4 tests:
$+10-10+20+20 = +40$.

Average is total/4, so Jenny's average is $40/4=10$ points higher.

Blake's 78 is irrelevant to the difference. Choices 15,20,25,40 confuse total with average.
ANSWER 9: A

Problem 10:
Compute $(.2)^3/(.02)^2$.

$(.2)^3 = 0.2\times0.2\times0.2 = 0.04\times0.2 = 0.008$.
$(.02)^2 = 0.02\times0.02 = 0.0004$.

Ratio $=0.008/0.0004 = 8/0.4 = 80/4 =20$.

Alternatively with powers of 10: $(2\times10^{-1})^3=8\times10^{-3}$, $(2\times10^{-2})^2=4\times10^{-4}$, ratio $=2\times10^{1}=20$.

So not .2, 2, 10, or 15.
ANSWER 10: E

Problem 1:
We are asked to evaluate the product of the 9 factors:
$$\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{10}\right)$$

First, simplify each factor:
$$1 - \frac{1}{2} = \frac{1}{2}$$
$$1 - \frac{1}{3} = \frac{2}{3}$$
$$1 - \frac{1}{4} = \frac{3}{4}$$
$$\vdots$$
$$1 - \frac{1}{10} = \frac{9}{10}$$

Now, multiply the fractions together:
$$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{9}{10}$$

Notice that this is a telescoping product where the denominator of each fraction cancels with the numerator of the following fraction:
$$\frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \cdots \times \frac{\cancel{9}}{10} = \frac{1}{10}$$

ANSWER 1: A

---

Problem 2:
We are asked to find the sum of the distinct prime integer divisors of $2016$.

First, find the prime factorization of $2016$:
- $2016 = 2 \times 1008$
- $1008 = 2 \times 504$
- $504 = 2 \times 252$
- $252 = 2 \times 126$
- $126 = 2 \times 63$
- $63 = 9 \times 7 = 3^2 \times 7$

Thus, the prime factorization of $2016$ is:
$$2016 = 2^5 \times 3^2 \times 7$$

The distinct prime divisors are $2$, $3$, and $7$. 
Their sum is:
$$2 + 3 + 7 = 12$$

ANSWER 2: B

---

Problem 3:
We are given that the population of Nisos Isles is $200$ in the year $1998$, and it triples every $25$ years. We want to find the year in which the population reaches approximately $6000$.

Let $n$ be the number of $25$-year periods that pass after $1998$. The population can be represented as:
$$P = 200 \times 3^n$$

We want $P \approx 6000$:
$$200 \times 3^n \approx 6000 \implies 3^n \approx \frac{6000}{200} = 30$$

Let's test powers of $3$:
- For $n = 1$: $3^1 = 3 \implies P = 600$
- For $n = 2$: $3^2 = 9 \implies P = 1800$
- For $n = 3$: $3^3 = 27 \implies P = 5400$
- For $n = 4$: $3^4 = 81 \implies P = 16200$

$5400$ is very close to $6000$, so about $n = 3$ periods of $25$ years are needed:
$$3 \times 25 = 75\text{ years}$$

Adding $75$ years to $1998$:
$$1998 + 75 = 2073 \approx 2075$$

ANSWER 3: B

---

Problem 4:
We are given that exactly $74\%$ of the people surveyed answered "yes." We need to find the minimum number of people who could have been surveyed.

Let $N$ be the total number of people surveyed. The number of people who answered "yes" is:
$$74\% \times N = \frac{74}{100} N = \frac{37}{50} N$$

Since the number of people who answered "yes" must be a whole number, $\frac{37}{50} N$ must be an integer. 
Since $37$ and $50$ share no common factors other than $1$ (they are coprime), $N$ must be a multiple of $50$. 

The smallest positive integer multiple of $50$ is $50$. 
With $N = 50$, exactly $\frac{37}{50} \times 50 = 37$ people answered "yes."

ANSWER 4: D

---

Problem 5:
Betty is multiplying $0.075$ by $2.56$.
She calculated $75 \times 256 = 19200$.

Count the total number of digits to the right of the decimal point in the original numbers:
- $0.075$ has $3$ decimal places.
- $2.56$ has $2$ decimal places.

The product must therefore have $3 + 2 = 5$ decimal places:
$$19200 \times 10^{-5} = 0.19200 = 0.192$$

ANSWER 5: B

---

Problem 6:
Julie's speech must last between one-half hour ($30$ minutes) and three-quarters of an hour ($45$ minutes). Her speaking rate is $150$ words per minute.

Calculate the minimum and maximum acceptable word counts:
- Minimum words: $30\text{ minutes} \times 150\text{ words/minute} = 4500\text{ words}$
- Maximum words: $45\text{ minutes} \times 150\text{ words/minute} = 6750\text{ words}$

An appropriate word count must lie in the range $[4500, 6750]$. 
Looking at the choices:
- A. $2250$ (too low)
- B. $3000$ (too low)
- C. $4200$ (too low)
- D. $4350$ (too low)
- E. $5650$ is between $4500$ and $6750$ ($\frac{5650}{150} \approx 37.7$ minutes).

ANSWER 6: E

---

Problem 7:
Ted's grandfather covered $2$ miles on each of the $3$ days. 
First, find the time he actually spent each day using $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
- **Monday:** $\frac{2\text{ miles}}{5\text{ mph}} = \frac{2}{5}\text{ hour} = \frac{2}{5} \times 60 = 24\text{ minutes}$.
- **Wednesday:** $\frac{2\text{ miles}}{3\text{ mph}} = \frac{2}{3}\text{ hour} = \frac{2}{3} \times 60 = 40\text{ minutes}$.
- **Friday:** $\frac{2\text{ miles}}{4\text{ mph}} = \frac{2}{4}\text{ hour} = \frac{1}{2} \times 60 = 30\text{ minutes}$.

Total actual time spent:
$$24 + 40 + 30 = 94\text{ minutes}$$

If he had walked at $4\text{ mph}$ every day, the time spent each day would be $\frac{2}{4}\text{ hour} = 30\text{ minutes}$, so for all $3$ days:
$$3 \times 30 = 90\text{ minutes}$$

He would have spent:
$$94 - 90 = 4\text{ minutes less}$$

ANSWER 7: D

---

Problem 8:
Let four consecutive odd integers be represented as:
$$n, \quad n+2, \quad n+4, \quad n+6$$
where $n$ is an odd integer.

Their sum is:
$$S = n + (n+2) + (n+4) + (n+6) = 4n + 12 = 4(n+3)$$

Since $n$ is an odd integer, $n+3$ must be an even integer, which can be written as $2k$ for some integer $k$.
Substituting this in gives:
$$S = 4(2k) = 8k$$

Therefore, the sum of four consecutive odd integers must be a multiple of $8$.
Let's check the given options:
- A. $16 = 8 \times 2$ ($1 + 3 + 5 + 7 = 16$)
- B. $40 = 8 \times 5$ ($7 + 9 + 11 + 13 = 40$)
- C. $72 = 8 \times 9$ ($15 + 17 + 19 + 21 = 72$)
- D. $100$: not divisible by $8$ ($100 / 8 = 12.5$)
- E. $200 = 8 \times 25$ ($47 + 49 + 51 + 53 = 200$)

Hence, $100$ cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

---

Problem 9:
Let Blake's scores on the four tests be $B_1, B_2, B_3, B_4$.
Jenny's scores are:
- First test: $B_1 + 10$
- Second test: $B_2 - 10$
- Third test: $B_3 + 20$
- Fourth test: $B_4 + 20$

The sum of Jenny's scores minus the sum of Blake's scores is:
$$(B_1 + 10 + B_2 - 10 + B_3 + 20 + B_4 + 20) - (B_1 + B_2 + B_3 + B_4) = 10 - 10 + 20 + 20 = 40\text{ points}$$

Since there are $4$ tests, the difference between Jenny's average and Blake's aver

# Problem 1:
We need the product of the 9 factors (1 − 1/n) for n = 2, 3, 4, …, 10.

Rewrite each factor: 1 − 1/n = (n−1)/n. So the product becomes

(1/2)(2/3)(3/4)(4/5)(5/6)(6/7)(7/8)(8/9)(9/10)

This **telescopes**: each numerator cancels the denominator before it (2 cancels 2, 3 cancels 3, …, 9 cancels 9), leaving only the first numerator and the last denominator:

Product = 1/10

Ruling out traps: every factor is less than 1, so the product must be *smaller* than any single factor — this eliminates D (10/11) and E (11/2) immediately. Choice C (1/2) would be the answer if only the first factor were used, and B (1/9) would result from miscounting the factors (stopping at 1 − 1/9). Since n runs from 2 to 10 inclusive, there are exactly 9 factors and the answer is 1/10.

ANSWER 1: A

---

# Problem 2:
Factor 2016 into primes by repeated division:

2016 = 2 × 1008 = 2² × 504 = 2³ × 252 = 2⁴ × 126 = 2⁵ × 63, and 63 = 3² × 7.

So 2016 = 2⁵ × 3² × 7.

The **distinct** prime divisors are just 2, 3, and 7 (the exponents don't matter — we count each prime once).

Sum = 2 + 3 + 7 = **12**

Ruling out traps: A (9) would come from forgetting the prime 3; E (63) is the odd part 3² × 7 of 2016, not a sum of primes; adding primes *with repetition* gives 2+2+2+2+2+3+3+7 = 23, which isn't even a choice. The correct distinct-prime sum is 12.

ANSWER 2: B

---

# Problem 3:
Starting at 200 people in 1998, the population triples every 25 years:

- 1998: 200
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16 200

We want about 6000. Check how many triplings are needed: 6000/200 = 30, and 3³ = 27 ≈ 30 (while 3⁴ = 81 is way too big). So about **three triplings ≈ 75 years** after 1998, i.e., around 2073.

Comparing to the choices: at 2073 the population is 5400 (off by only 600), while at 2098 it is 16 200 (off by over 10 000). The closest estimate offered is **2075**.

Ruling out traps: 2050 and 2100 bracket the answer but are farther from the true crossing point than 2075; 2125 and 2150 correspond to populations of tens of thousands.

ANSWER 3: B

---

# Problem 4:
If N people were asked and exactly 74% said yes, then 0.74N must be a whole number of people:

0.74N = (74/100)N = (37/50)N

Since 37 is prime and shares no factor with 50, the fraction 37/50 is fully reduced. For (37/50)N to be an integer, **50 must divide N**. The smallest such N is 50.

Check: with 50 people, 0.74 × 50 = 37 people said yes — exactly 74%. ✓

Ruling out the smaller choices: 10 people → 7.4 yes-votes; 20 → 14.8; 25 → 18.5 — none of these is a whole number of people, so each is impossible. Hence 50 is the fewest.

ANSWER 4: D

---

# Problem 5:
First, confirm the calculator display: entering 075 × 256 (no decimal points) gives 75 × 256 = 19 200. ✓ That matches.

Now place the decimal correctly. The rule for multiplying decimals: the product has as many decimal places as the two factors combined.

- 0.075 has **3** decimal places
- 2.56 has **2** decimal places
- Product has 3 + 2 = **5** decimal places

Start from 19 200 and move the decimal point 5 places to the left:

19 200 → 0.19200 = **0.192**

Ruling out traps: 192 (E) and 19.2 (D) come from shifting the point only 2 or 3 places; 0.0192 (A) from shifting 6 places. A quick sanity check: 0.075 × 2.56 should be a bit less than 0.1 × 2.56 = 0.256, and 0.192 fits perfectly.

ANSWER 5: B

---

# Problem 6:
Convert the time limits to minutes, then to words:

- Minimum time: 1/2 hour = 30 minutes → 150 × 30 = **4500 words**
- Maximum time: 3/4 hour = 45 minutes → 150 × 45 = **6750 words**

So the speech needs between 4500 and 6750 words.

Check each choice: 2250 words = 15 min (too short); 3000 = 20 min (too short); 4200 = 28 min (too short); 4350 = 29 min (just under a half hour — too short); 5650 = 5650/150 ≈ 37.7 minutes, which lies nicely between 30 and 45 minutes. ✓

Only 5650 works.

ANSWER 6: E

---

# Problem 7:
Use time = distance ÷ speed, and convert to minutes (1 hour = 60 min):

- Monday: 2 miles at 5 mph → 2/5 hour = 24 minutes
- Wednesday: 2 miles at 3 mph → 2/3 hour = 40 minutes
- Friday: 2 miles at 4 mph → 2/4 = 1/2 hour = 30 minutes

Actual total: 24 + 40 + 30 = **94 minutes**.

If he had always gone 4 mph: total distance 6 miles at 4 mph → 6/4 = 1.5 hours = 90 minutes.

Difference: 94 − 90 = **4 minutes less**.

Ruling out traps: note Monday would actually take *longer* at 4 mph (30 vs. 24 min), so you can't just assume each day saves time — Wednesday's big savings (40 → 30) must be combined with Monday's loss (24 → 30). Net saving = 10 − 6 = 4 minutes, matching the totals.

ANSWER 7: D

---

# Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6, where a is odd. Their sum is

a + (a+2) + (a+4) + (a+6) = 4a + 12 = 4(a + 3).

Since a is odd, a + 3 is **even**, so the sum equals 4 × (an even number), i.e., every such sum is **divisible by 8**.

Test divisibility by 8 for each choice:

- 16 ÷ 8 = 2 ✓ (indeed 1 + 3 + 5 + 7 = 16)
- 40 ÷ 8 = 5 ✓ (7 + 9 + 11 + 13 = 40)
- 72 ÷ 8 = 9 ✓ (15 + 17 + 19 + 21 = 72)
- 100 ÷ 8 = 12.5 ✗ — **not divisible by 8**
- 200 ÷ 8 = 25 ✓ (47 + 49 + 51 + 53 = 200)

So 100 cannot be written as such a sum. (Check directly: 100 = 4(a+3) forces a + 3 = 25, so a = 22 — but 22 is even, a contradiction.)

ANSWER 8: D

---

# Problem 9:
Blake's total over four tests: 4 × 78 = 312 points.

Jenny's scores compared to Blake's, test by test: +10, −10, +20, +20.

Total difference: +10 − 10 + 20 + 20 = **+40 points**.

Jenny's total: 312 + 40 = 352, so her average is 352 ÷ 4 = 88.

Jenny's average exceeds Blake's by 88 − 78 = **10 points**.

(Equivalently: the *average* difference is the total difference divided by the number of tests: 40 ÷ 4 = 10.)

Ruling out traps: E (40) is the total point difference — it forgets to divide by the 4 tests; C (20) might come from averaging only the +20's and ignoring the +10/−10 pair; B and D have no valid basis.

ANSWER 9: A

---

# Problem 10:
Compute the numerator and denominator separately:

- (0.2)³ = 

Problem 1:  
The product is  
\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots\left(1-\frac1{10}\right).
\]
Rewrite each factor:
\[
\frac12\cdot \frac23\cdot \frac34\cdot \frac45\cdots \frac9{10}.
\]
Everything cancels except the first numerator and the last denominator:
\[
\frac1{10}.
\]
So the answer is A.

ANSWER 1: A

Problem 2:  
We need the sum of the distinct prime factors of 2016. Factor:
\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot 252=2^4\cdot 126=2^5\cdot 63.
\]
Also,
\[
63=3^2\cdot 7.
\]
Thus
\[
2016=2^5\cdot 3^2\cdot 7.
\]
The distinct prime divisors are 2, 3, and 7. Their sum is
\[
2+3+7=12.
\]

ANSWER 2: B

Problem 3:  
The population triples every 25 years. Starting with 200 in 1998, after \(t\) years the population is approximately
\[
200\cdot 3^{t/25}.
\]
We want this to be about 6000:
\[
200\cdot 3^{t/25}=6000.
\]
Divide by 200:
\[
3^{t/25}=30.
\]
Now,
\[
3^3=27
\]
and
\[
3^4=81.
\]
So \(t/25\) is a little more than 3, meaning \(t\) is a little more than 75 years.  
\[
1998+75=2073,
\]
so the closest choice is 2075.

ANSWER 3: B

Problem 4:  
Exactly 74% answered “yes.” Write 74% as a fraction in lowest terms:
\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]
For the number of “yes” answers to be an integer, the total number of people asked must be a multiple of 50. The smallest such number among the choices is 50.  
Check: \(\frac{37}{50}=74\%\).

ANSWER 4: D

Problem 5:  
Betty forgot the decimal points, so she effectively multiplied
\[
75\cdot 256=19200.
\]
Now restore the decimal places.  
\(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places, so the correct product must have
\[
3+2=5
\]
decimal places. Therefore,
\[
0.075\cdot 2.56=\frac{19200}{100000}=0.192.
\]

ANSWER 5: B

Problem 6:  
The speech must last between one-half hour and three-quarters of an hour, which is between
\[
30\text{ minutes}
\]
and
\[
45\text{ minutes}.
\]
At 150 words per minute, the acceptable word counts are between
\[
150\cdot 30=4500
\]
and
\[
150\cdot 45=6750.
\]
Among the choices, only 5650 lies between 4500 and 6750.

ANSWER 6: E

Problem 7:  
Each day he went 2 miles.

Monday: speed 5 mph, time
\[
\frac{2}{5}\text{ hour}=24\text{ minutes}.
\]
Wednesday: speed 3 mph, time
\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]
Friday: speed 4 mph, time
\[
\frac{2}{4}\text{ hour}=30\text{ minutes}.
\]
Total time:
\[
24+40+30=94\text{ minutes}.
\]
If he had walked all 6 miles at 4 mph, the time would be
\[
\frac{6}{4}\text{ hour}=1.5\text{ hours}=90\text{ minutes}.
\]
The difference is
\[
94-90=4\text{ minutes}.
\]

ANSWER 7: D

Problem 8:  
Let four consecutive odd integers be
\[
2k+1,\;2k+3,\;2k+5,\;2k+7.
\]
Their sum is
\[
(2k+1)+(2k+3)+(2k+5)+(2k+7)=8k+16=8(k+2).
\]
So any sum of four consecutive odd integers must be divisible by 8.

Check the choices:
- 16 is divisible by 8.
- 40 is divisible by 8.
- 72 is divisible by 8.
- 100 is not divisible by 8.
- 200 is divisible by 8.

Therefore, 100 cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

Problem 9:  
Blake’s average was 78 on four tests, so his total score was
\[
4\cdot 78=312.
\]
Jenny’s scores differed from Blake’s by
\[
+10,\;-10,\;+20,\;+20.
\]
The total difference is
\[
10-10+20+20=40.
\]
So Jenny’s total score was 40 points higher than Blake’s. Since there are 4 tests, her average was higher by
\[
\frac{40}{4}=10.
\]

ANSWER 9: A

Problem 10:  
Compute:
\[
(0.2)^3=0.008
\]
and
\[
(0.02)^2=0.0004.
\]
Thus
\[
\frac{(0.2)^3}{(0.02)^2}=\frac{0.008}{0.0004}=20.
\]

ANSWER 10: E

Problem 1:
Each factor simplifies: (1 − 1/2) = 1/2, (1 − 1/3) = 2/3, …, (1 − 1/10) = 9/10.

Product = (1/2)(2/3)(3/4)⋯(9/10).

Everything telescopes: the numerator of each fraction cancels the denominator of the previous one, leaving 1/10.

(Choice B, 1/9, would come from miscounting the last denominator; the product is definitely less than 1/2, ruling out C, D, E.)

ANSWER 1: A

Problem 2:
Factor 2016: 2016 = 2 · 1008 = 2·2·504 = 2·2·2·252 = 2^5 · 63 = 2^5 · 3^2 · 7.

Distinct primes: 2, 3, 7. Sum = 2 + 3 + 7 = 12.

(9 = 2+7 misses 3; 16 = 2·5+... etc.; 49 and 63 come from adding non‑prime factors or all factors with multiplicity.)

ANSWER 2: B

Problem 3:
Start: 1998 → 200 people. Tripling every 25 years:
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16200

The population reaches about 6000 shortly after 2073, so the best estimate among the choices is 2075.

(2050 gives ~1800, 2100 gives ~16000 — both far off.)

ANSWER 3: B

Problem 4:
We need 74% of the number of people to be a whole number: 0.74n = 37n/50 must be an integer.

Since 37/50 is in lowest terms (37 is prime), n must be a multiple of 50. The smallest is n = 50 (then 37 people said yes).

Check the smaller options: 74% of 10 = 7.4, of 20 = 14.8, of 25 = 18.5 — none are whole numbers.

ANSWER 4: D

Problem 5:
Ignoring decimals: 75 × 256 = 19200, matching the display.

Count decimal places: 0.075 has 3, 2.56 has 2, total 5 decimal places.

So 0.075 × 2.56 = 19200 with the decimal point moved 5 places left = 0.19200 = 0.192.

Sanity check: 0.075 ≈ 0.08 and 2.56 ≈ 2.5, product ≈ 0.2. So 0.192 is right (not 0.0192 or 1.92).

ANSWER 5: B

Problem 6:
Time allowed: 30 minutes to 45 minutes.

At 150 words per minute:
- Minimum words = 150 × 30 = 4500
- Maximum words = 150 × 45 = 6750

We need a count between 4500 and 6750. Only 5650 qualifies.

(2250 would take 15 min, 3000 → 20 min, 4200 → 28 min, 4350 → 29 min — all too short.)

ANSWER 6: E

Problem 7:
Time = distance ÷ speed, with 2 miles each day.

- Monday at 5 mph: 2/5 h = 24 min
- Wednesday at 3 mph: 2/3 h = 40 min
- Friday at 4 mph: 2/4 h = 30 min

Total actual = 24 + 40 + 30 = 94 minutes.

If always 4 mph: 30 min each day → 3 × 30 = 90 minutes.

Difference: 94 − 90 = 4 minutes less.

ANSWER 7: D

Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6 with a odd.

Sum = 4a + 12 = 4(a + 3). Since a is odd, a + 3 is even, say a + 3 = 2k, so the sum = 8k — a multiple of 8.

Check the choices for divisibility by 8:
- 16 = 8·2 ✓ (1+3+5+7)
- 40 = 8·5 ✓ (7+9+11+13)
- 72 = 8·9 ✓ (15+17+19+21)
- 100 ÷ 8 = 12.5 ✗
- 200 = 8·25 ✓ (47+49+51+53)

So 100 cannot be written that way.

ANSWER 8: D

Problem 9:
Jenny's total differs from Blake's total by:
+10 − 10 + 20 + 20 = +40 points.

Average difference = 40 ÷ 4 = 10 points.

(40 is the total difference, not the average — that's the trap in choice E. Blake's 78 average is not even needed, though Jenny averages 88, which is under 100, so it's consistent.)

ANSWER 9: A

Problem 10:
(0.2)³ = 0.008 and (0.02)² = 0.0004.

0.008 ÷ 0.0004 = 80/4 = 20.

(Alternatively: (2×10⁻¹)³ / (2×10⁻²)² = 8×10⁻³ / (4×10⁻⁴) = 2 × 10¹ = 20.)

ANSWER 10: E

Problem 1:

We need to compute

\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots \left(1-\frac1{10}\right).
\]

Rewrite each factor:

\[
1-\frac12=\frac12,\qquad
1-\frac13=\frac23,\qquad
1-\frac14=\frac34,
\]

and so on. Thus the product is

\[
\frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdots\frac9{10}.
\]

Nearly everything cancels:

\[
\frac{\cancel1}{2}\cdot\frac{\cancel2}{3}\cdot\frac{\cancel3}{4}\cdots
\frac{\cancel9}{10}
=\frac1{10}.
\]

Choice B, \(\frac19\), might seem plausible if one thinks there are 9 factors, but the cancellation leaves the first numerator \(1\) and final denominator \(10\).

\[
\boxed{\frac1{10}}
\]

ANSWER 1: A

Problem 2:

We need the sum of the distinct prime divisors of \(2016\).

Factor \(2016\):

\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot252=2^4\cdot126=2^5\cdot63.
\]

Since

\[
63=3^2\cdot7,
\]

we have

\[
2016=2^5\cdot3^2\cdot7.
\]

The distinct prime divisors are

\[
2,\ 3,\ 7.
\]

Their sum is

\[
2+3+7=12.
\]

Do not add repeated prime factors such as all five 2s and both 3s; the question asks for distinct primes only.

\[
\boxed{12}
\]

ANSWER 2: B

Problem 3:

The population starts at \(200\) in 1998 and triples every 25 years.

We want it to be about \(6000\). Compare:

\[
\frac{6000}{200}=30.
\]

So the population must grow by a factor of about \(30\).

Powers of \(3\) are:

\[
3^1=3,\qquad 3^2=9,\qquad 3^3=27,\qquad 3^4=81.
\]

A factor of \(27\) is close to \(30\), so we need about 3 periods of 25 years:

\[
3\cdot25=75\text{ years}.
\]

Therefore the estimated year is

\[
1998+75=2073.
\]

The nearest choice is \(2075\).

Indeed, after 75 years the estimated population is

\[
200\cdot 27=5400,
\]

which is reasonably close to \(6000\). Waiting 100 years would give \(200\cdot81=16200\), much too large.

\[
\boxed{2075}
\]

ANSWER 3: B

Problem 4:

Exactly \(74\%\) of the people answered yes.

Convert \(74\%\) to a fraction:

\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]

For the number of yes answers to be a whole number, the total number surveyed must be a multiple of \(50\).

The smallest possible total is therefore

\[
50.
\]

For example,

\[
74\%\text{ of }50=\frac{74}{100}\cdot 50=37,
\]

which is an integer number of people.

Smaller choices such as 10, 20, and 25 do not work, since \(74\%\) of them is not a whole number.

\[
\boxed{50}
\]

ANSWER 4: D

Problem 5:

Without decimal points, Betty entered

\[
75\times256=19200.
\]

Now place the decimal points correctly.

The number \(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places. Therefore the product must have

\[
3+2=5
\]

decimal places.

So move the decimal point in \(19200\) five places left:

\[
19200\longrightarrow 0.19200=0.192.
\]

Thus

\[
0.075\times2.56=0.192.
\]

Choice A, \(0.0192\), has one too many decimal places, while \(1.92\) and larger answers are too large because \(0.075\) is less than \(0.1\).

\[
\boxed{0.192}
\]

ANSWER 5: B

Problem 6:

Julie’s speech must last between one-half hour and three-quarters of an hour.

Convert these times to minutes:

\[
\frac12\text{ hour}=30\text{ minutes},
\qquad
\frac34\text{ hour}=45\text{ minutes}.
\]

At \(150\) words per minute, the minimum appropriate word count is

\[
30\cdot150=4500.
\]

The maximum appropriate word count is

\[
45\cdot150=6750.
\]

So her speech should contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) and \(3000\) are too few.
- \(4200\) and \(4350\) are still below \(4500\).
- \(5650\) is between \(4500\) and \(6750\).

Therefore the appropriate word count is

\[
\boxed{5650}.
\]

ANSWER 6: E

Problem 7:

Grandfather traveled 2 miles on each of 3 days.

Find his actual treadmill time.

On Monday, he jogged at 5 mph:

\[
\frac{2}{5}\text{ hour}=0.4\text{ hour}=24\text{ minutes}.
\]

On Wednesday, he walked at 3 mph:

\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]

On Friday, he walked at 4 mph:

\[
\frac24\text{ hour}=\frac12\text{ hour}=30\text{ minutes}.
\]

So his actual total time was

\[
24+40+30=94\text{ minutes}.
\]

If he had walked at 4 mph on all three days, each 2-mile trip would take 30 minutes:

\[
3\cdot30=90\text{ minutes}.
\]

Thus he would have spent

\[
94-90=4
\]

fewer minutes on the treadmill.

The 5-mph jogging day saves time, but the 3-mph walking day costs more time; together the actual time is 4 minutes longer than always walking at 4 mph.

\[
\boxed{4}
\]

ANSWER 7: D

Problem 8:

Let the four consecutive odd integers be

\[
n,\ n+2,\ n+4,\ n+6,
\]

where \(n\) is odd.

Their sum is

\[
n+(n+2)+(n+4)+(n+6)=4n+12.
\]

Factor:

\[
4n+12=4(n+3).
\]

Since \(n\) is odd, \(n+3\) is even. Therefore \(4(n+3)\) must be divisible by \(8\).

So a number can be written as the sum of four consecutive odd integers only if it is divisible by \(8\).

Check the choices:

\[
16\div8=2,
\quad
40\div8=5,
\quad
72\div8=9,
\quad
200\div8=25,
\]

so all of these are divisible by 8.

But

\[
100\div8=12.5,
\]

so \(100\) is not divisible by 8. Therefore it cannot be the sum of four consecutive odd integers.

For example, \(16=1+3+5+7\), showing that the condition is possible for the other type of choices.

\[
\boxed{100}
\]

ANSWER 8: D

Problem 9:

Blake’s average is not actually needed, because we can compare Jenny’s scores directly to Blake’s.

Jenny’s score changes relative to Blake’s are:

- first test: \(+10\),
- second test: \(-10\),
- third test: \(+20\),
- fourth test: \(+20\).

Her total score exceeds Blake’s by

\[
10-10+20+20=40.
\]

Since there are four tests, her average exceeds Blake’s by

\[
\frac{40}{4}=10.
\]

The \(40\) is the difference in total points, not the difference in average points.

\[
\boxed{10}
\]

ANSWER 9: A

Problem 10:

Compute

\[
\frac{(0.2)^3}{(0.02)^2}.
\]

First,

\[
(0.2)^3=0.2\cdot0.2\cdot0.2=0.008.
\]

Next,

\[
(0.02)^2=0.02\cdot0.02=0.0004.
\]

Therefore,

\[
\frac{0.008}{0.0004}=20.
\]

Another way is to use powers of 10:

\[
0.2=2\cdot10^{-1},\qquad 0.02=2\cdot10^{-2}.
\]

Then

\[
\fra

Problem 1:
We are asked for the value of $(1-1/2)(1-1/3)\cdots (1-1/10)$.

Write each factor as a single fraction:
$1-1/2 = 1/2$, $1-1/3 = 2/3$, $1-1/4 = 3/4$, ..., $1-1/10 = 9/10$.

So the product is:
$$\frac12 \cdot \frac23 \cdot \frac34 \cdots \frac89 \cdot \frac9{10}.$$

This telescopes: the numerator 2 cancels denominator 2, numerator 3 cancels denominator 3, etc. All cancel except the first denominator 2? Let's see:
$$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10} = \frac{1}{10}.$$
Formally numerator $=1\cdot2\cdots9$, denominator $=2\cdot3\cdots10$, ratio $=1/10$.

Choices B $1/9$, C $1/2$, D $10/11$, E $11/2$ do not result from cancellation.
ANSWER 1: A

Problem 2:
We need the sum of the distinct prime divisors of 2016.

Factor 2016. It is even:
$2016 = 2\times1008 = 2\times2\times504 = 2^2\times504$.
$504 = 2\times252 = 2^2\times252$? Continue systematically:
$2016 /2 =1008$, $/2=504$, $/2=252$, $/2=126$, $/2=63$.
So $2016 = 2^5 \times 63$.
$63 = 7\times9 = 3^2\times7$.

Hence $2016 = 2^5 \cdot 3^2 \cdot 7^1$.
The distinct primes occurring are $2,3,7$.

Their sum is $2+3+7 = 12$.

This rules out A 9, C 16, D 49, E 63 which would come from missing a prime or including powers.
ANSWER 2: B

Problem 3:
In 1998 population is 200, triples every 25 years. When is it about 6000?

Ratio wanted: $6000/200 = 30$. So we need about 30 times growth.

Each 25 years multiply by 3:
After 25 yr: $200\times3=600$.
After 50 yr: $600\times3=1800$.
After 75 yr: $1800\times3=5400$.

5400 is about 6000. So about 3 triplings, i.e. about $3\times25=75$ years after 1998.

$1998+75 = 2073$, closest to 2075 among choices.

2050 is only ~2 triplings, 2100+ is too many.
ANSWER 3: B

Problem 4:
Some $n$ people asked, exactly $74\%$ said yes. Find smallest possible $n$.

Let $k$ be number saying yes. Then $k/n = 74\% = 74/100 = 37/50$.

So $k = 37n/50$ must be an integer. Since 37 and 50 are coprime (gcd=1), $n$ must be a multiple of 50 for $37n/50$ to be integer.

Smallest positive multiple is $n=50$, giving $k=37$.

$n=10,20,25$ give $7.4,14.8,18.5$ yes-votes, impossible. $n=100$ works but is not fewest.
ANSWER 4: D

Problem 5:
True product is $0.075 \times 2.56$. Entering $75 \times 256$ gave $19200$. What is correct with decimals?

$0.075$ has 3 decimal places: $0.075 = 75/1000$.
$2.56$ has 2 decimal places: $2.56 = 256/100$.

So product $= 75\times256 / (1000\times100) = 19200 / 100000$.

$19200/100000 = 0.192$.

In general move decimal 3+2=5 places left. .0192 would be 6 places, 1.92 would be 4 places, etc.
ANSWER 5: B

Problem 6:
Speech must be $1/2$ hour to $3/4$ hour at 150 words per minute. Which word count fits?

Convert hours to minutes: $1/2$ hr $=30$ min, $3/4$ hr $=45$ min.

At 150 wpm:
minimum words $=30\times150 = 4500$.
maximum words $=45\times150 = 6750$.

We need word count between 4500 and 6750 inclusive.

A 2250 is 15 min, B 3000 is 20 min, C 4200 is 28 min, D 4350 is 29 min — all too short. Only E 5650 satisfies $4500 \le 5650 \le 6750$.
ANSWER 6: E

Problem 7:
2 miles each day for 3 days. Speeds: Monday 5 mph, Wednesday 3 mph, Friday 4 mph. Compare total time to time if all 3 days at 4 mph. How many minutes saved?

Time = distance / speed.

Actual:
Monday: $2/5$ hr $=0.4$ hr $=0.4\times60=24$ min.
Wednesday: $2/3$ hr $= (2/3)\times60 =40$ min.
Friday: $2/4=1/2$ hr $=30$ min.
Total $=24+40+30=94$ min.

If always 4 mph: each day $30$ min, total $3\times30=90$ min.

Saving $=94-90=4$ minutes.
ANSWER 7: D

Problem 8:
Which integer cannot be sum of four consecutive odd integers?

Let the four be $n, n+2, n+4, n+6$ with $n$ odd.
Sum $=4n+12 =4(n+3)$.

Since $n$ is odd, $n+3$ is even, say $n+3=2m$. Then sum $=4\cdot2m=8m$, a multiple of 8. So any such sum must be divisible by 8.

Test:
16: $16/8=2$ possible; indeed $1+3+5+7=16$.
40: $40/8=5$ possible; indeed $7+9+11+13=40$.
72: $72/8=9$ possible; e.g. $15+17+19+21=72$.
200: $200/8=25$ possible; e.g. $47+49+51+53=200$.
100: $100/8=12.5$ not integer, so impossible.

Thus 100 cannot be written that way.
ANSWER 8: D

Problem 9:
Blake averaged 78 over four tests. Jenny vs Blake: +10, -10, +20, +20. By how much does Jenny's average exceed Blake's?

Only differences matter. Total points difference over 4 tests:
$+10-10+20+20 = +40$.

Average is total/4, so Jenny's average is $40/4=10$ points higher.

Blake's 78 is irrelevant to the difference. Choices 15,20,25,40 confuse total with average.
ANSWER 9: A

Problem 10:
Compute $(.2)^3/(.02)^2$.

$(.2)^3 = 0.2\times0.2\times0.2 = 0.04\times0.2 = 0.008$.
$(.02)^2 = 0.02\times0.02 = 0.0004$.

Ratio $=0.008/0.0004 = 8/0.4 = 80/4 =20$.

Alternatively with powers of 10: $(2\times10^{-1})^3=8\times10^{-3}$, $(2\times10^{-2})^2=4\times10^{-4}$, ratio $=2\times10^{1}=20$.

So not .2, 2, 10, or 15.
ANSWER 10: E

Problem 1:
We are asked to evaluate the product of the 9 factors:
$$\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{10}\right)$$

First, simplify each factor:
$$1 - \frac{1}{2} = \frac{1}{2}$$
$$1 - \frac{1}{3} = \frac{2}{3}$$
$$1 - \frac{1}{4} = \frac{3}{4}$$
$$\vdots$$
$$1 - \frac{1}{10} = \frac{9}{10}$$

Now, multiply the fractions together:
$$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{9}{10}$$

Notice that this is a telescoping product where the denominator of each fraction cancels with the numerator of the following fraction:
$$\frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \cdots \times \frac{\cancel{9}}{10} = \frac{1}{10}$$

ANSWER 1: A

---

Problem 2:
We are asked to find the sum of the distinct prime integer divisors of $2016$.

First, find the prime factorization of $2016$:
- $2016 = 2 \times 1008$
- $1008 = 2 \times 504$
- $504 = 2 \times 252$
- $252 = 2 \times 126$
- $126 = 2 \times 63$
- $63 = 9 \times 7 = 3^2 \times 7$

Thus, the prime factorization of $2016$ is:
$$2016 = 2^5 \times 3^2 \times 7$$

The distinct prime divisors are $2$, $3$, and $7$. 
Their sum is:
$$2 + 3 + 7 = 12$$

ANSWER 2: B

---

Problem 3:
We are given that the population of Nisos Isles is $200$ in the year $1998$, and it triples every $25$ years. We want to find the year in which the population reaches approximately $6000$.

Let $n$ be the number of $25$-year periods that pass after $1998$. The population can be represented as:
$$P = 200 \times 3^n$$

We want $P \approx 6000$:
$$200 \times 3^n \approx 6000 \implies 3^n \approx \frac{6000}{200} = 30$$

Let's test powers of $3$:
- For $n = 1$: $3^1 = 3 \implies P = 600$
- For $n = 2$: $3^2 = 9 \implies P = 1800$
- For $n = 3$: $3^3 = 27 \implies P = 5400$
- For $n = 4$: $3^4 = 81 \implies P = 16200$

$5400$ is very close to $6000$, so about $n = 3$ periods of $25$ years are needed:
$$3 \times 25 = 75\text{ years}$$

Adding $75$ years to $1998$:
$$1998 + 75 = 2073 \approx 2075$$

ANSWER 3: B

---

Problem 4:
We are given that exactly $74\%$ of the people surveyed answered "yes." We need to find the minimum number of people who could have been surveyed.

Let $N$ be the total number of people surveyed. The number of people who answered "yes" is:
$$74\% \times N = \frac{74}{100} N = \frac{37}{50} N$$

Since the number of people who answered "yes" must be a whole number, $\frac{37}{50} N$ must be an integer. 
Since $37$ and $50$ share no common factors other than $1$ (they are coprime), $N$ must be a multiple of $50$. 

The smallest positive integer multiple of $50$ is $50$. 
With $N = 50$, exactly $\frac{37}{50} \times 50 = 37$ people answered "yes."

ANSWER 4: D

---

Problem 5:
Betty is multiplying $0.075$ by $2.56$.
She calculated $75 \times 256 = 19200$.

Count the total number of digits to the right of the decimal point in the original numbers:
- $0.075$ has $3$ decimal places.
- $2.56$ has $2$ decimal places.

The product must therefore have $3 + 2 = 5$ decimal places:
$$19200 \times 10^{-5} = 0.19200 = 0.192$$

ANSWER 5: B

---

Problem 6:
Julie's speech must last between one-half hour ($30$ minutes) and three-quarters of an hour ($45$ minutes). Her speaking rate is $150$ words per minute.

Calculate the minimum and maximum acceptable word counts:
- Minimum words: $30\text{ minutes} \times 150\text{ words/minute} = 4500\text{ words}$
- Maximum words: $45\text{ minutes} \times 150\text{ words/minute} = 6750\text{ words}$

An appropriate word count must lie in the range $[4500, 6750]$. 
Looking at the choices:
- A. $2250$ (too low)
- B. $3000$ (too low)
- C. $4200$ (too low)
- D. $4350$ (too low)
- E. $5650$ is between $4500$ and $6750$ ($\frac{5650}{150} \approx 37.7$ minutes).

ANSWER 6: E

---

Problem 7:
Ted's grandfather covered $2$ miles on each of the $3$ days. 
First, find the time he actually spent each day using $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
- **Monday:** $\frac{2\text{ miles}}{5\text{ mph}} = \frac{2}{5}\text{ hour} = \frac{2}{5} \times 60 = 24\text{ minutes}$.
- **Wednesday:** $\frac{2\text{ miles}}{3\text{ mph}} = \frac{2}{3}\text{ hour} = \frac{2}{3} \times 60 = 40\text{ minutes}$.
- **Friday:** $\frac{2\text{ miles}}{4\text{ mph}} = \frac{2}{4}\text{ hour} = \frac{1}{2} \times 60 = 30\text{ minutes}$.

Total actual time spent:
$$24 + 40 + 30 = 94\text{ minutes}$$

If he had walked at $4\text{ mph}$ every day, the time spent each day would be $\frac{2}{4}\text{ hour} = 30\text{ minutes}$, so for all $3$ days:
$$3 \times 30 = 90\text{ minutes}$$

He would have spent:
$$94 - 90 = 4\text{ minutes less}$$

ANSWER 7: D

---

Problem 8:
Let four consecutive odd integers be represented as:
$$n, \quad n+2, \quad n+4, \quad n+6$$
where $n$ is an odd integer.

Their sum is:
$$S = n + (n+2) + (n+4) + (n+6) = 4n + 12 = 4(n+3)$$

Since $n$ is an odd integer, $n+3$ must be an even integer, which can be written as $2k$ for some integer $k$.
Substituting this in gives:
$$S = 4(2k) = 8k$$

Therefore, the sum of four consecutive odd integers must be a multiple of $8$.
Let's check the given options:
- A. $16 = 8 \times 2$ ($1 + 3 + 5 + 7 = 16$)
- B. $40 = 8 \times 5$ ($7 + 9 + 11 + 13 = 40$)
- C. $72 = 8 \times 9$ ($15 + 17 + 19 + 21 = 72$)
- D. $100$: not divisible by $8$ ($100 / 8 = 12.5$)
- E. $200 = 8 \times 25$ ($47 + 49 + 51 + 53 = 200$)

Hence, $100$ cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

---

Problem 9:
Let Blake's scores on the four tests be $B_1, B_2, B_3, B_4$.
Jenny's scores are:
- First test: $B_1 + 10$
- Second test: $B_2 - 10$
- Third test: $B_3 + 20$
- Fourth test: $B_4 + 20$

The sum of Jenny's scores minus the sum of Blake's scores is:
$$(B_1 + 10 + B_2 - 10 + B_3 + 20 + B_4 + 20) - (B_1 + B_2 + B_3 + B_4) = 10 - 10 + 20 + 20 = 40\text{ points}$$

Since there are $4$ tests, the difference between Jenny's average and Blake's aver

# Problem 1:
We need the product of the 9 factors (1 − 1/n) for n = 2, 3, 4, …, 10.

Rewrite each factor: 1 − 1/n = (n−1)/n. So the product becomes

(1/2)(2/3)(3/4)(4/5)(5/6)(6/7)(7/8)(8/9)(9/10)

This **telescopes**: each numerator cancels the denominator before it (2 cancels 2, 3 cancels 3, …, 9 cancels 9), leaving only the first numerator and the last denominator:

Product = 1/10

Ruling out traps: every factor is less than 1, so the product must be *smaller* than any single factor — this eliminates D (10/11) and E (11/2) immediately. Choice C (1/2) would be the answer if only the first factor were used, and B (1/9) would result from miscounting the factors (stopping at 1 − 1/9). Since n runs from 2 to 10 inclusive, there are exactly 9 factors and the answer is 1/10.

ANSWER 1: A

---

# Problem 2:
Factor 2016 into primes by repeated division:

2016 = 2 × 1008 = 2² × 504 = 2³ × 252 = 2⁴ × 126 = 2⁵ × 63, and 63 = 3² × 7.

So 2016 = 2⁵ × 3² × 7.

The **distinct** prime divisors are just 2, 3, and 7 (the exponents don't matter — we count each prime once).

Sum = 2 + 3 + 7 = **12**

Ruling out traps: A (9) would come from forgetting the prime 3; E (63) is the odd part 3² × 7 of 2016, not a sum of primes; adding primes *with repetition* gives 2+2+2+2+2+3+3+7 = 23, which isn't even a choice. The correct distinct-prime sum is 12.

ANSWER 2: B

---

# Problem 3:
Starting at 200 people in 1998, the population triples every 25 years:

- 1998: 200
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16 200

We want about 6000. Check how many triplings are needed: 6000/200 = 30, and 3³ = 27 ≈ 30 (while 3⁴ = 81 is way too big). So about **three triplings ≈ 75 years** after 1998, i.e., around 2073.

Comparing to the choices: at 2073 the population is 5400 (off by only 600), while at 2098 it is 16 200 (off by over 10 000). The closest estimate offered is **2075**.

Ruling out traps: 2050 and 2100 bracket the answer but are farther from the true crossing point than 2075; 2125 and 2150 correspond to populations of tens of thousands.

ANSWER 3: B

---

# Problem 4:
If N people were asked and exactly 74% said yes, then 0.74N must be a whole number of people:

0.74N = (74/100)N = (37/50)N

Since 37 is prime and shares no factor with 50, the fraction 37/50 is fully reduced. For (37/50)N to be an integer, **50 must divide N**. The smallest such N is 50.

Check: with 50 people, 0.74 × 50 = 37 people said yes — exactly 74%. ✓

Ruling out the smaller choices: 10 people → 7.4 yes-votes; 20 → 14.8; 25 → 18.5 — none of these is a whole number of people, so each is impossible. Hence 50 is the fewest.

ANSWER 4: D

---

# Problem 5:
First, confirm the calculator display: entering 075 × 256 (no decimal points) gives 75 × 256 = 19 200. ✓ That matches.

Now place the decimal correctly. The rule for multiplying decimals: the product has as many decimal places as the two factors combined.

- 0.075 has **3** decimal places
- 2.56 has **2** decimal places
- Product has 3 + 2 = **5** decimal places

Start from 19 200 and move the decimal point 5 places to the left:

19 200 → 0.19200 = **0.192**

Ruling out traps: 192 (E) and 19.2 (D) come from shifting the point only 2 or 3 places; 0.0192 (A) from shifting 6 places. A quick sanity check: 0.075 × 2.56 should be a bit less than 0.1 × 2.56 = 0.256, and 0.192 fits perfectly.

ANSWER 5: B

---

# Problem 6:
Convert the time limits to minutes, then to words:

- Minimum time: 1/2 hour = 30 minutes → 150 × 30 = **4500 words**
- Maximum time: 3/4 hour = 45 minutes → 150 × 45 = **6750 words**

So the speech needs between 4500 and 6750 words.

Check each choice: 2250 words = 15 min (too short); 3000 = 20 min (too short); 4200 = 28 min (too short); 4350 = 29 min (just under a half hour — too short); 5650 = 5650/150 ≈ 37.7 minutes, which lies nicely between 30 and 45 minutes. ✓

Only 5650 works.

ANSWER 6: E

---

# Problem 7:
Use time = distance ÷ speed, and convert to minutes (1 hour = 60 min):

- Monday: 2 miles at 5 mph → 2/5 hour = 24 minutes
- Wednesday: 2 miles at 3 mph → 2/3 hour = 40 minutes
- Friday: 2 miles at 4 mph → 2/4 = 1/2 hour = 30 minutes

Actual total: 24 + 40 + 30 = **94 minutes**.

If he had always gone 4 mph: total distance 6 miles at 4 mph → 6/4 = 1.5 hours = 90 minutes.

Difference: 94 − 90 = **4 minutes less**.

Ruling out traps: note Monday would actually take *longer* at 4 mph (30 vs. 24 min), so you can't just assume each day saves time — Wednesday's big savings (40 → 30) must be combined with Monday's loss (24 → 30). Net saving = 10 − 6 = 4 minutes, matching the totals.

ANSWER 7: D

---

# Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6, where a is odd. Their sum is

a + (a+2) + (a+4) + (a+6) = 4a + 12 = 4(a + 3).

Since a is odd, a + 3 is **even**, so the sum equals 4 × (an even number), i.e., every such sum is **divisible by 8**.

Test divisibility by 8 for each choice:

- 16 ÷ 8 = 2 ✓ (indeed 1 + 3 + 5 + 7 = 16)
- 40 ÷ 8 = 5 ✓ (7 + 9 + 11 + 13 = 40)
- 72 ÷ 8 = 9 ✓ (15 + 17 + 19 + 21 = 72)
- 100 ÷ 8 = 12.5 ✗ — **not divisible by 8**
- 200 ÷ 8 = 25 ✓ (47 + 49 + 51 + 53 = 200)

So 100 cannot be written as such a sum. (Check directly: 100 = 4(a+3) forces a + 3 = 25, so a = 22 — but 22 is even, a contradiction.)

ANSWER 8: D

---

# Problem 9:
Blake's total over four tests: 4 × 78 = 312 points.

Jenny's scores compared to Blake's, test by test: +10, −10, +20, +20.

Total difference: +10 − 10 + 20 + 20 = **+40 points**.

Jenny's total: 312 + 40 = 352, so her average is 352 ÷ 4 = 88.

Jenny's average exceeds Blake's by 88 − 78 = **10 points**.

(Equivalently: the *average* difference is the total difference divided by the number of tests: 40 ÷ 4 = 10.)

Ruling out traps: E (40) is the total point difference — it forgets to divide by the 4 tests; C (20) might come from averaging only the +20's and ignoring the +10/−10 pair; B and D have no valid basis.

ANSWER 9: A

---

# Problem 10:
Compute the numerator and denominator separately:

- (0.2)³ = 

Problem 1:  
The product is  
\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots\left(1-\frac1{10}\right).
\]
Rewrite each factor:
\[
\frac12\cdot \frac23\cdot \frac34\cdot \frac45\cdots \frac9{10}.
\]
Everything cancels except the first numerator and the last denominator:
\[
\frac1{10}.
\]
So the answer is A.

ANSWER 1: A

Problem 2:  
We need the sum of the distinct prime factors of 2016. Factor:
\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot 252=2^4\cdot 126=2^5\cdot 63.
\]
Also,
\[
63=3^2\cdot 7.
\]
Thus
\[
2016=2^5\cdot 3^2\cdot 7.
\]
The distinct prime divisors are 2, 3, and 7. Their sum is
\[
2+3+7=12.
\]

ANSWER 2: B

Problem 3:  
The population triples every 25 years. Starting with 200 in 1998, after \(t\) years the population is approximately
\[
200\cdot 3^{t/25}.
\]
We want this to be about 6000:
\[
200\cdot 3^{t/25}=6000.
\]
Divide by 200:
\[
3^{t/25}=30.
\]
Now,
\[
3^3=27
\]
and
\[
3^4=81.
\]
So \(t/25\) is a little more than 3, meaning \(t\) is a little more than 75 years.  
\[
1998+75=2073,
\]
so the closest choice is 2075.

ANSWER 3: B

Problem 4:  
Exactly 74% answered “yes.” Write 74% as a fraction in lowest terms:
\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]
For the number of “yes” answers to be an integer, the total number of people asked must be a multiple of 50. The smallest such number among the choices is 50.  
Check: \(\frac{37}{50}=74\%\).

ANSWER 4: D

Problem 5:  
Betty forgot the decimal points, so she effectively multiplied
\[
75\cdot 256=19200.
\]
Now restore the decimal places.  
\(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places, so the correct product must have
\[
3+2=5
\]
decimal places. Therefore,
\[
0.075\cdot 2.56=\frac{19200}{100000}=0.192.
\]

ANSWER 5: B

Problem 6:  
The speech must last between one-half hour and three-quarters of an hour, which is between
\[
30\text{ minutes}
\]
and
\[
45\text{ minutes}.
\]
At 150 words per minute, the acceptable word counts are between
\[
150\cdot 30=4500
\]
and
\[
150\cdot 45=6750.
\]
Among the choices, only 5650 lies between 4500 and 6750.

ANSWER 6: E

Problem 7:  
Each day he went 2 miles.

Monday: speed 5 mph, time
\[
\frac{2}{5}\text{ hour}=24\text{ minutes}.
\]
Wednesday: speed 3 mph, time
\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]
Friday: speed 4 mph, time
\[
\frac{2}{4}\text{ hour}=30\text{ minutes}.
\]
Total time:
\[
24+40+30=94\text{ minutes}.
\]
If he had walked all 6 miles at 4 mph, the time would be
\[
\frac{6}{4}\text{ hour}=1.5\text{ hours}=90\text{ minutes}.
\]
The difference is
\[
94-90=4\text{ minutes}.
\]

ANSWER 7: D

Problem 8:  
Let four consecutive odd integers be
\[
2k+1,\;2k+3,\;2k+5,\;2k+7.
\]
Their sum is
\[
(2k+1)+(2k+3)+(2k+5)+(2k+7)=8k+16=8(k+2).
\]
So any sum of four consecutive odd integers must be divisible by 8.

Check the choices:
- 16 is divisible by 8.
- 40 is divisible by 8.
- 72 is divisible by 8.
- 100 is not divisible by 8.
- 200 is divisible by 8.

Therefore, 100 cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

Problem 9:  
Blake’s average was 78 on four tests, so his total score was
\[
4\cdot 78=312.
\]
Jenny’s scores differed from Blake’s by
\[
+10,\;-10,\;+20,\;+20.
\]
The total difference is
\[
10-10+20+20=40.
\]
So Jenny’s total score was 40 points higher than Blake’s. Since there are 4 tests, her average was higher by
\[
\frac{40}{4}=10.
\]

ANSWER 9: A

Problem 10:  
Compute:
\[
(0.2)^3=0.008
\]
and
\[
(0.02)^2=0.0004.
\]
Thus
\[
\frac{(0.2)^3}{(0.02)^2}=\frac{0.008}{0.0004}=20.
\]

ANSWER 10: E

Problem 1:
Each factor simplifies: (1 − 1/2) = 1/2, (1 − 1/3) = 2/3, …, (1 − 1/10) = 9/10.

Product = (1/2)(2/3)(3/4)⋯(9/10).

Everything telescopes: the numerator of each fraction cancels the denominator of the previous one, leaving 1/10.

(Choice B, 1/9, would come from miscounting the last denominator; the product is definitely less than 1/2, ruling out C, D, E.)

ANSWER 1: A

Problem 2:
Factor 2016: 2016 = 2 · 1008 = 2·2·504 = 2·2·2·252 = 2^5 · 63 = 2^5 · 3^2 · 7.

Distinct primes: 2, 3, 7. Sum = 2 + 3 + 7 = 12.

(9 = 2+7 misses 3; 16 = 2·5+... etc.; 49 and 63 come from adding non‑prime factors or all factors with multiplicity.)

ANSWER 2: B

Problem 3:
Start: 1998 → 200 people. Tripling every 25 years:
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16200

The population reaches about 6000 shortly after 2073, so the best estimate among the choices is 2075.

(2050 gives ~1800, 2100 gives ~16000 — both far off.)

ANSWER 3: B

Problem 4:
We need 74% of the number of people to be a whole number: 0.74n = 37n/50 must be an integer.

Since 37/50 is in lowest terms (37 is prime), n must be a multiple of 50. The smallest is n = 50 (then 37 people said yes).

Check the smaller options: 74% of 10 = 7.4, of 20 = 14.8, of 25 = 18.5 — none are whole numbers.

ANSWER 4: D

Problem 5:
Ignoring decimals: 75 × 256 = 19200, matching the display.

Count decimal places: 0.075 has 3, 2.56 has 2, total 5 decimal places.

So 0.075 × 2.56 = 19200 with the decimal point moved 5 places left = 0.19200 = 0.192.

Sanity check: 0.075 ≈ 0.08 and 2.56 ≈ 2.5, product ≈ 0.2. So 0.192 is right (not 0.0192 or 1.92).

ANSWER 5: B

Problem 6:
Time allowed: 30 minutes to 45 minutes.

At 150 words per minute:
- Minimum words = 150 × 30 = 4500
- Maximum words = 150 × 45 = 6750

We need a count between 4500 and 6750. Only 5650 qualifies.

(2250 would take 15 min, 3000 → 20 min, 4200 → 28 min, 4350 → 29 min — all too short.)

ANSWER 6: E

Problem 7:
Time = distance ÷ speed, with 2 miles each day.

- Monday at 5 mph: 2/5 h = 24 min
- Wednesday at 3 mph: 2/3 h = 40 min
- Friday at 4 mph: 2/4 h = 30 min

Total actual = 24 + 40 + 30 = 94 minutes.

If always 4 mph: 30 min each day → 3 × 30 = 90 minutes.

Difference: 94 − 90 = 4 minutes less.

ANSWER 7: D

Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6 with a odd.

Sum = 4a + 12 = 4(a + 3). Since a is odd, a + 3 is even, say a + 3 = 2k, so the sum = 8k — a multiple of 8.

Check the choices for divisibility by 8:
- 16 = 8·2 ✓ (1+3+5+7)
- 40 = 8·5 ✓ (7+9+11+13)
- 72 = 8·9 ✓ (15+17+19+21)
- 100 ÷ 8 = 12.5 ✗
- 200 = 8·25 ✓ (47+49+51+53)

So 100 cannot be written that way.

ANSWER 8: D

Problem 9:
Jenny's total differs from Blake's total by:
+10 − 10 + 20 + 20 = +40 points.

Average difference = 40 ÷ 4 = 10 points.

(40 is the total difference, not the average — that's the trap in choice E. Blake's 78 average is not even needed, though Jenny averages 88, which is under 100, so it's consistent.)

ANSWER 9: A

Problem 10:
(0.2)³ = 0.008 and (0.02)² = 0.0004.

0.008 ÷ 0.0004 = 80/4 = 20.

(Alternatively: (2×10⁻¹)³ / (2×10⁻²)² = 8×10⁻³ / (4×10⁻⁴) = 2 × 10¹ = 20.)

ANSWER 10: E

Problem 1:

We need to compute

\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots \left(1-\frac1{10}\right).
\]

Rewrite each factor:

\[
1-\frac12=\frac12,\qquad
1-\frac13=\frac23,\qquad
1-\frac14=\frac34,
\]

and so on. Thus the product is

\[
\frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdots\frac9{10}.
\]

Nearly everything cancels:

\[
\frac{\cancel1}{2}\cdot\frac{\cancel2}{3}\cdot\frac{\cancel3}{4}\cdots
\frac{\cancel9}{10}
=\frac1{10}.
\]

Choice B, \(\frac19\), might seem plausible if one thinks there are 9 factors, but the cancellation leaves the first numerator \(1\) and final denominator \(10\).

\[
\boxed{\frac1{10}}
\]

ANSWER 1: A

Problem 2:

We need the sum of the distinct prime divisors of \(2016\).

Factor \(2016\):

\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot252=2^4\cdot126=2^5\cdot63.
\]

Since

\[
63=3^2\cdot7,
\]

we have

\[
2016=2^5\cdot3^2\cdot7.
\]

The distinct prime divisors are

\[
2,\ 3,\ 7.
\]

Their sum is

\[
2+3+7=12.
\]

Do not add repeated prime factors such as all five 2s and both 3s; the question asks for distinct primes only.

\[
\boxed{12}
\]

ANSWER 2: B

Problem 3:

The population starts at \(200\) in 1998 and triples every 25 years.

We want it to be about \(6000\). Compare:

\[
\frac{6000}{200}=30.
\]

So the population must grow by a factor of about \(30\).

Powers of \(3\) are:

\[
3^1=3,\qquad 3^2=9,\qquad 3^3=27,\qquad 3^4=81.
\]

A factor of \(27\) is close to \(30\), so we need about 3 periods of 25 years:

\[
3\cdot25=75\text{ years}.
\]

Therefore the estimated year is

\[
1998+75=2073.
\]

The nearest choice is \(2075\).

Indeed, after 75 years the estimated population is

\[
200\cdot 27=5400,
\]

which is reasonably close to \(6000\). Waiting 100 years would give \(200\cdot81=16200\), much too large.

\[
\boxed{2075}
\]

ANSWER 3: B

Problem 4:

Exactly \(74\%\) of the people answered yes.

Convert \(74\%\) to a fraction:

\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]

For the number of yes answers to be a whole number, the total number surveyed must be a multiple of \(50\).

The smallest possible total is therefore

\[
50.
\]

For example,

\[
74\%\text{ of }50=\frac{74}{100}\cdot 50=37,
\]

which is an integer number of people.

Smaller choices such as 10, 20, and 25 do not work, since \(74\%\) of them is not a whole number.

\[
\boxed{50}
\]

ANSWER 4: D

Problem 5:

Without decimal points, Betty entered

\[
75\times256=19200.
\]

Now place the decimal points correctly.

The number \(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places. Therefore the product must have

\[
3+2=5
\]

decimal places.

So move the decimal point in \(19200\) five places left:

\[
19200\longrightarrow 0.19200=0.192.
\]

Thus

\[
0.075\times2.56=0.192.
\]

Choice A, \(0.0192\), has one too many decimal places, while \(1.92\) and larger answers are too large because \(0.075\) is less than \(0.1\).

\[
\boxed{0.192}
\]

ANSWER 5: B

Problem 6:

Julie’s speech must last between one-half hour and three-quarters of an hour.

Convert these times to minutes:

\[
\frac12\text{ hour}=30\text{ minutes},
\qquad
\frac34\text{ hour}=45\text{ minutes}.
\]

At \(150\) words per minute, the minimum appropriate word count is

\[
30\cdot150=4500.
\]

The maximum appropriate word count is

\[
45\cdot150=6750.
\]

So her speech should contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) and \(3000\) are too few.
- \(4200\) and \(4350\) are still below \(4500\).
- \(5650\) is between \(4500\) and \(6750\).

Therefore the appropriate word count is

\[
\boxed{5650}.
\]

ANSWER 6: E

Problem 7:

Grandfather traveled 2 miles on each of 3 days.

Find his actual treadmill time.

On Monday, he jogged at 5 mph:

\[
\frac{2}{5}\text{ hour}=0.4\text{ hour}=24\text{ minutes}.
\]

On Wednesday, he walked at 3 mph:

\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]

On Friday, he walked at 4 mph:

\[
\frac24\text{ hour}=\frac12\text{ hour}=30\text{ minutes}.
\]

So his actual total time was

\[
24+40+30=94\text{ minutes}.
\]

If he had walked at 4 mph on all three days, each 2-mile trip would take 30 minutes:

\[
3\cdot30=90\text{ minutes}.
\]

Thus he would have spent

\[
94-90=4
\]

fewer minutes on the treadmill.

The 5-mph jogging day saves time, but the 3-mph walking day costs more time; together the actual time is 4 minutes longer than always walking at 4 mph.

\[
\boxed{4}
\]

ANSWER 7: D

Problem 8:

Let the four consecutive odd integers be

\[
n,\ n+2,\ n+4,\ n+6,
\]

where \(n\) is odd.

Their sum is

\[
n+(n+2)+(n+4)+(n+6)=4n+12.
\]

Factor:

\[
4n+12=4(n+3).
\]

Since \(n\) is odd, \(n+3\) is even. Therefore \(4(n+3)\) must be divisible by \(8\).

So a number can be written as the sum of four consecutive odd integers only if it is divisible by \(8\).

Check the choices:

\[
16\div8=2,
\quad
40\div8=5,
\quad
72\div8=9,
\quad
200\div8=25,
\]

so all of these are divisible by 8.

But

\[
100\div8=12.5,
\]

so \(100\) is not divisible by 8. Therefore it cannot be the sum of four consecutive odd integers.

For example, \(16=1+3+5+7\), showing that the condition is possible for the other type of choices.

\[
\boxed{100}
\]

ANSWER 8: D

Problem 9:

Blake’s average is not actually needed, because we can compare Jenny’s scores directly to Blake’s.

Jenny’s score changes relative to Blake’s are:

- first test: \(+10\),
- second test: \(-10\),
- third test: \(+20\),
- fourth test: \(+20\).

Her total score exceeds Blake’s by

\[
10-10+20+20=40.
\]

Since there are four tests, her average exceeds Blake’s by

\[
\frac{40}{4}=10.
\]

The \(40\) is the difference in total points, not the difference in average points.

\[
\boxed{10}
\]

ANSWER 9: A

Problem 10:

Compute

\[
\frac{(0.2)^3}{(0.02)^2}.
\]

First,

\[
(0.2)^3=0.2\cdot0.2\cdot0.2=0.008.
\]

Next,

\[
(0.02)^2=0.02\cdot0.02=0.0004.
\]

Therefore,

\[
\frac{0.008}{0.0004}=20.
\]

Another way is to use powers of 10:

\[
0.2=2\cdot10^{-1},\qquad 0.02=2\cdot10^{-2}.
\]

Then

\[
\fra

Problem 1:
We are asked for the value of $(1-1/2)(1-1/3)\cdots (1-1/10)$.

Write each factor as a single fraction:
$1-1/2 = 1/2$, $1-1/3 = 2/3$, $1-1/4 = 3/4$, ..., $1-1/10 = 9/10$.

So the product is:
$$\frac12 \cdot \frac23 \cdot \frac34 \cdots \frac89 \cdot \frac9{10}.$$

This telescopes: the numerator 2 cancels denominator 2, numerator 3 cancels denominator 3, etc. All cancel except the first denominator 2? Let's see:
$$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10} = \frac{1}{10}.$$
Formally numerator $=1\cdot2\cdots9$, denominator $=2\cdot3\cdots10$, ratio $=1/10$.

Choices B $1/9$, C $1/2$, D $10/11$, E $11/2$ do not result from cancellation.
ANSWER 1: A

Problem 2:
We need the sum of the distinct prime divisors of 2016.

Factor 2016. It is even:
$2016 = 2\times1008 = 2\times2\times504 = 2^2\times504$.
$504 = 2\times252 = 2^2\times252$? Continue systematically:
$2016 /2 =1008$, $/2=504$, $/2=252$, $/2=126$, $/2=63$.
So $2016 = 2^5 \times 63$.
$63 = 7\times9 = 3^2\times7$.

Hence $2016 = 2^5 \cdot 3^2 \cdot 7^1$.
The distinct primes occurring are $2,3,7$.

Their sum is $2+3+7 = 12$.

This rules out A 9, C 16, D 49, E 63 which would come from missing a prime or including powers.
ANSWER 2: B

Problem 3:
In 1998 population is 200, triples every 25 years. When is it about 6000?

Ratio wanted: $6000/200 = 30$. So we need about 30 times growth.

Each 25 years multiply by 3:
After 25 yr: $200\times3=600$.
After 50 yr: $600\times3=1800$.
After 75 yr: $1800\times3=5400$.

5400 is about 6000. So about 3 triplings, i.e. about $3\times25=75$ years after 1998.

$1998+75 = 2073$, closest to 2075 among choices.

2050 is only ~2 triplings, 2100+ is too many.
ANSWER 3: B

Problem 4:
Some $n$ people asked, exactly $74\%$ said yes. Find smallest possible $n$.

Let $k$ be number saying yes. Then $k/n = 74\% = 74/100 = 37/50$.

So $k = 37n/50$ must be an integer. Since 37 and 50 are coprime (gcd=1), $n$ must be a multiple of 50 for $37n/50$ to be integer.

Smallest positive multiple is $n=50$, giving $k=37$.

$n=10,20,25$ give $7.4,14.8,18.5$ yes-votes, impossible. $n=100$ works but is not fewest.
ANSWER 4: D

Problem 5:
True product is $0.075 \times 2.56$. Entering $75 \times 256$ gave $19200$. What is correct with decimals?

$0.075$ has 3 decimal places: $0.075 = 75/1000$.
$2.56$ has 2 decimal places: $2.56 = 256/100$.

So product $= 75\times256 / (1000\times100) = 19200 / 100000$.

$19200/100000 = 0.192$.

In general move decimal 3+2=5 places left. .0192 would be 6 places, 1.92 would be 4 places, etc.
ANSWER 5: B

Problem 6:
Speech must be $1/2$ hour to $3/4$ hour at 150 words per minute. Which word count fits?

Convert hours to minutes: $1/2$ hr $=30$ min, $3/4$ hr $=45$ min.

At 150 wpm:
minimum words $=30\times150 = 4500$.
maximum words $=45\times150 = 6750$.

We need word count between 4500 and 6750 inclusive.

A 2250 is 15 min, B 3000 is 20 min, C 4200 is 28 min, D 4350 is 29 min — all too short. Only E 5650 satisfies $4500 \le 5650 \le 6750$.
ANSWER 6: E

Problem 7:
2 miles each day for 3 days. Speeds: Monday 5 mph, Wednesday 3 mph, Friday 4 mph. Compare total time to time if all 3 days at 4 mph. How many minutes saved?

Time = distance / speed.

Actual:
Monday: $2/5$ hr $=0.4$ hr $=0.4\times60=24$ min.
Wednesday: $2/3$ hr $= (2/3)\times60 =40$ min.
Friday: $2/4=1/2$ hr $=30$ min.
Total $=24+40+30=94$ min.

If always 4 mph: each day $30$ min, total $3\times30=90$ min.

Saving $=94-90=4$ minutes.
ANSWER 7: D

Problem 8:
Which integer cannot be sum of four consecutive odd integers?

Let the four be $n, n+2, n+4, n+6$ with $n$ odd.
Sum $=4n+12 =4(n+3)$.

Since $n$ is odd, $n+3$ is even, say $n+3=2m$. Then sum $=4\cdot2m=8m$, a multiple of 8. So any such sum must be divisible by 8.

Test:
16: $16/8=2$ possible; indeed $1+3+5+7=16$.
40: $40/8=5$ possible; indeed $7+9+11+13=40$.
72: $72/8=9$ possible; e.g. $15+17+19+21=72$.
200: $200/8=25$ possible; e.g. $47+49+51+53=200$.
100: $100/8=12.5$ not integer, so impossible.

Thus 100 cannot be written that way.
ANSWER 8: D

Problem 9:
Blake averaged 78 over four tests. Jenny vs Blake: +10, -10, +20, +20. By how much does Jenny's average exceed Blake's?

Only differences matter. Total points difference over 4 tests:
$+10-10+20+20 = +40$.

Average is total/4, so Jenny's average is $40/4=10$ points higher.

Blake's 78 is irrelevant to the difference. Choices 15,20,25,40 confuse total with average.
ANSWER 9: A

Problem 10:
Compute $(.2)^3/(.02)^2$.

$(.2)^3 = 0.2\times0.2\times0.2 = 0.04\times0.2 = 0.008$.
$(.02)^2 = 0.02\times0.02 = 0.0004$.

Ratio $=0.008/0.0004 = 8/0.4 = 80/4 =20$.

Alternatively with powers of 10: $(2\times10^{-1})^3=8\times10^{-3}$, $(2\times10^{-2})^2=4\times10^{-4}$, ratio $=2\times10^{1}=20$.

So not .2, 2, 10, or 15.
ANSWER 10: E

Problem 1:
We are asked to evaluate the product of the 9 factors:
$$\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{10}\right)$$

First, simplify each factor:
$$1 - \frac{1}{2} = \frac{1}{2}$$
$$1 - \frac{1}{3} = \frac{2}{3}$$
$$1 - \frac{1}{4} = \frac{3}{4}$$
$$\vdots$$
$$1 - \frac{1}{10} = \frac{9}{10}$$

Now, multiply the fractions together:
$$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{9}{10}$$

Notice that this is a telescoping product where the denominator of each fraction cancels with the numerator of the following fraction:
$$\frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \cdots \times \frac{\cancel{9}}{10} = \frac{1}{10}$$

ANSWER 1: A

---

Problem 2:
We are asked to find the sum of the distinct prime integer divisors of $2016$.

First, find the prime factorization of $2016$:
- $2016 = 2 \times 1008$
- $1008 = 2 \times 504$
- $504 = 2 \times 252$
- $252 = 2 \times 126$
- $126 = 2 \times 63$
- $63 = 9 \times 7 = 3^2 \times 7$

Thus, the prime factorization of $2016$ is:
$$2016 = 2^5 \times 3^2 \times 7$$

The distinct prime divisors are $2$, $3$, and $7$. 
Their sum is:
$$2 + 3 + 7 = 12$$

ANSWER 2: B

---

Problem 3:
We are given that the population of Nisos Isles is $200$ in the year $1998$, and it triples every $25$ years. We want to find the year in which the population reaches approximately $6000$.

Let $n$ be the number of $25$-year periods that pass after $1998$. The population can be represented as:
$$P = 200 \times 3^n$$

We want $P \approx 6000$:
$$200 \times 3^n \approx 6000 \implies 3^n \approx \frac{6000}{200} = 30$$

Let's test powers of $3$:
- For $n = 1$: $3^1 = 3 \implies P = 600$
- For $n = 2$: $3^2 = 9 \implies P = 1800$
- For $n = 3$: $3^3 = 27 \implies P = 5400$
- For $n = 4$: $3^4 = 81 \implies P = 16200$

$5400$ is very close to $6000$, so about $n = 3$ periods of $25$ years are needed:
$$3 \times 25 = 75\text{ years}$$

Adding $75$ years to $1998$:
$$1998 + 75 = 2073 \approx 2075$$

ANSWER 3: B

---

Problem 4:
We are given that exactly $74\%$ of the people surveyed answered "yes." We need to find the minimum number of people who could have been surveyed.

Let $N$ be the total number of people surveyed. The number of people who answered "yes" is:
$$74\% \times N = \frac{74}{100} N = \frac{37}{50} N$$

Since the number of people who answered "yes" must be a whole number, $\frac{37}{50} N$ must be an integer. 
Since $37$ and $50$ share no common factors other than $1$ (they are coprime), $N$ must be a multiple of $50$. 

The smallest positive integer multiple of $50$ is $50$. 
With $N = 50$, exactly $\frac{37}{50} \times 50 = 37$ people answered "yes."

ANSWER 4: D

---

Problem 5:
Betty is multiplying $0.075$ by $2.56$.
She calculated $75 \times 256 = 19200$.

Count the total number of digits to the right of the decimal point in the original numbers:
- $0.075$ has $3$ decimal places.
- $2.56$ has $2$ decimal places.

The product must therefore have $3 + 2 = 5$ decimal places:
$$19200 \times 10^{-5} = 0.19200 = 0.192$$

ANSWER 5: B

---

Problem 6:
Julie's speech must last between one-half hour ($30$ minutes) and three-quarters of an hour ($45$ minutes). Her speaking rate is $150$ words per minute.

Calculate the minimum and maximum acceptable word counts:
- Minimum words: $30\text{ minutes} \times 150\text{ words/minute} = 4500\text{ words}$
- Maximum words: $45\text{ minutes} \times 150\text{ words/minute} = 6750\text{ words}$

An appropriate word count must lie in the range $[4500, 6750]$. 
Looking at the choices:
- A. $2250$ (too low)
- B. $3000$ (too low)
- C. $4200$ (too low)
- D. $4350$ (too low)
- E. $5650$ is between $4500$ and $6750$ ($\frac{5650}{150} \approx 37.7$ minutes).

ANSWER 6: E

---

Problem 7:
Ted's grandfather covered $2$ miles on each of the $3$ days. 
First, find the time he actually spent each day using $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
- **Monday:** $\frac{2\text{ miles}}{5\text{ mph}} = \frac{2}{5}\text{ hour} = \frac{2}{5} \times 60 = 24\text{ minutes}$.
- **Wednesday:** $\frac{2\text{ miles}}{3\text{ mph}} = \frac{2}{3}\text{ hour} = \frac{2}{3} \times 60 = 40\text{ minutes}$.
- **Friday:** $\frac{2\text{ miles}}{4\text{ mph}} = \frac{2}{4}\text{ hour} = \frac{1}{2} \times 60 = 30\text{ minutes}$.

Total actual time spent:
$$24 + 40 + 30 = 94\text{ minutes}$$

If he had walked at $4\text{ mph}$ every day, the time spent each day would be $\frac{2}{4}\text{ hour} = 30\text{ minutes}$, so for all $3$ days:
$$3 \times 30 = 90\text{ minutes}$$

He would have spent:
$$94 - 90 = 4\text{ minutes less}$$

ANSWER 7: D

---

Problem 8:
Let four consecutive odd integers be represented as:
$$n, \quad n+2, \quad n+4, \quad n+6$$
where $n$ is an odd integer.

Their sum is:
$$S = n + (n+2) + (n+4) + (n+6) = 4n + 12 = 4(n+3)$$

Since $n$ is an odd integer, $n+3$ must be an even integer, which can be written as $2k$ for some integer $k$.
Substituting this in gives:
$$S = 4(2k) = 8k$$

Therefore, the sum of four consecutive odd integers must be a multiple of $8$.
Let's check the given options:
- A. $16 = 8 \times 2$ ($1 + 3 + 5 + 7 = 16$)
- B. $40 = 8 \times 5$ ($7 + 9 + 11 + 13 = 40$)
- C. $72 = 8 \times 9$ ($15 + 17 + 19 + 21 = 72$)
- D. $100$: not divisible by $8$ ($100 / 8 = 12.5$)
- E. $200 = 8 \times 25$ ($47 + 49 + 51 + 53 = 200$)

Hence, $100$ cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

---

Problem 9:
Let Blake's scores on the four tests be $B_1, B_2, B_3, B_4$.
Jenny's scores are:
- First test: $B_1 + 10$
- Second test: $B_2 - 10$
- Third test: $B_3 + 20$
- Fourth test: $B_4 + 20$

The sum of Jenny's scores minus the sum of Blake's scores is:
$$(B_1 + 10 + B_2 - 10 + B_3 + 20 + B_4 + 20) - (B_1 + B_2 + B_3 + B_4) = 10 - 10 + 20 + 20 = 40\text{ points}$$

Since there are $4$ tests, the difference between Jenny's average and Blake's aver

# Problem 1:
We need the product of the 9 factors (1 − 1/n) for n = 2, 3, 4, …, 10.

Rewrite each factor: 1 − 1/n = (n−1)/n. So the product becomes

(1/2)(2/3)(3/4)(4/5)(5/6)(6/7)(7/8)(8/9)(9/10)

This **telescopes**: each numerator cancels the denominator before it (2 cancels 2, 3 cancels 3, …, 9 cancels 9), leaving only the first numerator and the last denominator:

Product = 1/10

Ruling out traps: every factor is less than 1, so the product must be *smaller* than any single factor — this eliminates D (10/11) and E (11/2) immediately. Choice C (1/2) would be the answer if only the first factor were used, and B (1/9) would result from miscounting the factors (stopping at 1 − 1/9). Since n runs from 2 to 10 inclusive, there are exactly 9 factors and the answer is 1/10.

ANSWER 1: A

---

# Problem 2:
Factor 2016 into primes by repeated division:

2016 = 2 × 1008 = 2² × 504 = 2³ × 252 = 2⁴ × 126 = 2⁵ × 63, and 63 = 3² × 7.

So 2016 = 2⁵ × 3² × 7.

The **distinct** prime divisors are just 2, 3, and 7 (the exponents don't matter — we count each prime once).

Sum = 2 + 3 + 7 = **12**

Ruling out traps: A (9) would come from forgetting the prime 3; E (63) is the odd part 3² × 7 of 2016, not a sum of primes; adding primes *with repetition* gives 2+2+2+2+2+3+3+7 = 23, which isn't even a choice. The correct distinct-prime sum is 12.

ANSWER 2: B

---

# Problem 3:
Starting at 200 people in 1998, the population triples every 25 years:

- 1998: 200
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16 200

We want about 6000. Check how many triplings are needed: 6000/200 = 30, and 3³ = 27 ≈ 30 (while 3⁴ = 81 is way too big). So about **three triplings ≈ 75 years** after 1998, i.e., around 2073.

Comparing to the choices: at 2073 the population is 5400 (off by only 600), while at 2098 it is 16 200 (off by over 10 000). The closest estimate offered is **2075**.

Ruling out traps: 2050 and 2100 bracket the answer but are farther from the true crossing point than 2075; 2125 and 2150 correspond to populations of tens of thousands.

ANSWER 3: B

---

# Problem 4:
If N people were asked and exactly 74% said yes, then 0.74N must be a whole number of people:

0.74N = (74/100)N = (37/50)N

Since 37 is prime and shares no factor with 50, the fraction 37/50 is fully reduced. For (37/50)N to be an integer, **50 must divide N**. The smallest such N is 50.

Check: with 50 people, 0.74 × 50 = 37 people said yes — exactly 74%. ✓

Ruling out the smaller choices: 10 people → 7.4 yes-votes; 20 → 14.8; 25 → 18.5 — none of these is a whole number of people, so each is impossible. Hence 50 is the fewest.

ANSWER 4: D

---

# Problem 5:
First, confirm the calculator display: entering 075 × 256 (no decimal points) gives 75 × 256 = 19 200. ✓ That matches.

Now place the decimal correctly. The rule for multiplying decimals: the product has as many decimal places as the two factors combined.

- 0.075 has **3** decimal places
- 2.56 has **2** decimal places
- Product has 3 + 2 = **5** decimal places

Start from 19 200 and move the decimal point 5 places to the left:

19 200 → 0.19200 = **0.192**

Ruling out traps: 192 (E) and 19.2 (D) come from shifting the point only 2 or 3 places; 0.0192 (A) from shifting 6 places. A quick sanity check: 0.075 × 2.56 should be a bit less than 0.1 × 2.56 = 0.256, and 0.192 fits perfectly.

ANSWER 5: B

---

# Problem 6:
Convert the time limits to minutes, then to words:

- Minimum time: 1/2 hour = 30 minutes → 150 × 30 = **4500 words**
- Maximum time: 3/4 hour = 45 minutes → 150 × 45 = **6750 words**

So the speech needs between 4500 and 6750 words.

Check each choice: 2250 words = 15 min (too short); 3000 = 20 min (too short); 4200 = 28 min (too short); 4350 = 29 min (just under a half hour — too short); 5650 = 5650/150 ≈ 37.7 minutes, which lies nicely between 30 and 45 minutes. ✓

Only 5650 works.

ANSWER 6: E

---

# Problem 7:
Use time = distance ÷ speed, and convert to minutes (1 hour = 60 min):

- Monday: 2 miles at 5 mph → 2/5 hour = 24 minutes
- Wednesday: 2 miles at 3 mph → 2/3 hour = 40 minutes
- Friday: 2 miles at 4 mph → 2/4 = 1/2 hour = 30 minutes

Actual total: 24 + 40 + 30 = **94 minutes**.

If he had always gone 4 mph: total distance 6 miles at 4 mph → 6/4 = 1.5 hours = 90 minutes.

Difference: 94 − 90 = **4 minutes less**.

Ruling out traps: note Monday would actually take *longer* at 4 mph (30 vs. 24 min), so you can't just assume each day saves time — Wednesday's big savings (40 → 30) must be combined with Monday's loss (24 → 30). Net saving = 10 − 6 = 4 minutes, matching the totals.

ANSWER 7: D

---

# Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6, where a is odd. Their sum is

a + (a+2) + (a+4) + (a+6) = 4a + 12 = 4(a + 3).

Since a is odd, a + 3 is **even**, so the sum equals 4 × (an even number), i.e., every such sum is **divisible by 8**.

Test divisibility by 8 for each choice:

- 16 ÷ 8 = 2 ✓ (indeed 1 + 3 + 5 + 7 = 16)
- 40 ÷ 8 = 5 ✓ (7 + 9 + 11 + 13 = 40)
- 72 ÷ 8 = 9 ✓ (15 + 17 + 19 + 21 = 72)
- 100 ÷ 8 = 12.5 ✗ — **not divisible by 8**
- 200 ÷ 8 = 25 ✓ (47 + 49 + 51 + 53 = 200)

So 100 cannot be written as such a sum. (Check directly: 100 = 4(a+3) forces a + 3 = 25, so a = 22 — but 22 is even, a contradiction.)

ANSWER 8: D

---

# Problem 9:
Blake's total over four tests: 4 × 78 = 312 points.

Jenny's scores compared to Blake's, test by test: +10, −10, +20, +20.

Total difference: +10 − 10 + 20 + 20 = **+40 points**.

Jenny's total: 312 + 40 = 352, so her average is 352 ÷ 4 = 88.

Jenny's average exceeds Blake's by 88 − 78 = **10 points**.

(Equivalently: the *average* difference is the total difference divided by the number of tests: 40 ÷ 4 = 10.)

Ruling out traps: E (40) is the total point difference — it forgets to divide by the 4 tests; C (20) might come from averaging only the +20's and ignoring the +10/−10 pair; B and D have no valid basis.

ANSWER 9: A

---

# Problem 10:
Compute the numerator and denominator separately:

- (0.2)³ = 

Problem 1:  
The product is  
\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots\left(1-\frac1{10}\right).
\]
Rewrite each factor:
\[
\frac12\cdot \frac23\cdot \frac34\cdot \frac45\cdots \frac9{10}.
\]
Everything cancels except the first numerator and the last denominator:
\[
\frac1{10}.
\]
So the answer is A.

ANSWER 1: A

Problem 2:  
We need the sum of the distinct prime factors of 2016. Factor:
\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot 252=2^4\cdot 126=2^5\cdot 63.
\]
Also,
\[
63=3^2\cdot 7.
\]
Thus
\[
2016=2^5\cdot 3^2\cdot 7.
\]
The distinct prime divisors are 2, 3, and 7. Their sum is
\[
2+3+7=12.
\]

ANSWER 2: B

Problem 3:  
The population triples every 25 years. Starting with 200 in 1998, after \(t\) years the population is approximately
\[
200\cdot 3^{t/25}.
\]
We want this to be about 6000:
\[
200\cdot 3^{t/25}=6000.
\]
Divide by 200:
\[
3^{t/25}=30.
\]
Now,
\[
3^3=27
\]
and
\[
3^4=81.
\]
So \(t/25\) is a little more than 3, meaning \(t\) is a little more than 75 years.  
\[
1998+75=2073,
\]
so the closest choice is 2075.

ANSWER 3: B

Problem 4:  
Exactly 74% answered “yes.” Write 74% as a fraction in lowest terms:
\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]
For the number of “yes” answers to be an integer, the total number of people asked must be a multiple of 50. The smallest such number among the choices is 50.  
Check: \(\frac{37}{50}=74\%\).

ANSWER 4: D

Problem 5:  
Betty forgot the decimal points, so she effectively multiplied
\[
75\cdot 256=19200.
\]
Now restore the decimal places.  
\(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places, so the correct product must have
\[
3+2=5
\]
decimal places. Therefore,
\[
0.075\cdot 2.56=\frac{19200}{100000}=0.192.
\]

ANSWER 5: B

Problem 6:  
The speech must last between one-half hour and three-quarters of an hour, which is between
\[
30\text{ minutes}
\]
and
\[
45\text{ minutes}.
\]
At 150 words per minute, the acceptable word counts are between
\[
150\cdot 30=4500
\]
and
\[
150\cdot 45=6750.
\]
Among the choices, only 5650 lies between 4500 and 6750.

ANSWER 6: E

Problem 7:  
Each day he went 2 miles.

Monday: speed 5 mph, time
\[
\frac{2}{5}\text{ hour}=24\text{ minutes}.
\]
Wednesday: speed 3 mph, time
\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]
Friday: speed 4 mph, time
\[
\frac{2}{4}\text{ hour}=30\text{ minutes}.
\]
Total time:
\[
24+40+30=94\text{ minutes}.
\]
If he had walked all 6 miles at 4 mph, the time would be
\[
\frac{6}{4}\text{ hour}=1.5\text{ hours}=90\text{ minutes}.
\]
The difference is
\[
94-90=4\text{ minutes}.
\]

ANSWER 7: D

Problem 8:  
Let four consecutive odd integers be
\[
2k+1,\;2k+3,\;2k+5,\;2k+7.
\]
Their sum is
\[
(2k+1)+(2k+3)+(2k+5)+(2k+7)=8k+16=8(k+2).
\]
So any sum of four consecutive odd integers must be divisible by 8.

Check the choices:
- 16 is divisible by 8.
- 40 is divisible by 8.
- 72 is divisible by 8.
- 100 is not divisible by 8.
- 200 is divisible by 8.

Therefore, 100 cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

Problem 9:  
Blake’s average was 78 on four tests, so his total score was
\[
4\cdot 78=312.
\]
Jenny’s scores differed from Blake’s by
\[
+10,\;-10,\;+20,\;+20.
\]
The total difference is
\[
10-10+20+20=40.
\]
So Jenny’s total score was 40 points higher than Blake’s. Since there are 4 tests, her average was higher by
\[
\frac{40}{4}=10.
\]

ANSWER 9: A

Problem 10:  
Compute:
\[
(0.2)^3=0.008
\]
and
\[
(0.02)^2=0.0004.
\]
Thus
\[
\frac{(0.2)^3}{(0.02)^2}=\frac{0.008}{0.0004}=20.
\]

ANSWER 10: E

Problem 1:
Each factor simplifies: (1 − 1/2) = 1/2, (1 − 1/3) = 2/3, …, (1 − 1/10) = 9/10.

Product = (1/2)(2/3)(3/4)⋯(9/10).

Everything telescopes: the numerator of each fraction cancels the denominator of the previous one, leaving 1/10.

(Choice B, 1/9, would come from miscounting the last denominator; the product is definitely less than 1/2, ruling out C, D, E.)

ANSWER 1: A

Problem 2:
Factor 2016: 2016 = 2 · 1008 = 2·2·504 = 2·2·2·252 = 2^5 · 63 = 2^5 · 3^2 · 7.

Distinct primes: 2, 3, 7. Sum = 2 + 3 + 7 = 12.

(9 = 2+7 misses 3; 16 = 2·5+... etc.; 49 and 63 come from adding non‑prime factors or all factors with multiplicity.)

ANSWER 2: B

Problem 3:
Start: 1998 → 200 people. Tripling every 25 years:
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16200

The population reaches about 6000 shortly after 2073, so the best estimate among the choices is 2075.

(2050 gives ~1800, 2100 gives ~16000 — both far off.)

ANSWER 3: B

Problem 4:
We need 74% of the number of people to be a whole number: 0.74n = 37n/50 must be an integer.

Since 37/50 is in lowest terms (37 is prime), n must be a multiple of 50. The smallest is n = 50 (then 37 people said yes).

Check the smaller options: 74% of 10 = 7.4, of 20 = 14.8, of 25 = 18.5 — none are whole numbers.

ANSWER 4: D

Problem 5:
Ignoring decimals: 75 × 256 = 19200, matching the display.

Count decimal places: 0.075 has 3, 2.56 has 2, total 5 decimal places.

So 0.075 × 2.56 = 19200 with the decimal point moved 5 places left = 0.19200 = 0.192.

Sanity check: 0.075 ≈ 0.08 and 2.56 ≈ 2.5, product ≈ 0.2. So 0.192 is right (not 0.0192 or 1.92).

ANSWER 5: B

Problem 6:
Time allowed: 30 minutes to 45 minutes.

At 150 words per minute:
- Minimum words = 150 × 30 = 4500
- Maximum words = 150 × 45 = 6750

We need a count between 4500 and 6750. Only 5650 qualifies.

(2250 would take 15 min, 3000 → 20 min, 4200 → 28 min, 4350 → 29 min — all too short.)

ANSWER 6: E

Problem 7:
Time = distance ÷ speed, with 2 miles each day.

- Monday at 5 mph: 2/5 h = 24 min
- Wednesday at 3 mph: 2/3 h = 40 min
- Friday at 4 mph: 2/4 h = 30 min

Total actual = 24 + 40 + 30 = 94 minutes.

If always 4 mph: 30 min each day → 3 × 30 = 90 minutes.

Difference: 94 − 90 = 4 minutes less.

ANSWER 7: D

Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6 with a odd.

Sum = 4a + 12 = 4(a + 3). Since a is odd, a + 3 is even, say a + 3 = 2k, so the sum = 8k — a multiple of 8.

Check the choices for divisibility by 8:
- 16 = 8·2 ✓ (1+3+5+7)
- 40 = 8·5 ✓ (7+9+11+13)
- 72 = 8·9 ✓ (15+17+19+21)
- 100 ÷ 8 = 12.5 ✗
- 200 = 8·25 ✓ (47+49+51+53)

So 100 cannot be written that way.

ANSWER 8: D

Problem 9:
Jenny's total differs from Blake's total by:
+10 − 10 + 20 + 20 = +40 points.

Average difference = 40 ÷ 4 = 10 points.

(40 is the total difference, not the average — that's the trap in choice E. Blake's 78 average is not even needed, though Jenny averages 88, which is under 100, so it's consistent.)

ANSWER 9: A

Problem 10:
(0.2)³ = 0.008 and (0.02)² = 0.0004.

0.008 ÷ 0.0004 = 80/4 = 20.

(Alternatively: (2×10⁻¹)³ / (2×10⁻²)² = 8×10⁻³ / (4×10⁻⁴) = 2 × 10¹ = 20.)

ANSWER 10: E

Problem 1:

We need to compute

\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots \left(1-\frac1{10}\right).
\]

Rewrite each factor:

\[
1-\frac12=\frac12,\qquad
1-\frac13=\frac23,\qquad
1-\frac14=\frac34,
\]

and so on. Thus the product is

\[
\frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdots\frac9{10}.
\]

Nearly everything cancels:

\[
\frac{\cancel1}{2}\cdot\frac{\cancel2}{3}\cdot\frac{\cancel3}{4}\cdots
\frac{\cancel9}{10}
=\frac1{10}.
\]

Choice B, \(\frac19\), might seem plausible if one thinks there are 9 factors, but the cancellation leaves the first numerator \(1\) and final denominator \(10\).

\[
\boxed{\frac1{10}}
\]

ANSWER 1: A

Problem 2:

We need the sum of the distinct prime divisors of \(2016\).

Factor \(2016\):

\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot252=2^4\cdot126=2^5\cdot63.
\]

Since

\[
63=3^2\cdot7,
\]

we have

\[
2016=2^5\cdot3^2\cdot7.
\]

The distinct prime divisors are

\[
2,\ 3,\ 7.
\]

Their sum is

\[
2+3+7=12.
\]

Do not add repeated prime factors such as all five 2s and both 3s; the question asks for distinct primes only.

\[
\boxed{12}
\]

ANSWER 2: B

Problem 3:

The population starts at \(200\) in 1998 and triples every 25 years.

We want it to be about \(6000\). Compare:

\[
\frac{6000}{200}=30.
\]

So the population must grow by a factor of about \(30\).

Powers of \(3\) are:

\[
3^1=3,\qquad 3^2=9,\qquad 3^3=27,\qquad 3^4=81.
\]

A factor of \(27\) is close to \(30\), so we need about 3 periods of 25 years:

\[
3\cdot25=75\text{ years}.
\]

Therefore the estimated year is

\[
1998+75=2073.
\]

The nearest choice is \(2075\).

Indeed, after 75 years the estimated population is

\[
200\cdot 27=5400,
\]

which is reasonably close to \(6000\). Waiting 100 years would give \(200\cdot81=16200\), much too large.

\[
\boxed{2075}
\]

ANSWER 3: B

Problem 4:

Exactly \(74\%\) of the people answered yes.

Convert \(74\%\) to a fraction:

\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]

For the number of yes answers to be a whole number, the total number surveyed must be a multiple of \(50\).

The smallest possible total is therefore

\[
50.
\]

For example,

\[
74\%\text{ of }50=\frac{74}{100}\cdot 50=37,
\]

which is an integer number of people.

Smaller choices such as 10, 20, and 25 do not work, since \(74\%\) of them is not a whole number.

\[
\boxed{50}
\]

ANSWER 4: D

Problem 5:

Without decimal points, Betty entered

\[
75\times256=19200.
\]

Now place the decimal points correctly.

The number \(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places. Therefore the product must have

\[
3+2=5
\]

decimal places.

So move the decimal point in \(19200\) five places left:

\[
19200\longrightarrow 0.19200=0.192.
\]

Thus

\[
0.075\times2.56=0.192.
\]

Choice A, \(0.0192\), has one too many decimal places, while \(1.92\) and larger answers are too large because \(0.075\) is less than \(0.1\).

\[
\boxed{0.192}
\]

ANSWER 5: B

Problem 6:

Julie’s speech must last between one-half hour and three-quarters of an hour.

Convert these times to minutes:

\[
\frac12\text{ hour}=30\text{ minutes},
\qquad
\frac34\text{ hour}=45\text{ minutes}.
\]

At \(150\) words per minute, the minimum appropriate word count is

\[
30\cdot150=4500.
\]

The maximum appropriate word count is

\[
45\cdot150=6750.
\]

So her speech should contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) and \(3000\) are too few.
- \(4200\) and \(4350\) are still below \(4500\).
- \(5650\) is between \(4500\) and \(6750\).

Therefore the appropriate word count is

\[
\boxed{5650}.
\]

ANSWER 6: E

Problem 7:

Grandfather traveled 2 miles on each of 3 days.

Find his actual treadmill time.

On Monday, he jogged at 5 mph:

\[
\frac{2}{5}\text{ hour}=0.4\text{ hour}=24\text{ minutes}.
\]

On Wednesday, he walked at 3 mph:

\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]

On Friday, he walked at 4 mph:

\[
\frac24\text{ hour}=\frac12\text{ hour}=30\text{ minutes}.
\]

So his actual total time was

\[
24+40+30=94\text{ minutes}.
\]

If he had walked at 4 mph on all three days, each 2-mile trip would take 30 minutes:

\[
3\cdot30=90\text{ minutes}.
\]

Thus he would have spent

\[
94-90=4
\]

fewer minutes on the treadmill.

The 5-mph jogging day saves time, but the 3-mph walking day costs more time; together the actual time is 4 minutes longer than always walking at 4 mph.

\[
\boxed{4}
\]

ANSWER 7: D

Problem 8:

Let the four consecutive odd integers be

\[
n,\ n+2,\ n+4,\ n+6,
\]

where \(n\) is odd.

Their sum is

\[
n+(n+2)+(n+4)+(n+6)=4n+12.
\]

Factor:

\[
4n+12=4(n+3).
\]

Since \(n\) is odd, \(n+3\) is even. Therefore \(4(n+3)\) must be divisible by \(8\).

So a number can be written as the sum of four consecutive odd integers only if it is divisible by \(8\).

Check the choices:

\[
16\div8=2,
\quad
40\div8=5,
\quad
72\div8=9,
\quad
200\div8=25,
\]

so all of these are divisible by 8.

But

\[
100\div8=12.5,
\]

so \(100\) is not divisible by 8. Therefore it cannot be the sum of four consecutive odd integers.

For example, \(16=1+3+5+7\), showing that the condition is possible for the other type of choices.

\[
\boxed{100}
\]

ANSWER 8: D

Problem 9:

Blake’s average is not actually needed, because we can compare Jenny’s scores directly to Blake’s.

Jenny’s score changes relative to Blake’s are:

- first test: \(+10\),
- second test: \(-10\),
- third test: \(+20\),
- fourth test: \(+20\).

Her total score exceeds Blake’s by

\[
10-10+20+20=40.
\]

Since there are four tests, her average exceeds Blake’s by

\[
\frac{40}{4}=10.
\]

The \(40\) is the difference in total points, not the difference in average points.

\[
\boxed{10}
\]

ANSWER 9: A

Problem 10:

Compute

\[
\frac{(0.2)^3}{(0.02)^2}.
\]

First,

\[
(0.2)^3=0.2\cdot0.2\cdot0.2=0.008.
\]

Next,

\[
(0.02)^2=0.02\cdot0.02=0.0004.
\]

Therefore,

\[
\frac{0.008}{0.0004}=20.
\]

Another way is to use powers of 10:

\[
0.2=2\cdot10^{-1},\qquad 0.02=2\cdot10^{-2}.
\]

Then

\[
\fra

Problem 1:
We are asked for the value of $(1-1/2)(1-1/3)\cdots (1-1/10)$.

Write each factor as a single fraction:
$1-1/2 = 1/2$, $1-1/3 = 2/3$, $1-1/4 = 3/4$, ..., $1-1/10 = 9/10$.

So the product is:
$$\frac12 \cdot \frac23 \cdot \frac34 \cdots \frac89 \cdot \frac9{10}.$$

This telescopes: the numerator 2 cancels denominator 2, numerator 3 cancels denominator 3, etc. All cancel except the first denominator 2? Let's see:
$$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10} = \frac{1}{10}.$$
Formally numerator $=1\cdot2\cdots9$, denominator $=2\cdot3\cdots10$, ratio $=1/10$.

Choices B $1/9$, C $1/2$, D $10/11$, E $11/2$ do not result from cancellation.
ANSWER 1: A

Problem 2:
We need the sum of the distinct prime divisors of 2016.

Factor 2016. It is even:
$2016 = 2\times1008 = 2\times2\times504 = 2^2\times504$.
$504 = 2\times252 = 2^2\times252$? Continue systematically:
$2016 /2 =1008$, $/2=504$, $/2=252$, $/2=126$, $/2=63$.
So $2016 = 2^5 \times 63$.
$63 = 7\times9 = 3^2\times7$.

Hence $2016 = 2^5 \cdot 3^2 \cdot 7^1$.
The distinct primes occurring are $2,3,7$.

Their sum is $2+3+7 = 12$.

This rules out A 9, C 16, D 49, E 63 which would come from missing a prime or including powers.
ANSWER 2: B

Problem 3:
In 1998 population is 200, triples every 25 years. When is it about 6000?

Ratio wanted: $6000/200 = 30$. So we need about 30 times growth.

Each 25 years multiply by 3:
After 25 yr: $200\times3=600$.
After 50 yr: $600\times3=1800$.
After 75 yr: $1800\times3=5400$.

5400 is about 6000. So about 3 triplings, i.e. about $3\times25=75$ years after 1998.

$1998+75 = 2073$, closest to 2075 among choices.

2050 is only ~2 triplings, 2100+ is too many.
ANSWER 3: B

Problem 4:
Some $n$ people asked, exactly $74\%$ said yes. Find smallest possible $n$.

Let $k$ be number saying yes. Then $k/n = 74\% = 74/100 = 37/50$.

So $k = 37n/50$ must be an integer. Since 37 and 50 are coprime (gcd=1), $n$ must be a multiple of 50 for $37n/50$ to be integer.

Smallest positive multiple is $n=50$, giving $k=37$.

$n=10,20,25$ give $7.4,14.8,18.5$ yes-votes, impossible. $n=100$ works but is not fewest.
ANSWER 4: D

Problem 5:
True product is $0.075 \times 2.56$. Entering $75 \times 256$ gave $19200$. What is correct with decimals?

$0.075$ has 3 decimal places: $0.075 = 75/1000$.
$2.56$ has 2 decimal places: $2.56 = 256/100$.

So product $= 75\times256 / (1000\times100) = 19200 / 100000$.

$19200/100000 = 0.192$.

In general move decimal 3+2=5 places left. .0192 would be 6 places, 1.92 would be 4 places, etc.
ANSWER 5: B

Problem 6:
Speech must be $1/2$ hour to $3/4$ hour at 150 words per minute. Which word count fits?

Convert hours to minutes: $1/2$ hr $=30$ min, $3/4$ hr $=45$ min.

At 150 wpm:
minimum words $=30\times150 = 4500$.
maximum words $=45\times150 = 6750$.

We need word count between 4500 and 6750 inclusive.

A 2250 is 15 min, B 3000 is 20 min, C 4200 is 28 min, D 4350 is 29 min — all too short. Only E 5650 satisfies $4500 \le 5650 \le 6750$.
ANSWER 6: E

Problem 7:
2 miles each day for 3 days. Speeds: Monday 5 mph, Wednesday 3 mph, Friday 4 mph. Compare total time to time if all 3 days at 4 mph. How many minutes saved?

Time = distance / speed.

Actual:
Monday: $2/5$ hr $=0.4$ hr $=0.4\times60=24$ min.
Wednesday: $2/3$ hr $= (2/3)\times60 =40$ min.
Friday: $2/4=1/2$ hr $=30$ min.
Total $=24+40+30=94$ min.

If always 4 mph: each day $30$ min, total $3\times30=90$ min.

Saving $=94-90=4$ minutes.
ANSWER 7: D

Problem 8:
Which integer cannot be sum of four consecutive odd integers?

Let the four be $n, n+2, n+4, n+6$ with $n$ odd.
Sum $=4n+12 =4(n+3)$.

Since $n$ is odd, $n+3$ is even, say $n+3=2m$. Then sum $=4\cdot2m=8m$, a multiple of 8. So any such sum must be divisible by 8.

Test:
16: $16/8=2$ possible; indeed $1+3+5+7=16$.
40: $40/8=5$ possible; indeed $7+9+11+13=40$.
72: $72/8=9$ possible; e.g. $15+17+19+21=72$.
200: $200/8=25$ possible; e.g. $47+49+51+53=200$.
100: $100/8=12.5$ not integer, so impossible.

Thus 100 cannot be written that way.
ANSWER 8: D

Problem 9:
Blake averaged 78 over four tests. Jenny vs Blake: +10, -10, +20, +20. By how much does Jenny's average exceed Blake's?

Only differences matter. Total points difference over 4 tests:
$+10-10+20+20 = +40$.

Average is total/4, so Jenny's average is $40/4=10$ points higher.

Blake's 78 is irrelevant to the difference. Choices 15,20,25,40 confuse total with average.
ANSWER 9: A

Problem 10:
Compute $(.2)^3/(.02)^2$.

$(.2)^3 = 0.2\times0.2\times0.2 = 0.04\times0.2 = 0.008$.
$(.02)^2 = 0.02\times0.02 = 0.0004$.

Ratio $=0.008/0.0004 = 8/0.4 = 80/4 =20$.

Alternatively with powers of 10: $(2\times10^{-1})^3=8\times10^{-3}$, $(2\times10^{-2})^2=4\times10^{-4}$, ratio $=2\times10^{1}=20$.

So not .2, 2, 10, or 15.
ANSWER 10: E

Problem 1:
We are asked to evaluate the product of the 9 factors:
$$\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{10}\right)$$

First, simplify each factor:
$$1 - \frac{1}{2} = \frac{1}{2}$$
$$1 - \frac{1}{3} = \frac{2}{3}$$
$$1 - \frac{1}{4} = \frac{3}{4}$$
$$\vdots$$
$$1 - \frac{1}{10} = \frac{9}{10}$$

Now, multiply the fractions together:
$$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{9}{10}$$

Notice that this is a telescoping product where the denominator of each fraction cancels with the numerator of the following fraction:
$$\frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \cdots \times \frac{\cancel{9}}{10} = \frac{1}{10}$$

ANSWER 1: A

---

Problem 2:
We are asked to find the sum of the distinct prime integer divisors of $2016$.

First, find the prime factorization of $2016$:
- $2016 = 2 \times 1008$
- $1008 = 2 \times 504$
- $504 = 2 \times 252$
- $252 = 2 \times 126$
- $126 = 2 \times 63$
- $63 = 9 \times 7 = 3^2 \times 7$

Thus, the prime factorization of $2016$ is:
$$2016 = 2^5 \times 3^2 \times 7$$

The distinct prime divisors are $2$, $3$, and $7$. 
Their sum is:
$$2 + 3 + 7 = 12$$

ANSWER 2: B

---

Problem 3:
We are given that the population of Nisos Isles is $200$ in the year $1998$, and it triples every $25$ years. We want to find the year in which the population reaches approximately $6000$.

Let $n$ be the number of $25$-year periods that pass after $1998$. The population can be represented as:
$$P = 200 \times 3^n$$

We want $P \approx 6000$:
$$200 \times 3^n \approx 6000 \implies 3^n \approx \frac{6000}{200} = 30$$

Let's test powers of $3$:
- For $n = 1$: $3^1 = 3 \implies P = 600$
- For $n = 2$: $3^2 = 9 \implies P = 1800$
- For $n = 3$: $3^3 = 27 \implies P = 5400$
- For $n = 4$: $3^4 = 81 \implies P = 16200$

$5400$ is very close to $6000$, so about $n = 3$ periods of $25$ years are needed:
$$3 \times 25 = 75\text{ years}$$

Adding $75$ years to $1998$:
$$1998 + 75 = 2073 \approx 2075$$

ANSWER 3: B

---

Problem 4:
We are given that exactly $74\%$ of the people surveyed answered "yes." We need to find the minimum number of people who could have been surveyed.

Let $N$ be the total number of people surveyed. The number of people who answered "yes" is:
$$74\% \times N = \frac{74}{100} N = \frac{37}{50} N$$

Since the number of people who answered "yes" must be a whole number, $\frac{37}{50} N$ must be an integer. 
Since $37$ and $50$ share no common factors other than $1$ (they are coprime), $N$ must be a multiple of $50$. 

The smallest positive integer multiple of $50$ is $50$. 
With $N = 50$, exactly $\frac{37}{50} \times 50 = 37$ people answered "yes."

ANSWER 4: D

---

Problem 5:
Betty is multiplying $0.075$ by $2.56$.
She calculated $75 \times 256 = 19200$.

Count the total number of digits to the right of the decimal point in the original numbers:
- $0.075$ has $3$ decimal places.
- $2.56$ has $2$ decimal places.

The product must therefore have $3 + 2 = 5$ decimal places:
$$19200 \times 10^{-5} = 0.19200 = 0.192$$

ANSWER 5: B

---

Problem 6:
Julie's speech must last between one-half hour ($30$ minutes) and three-quarters of an hour ($45$ minutes). Her speaking rate is $150$ words per minute.

Calculate the minimum and maximum acceptable word counts:
- Minimum words: $30\text{ minutes} \times 150\text{ words/minute} = 4500\text{ words}$
- Maximum words: $45\text{ minutes} \times 150\text{ words/minute} = 6750\text{ words}$

An appropriate word count must lie in the range $[4500, 6750]$. 
Looking at the choices:
- A. $2250$ (too low)
- B. $3000$ (too low)
- C. $4200$ (too low)
- D. $4350$ (too low)
- E. $5650$ is between $4500$ and $6750$ ($\frac{5650}{150} \approx 37.7$ minutes).

ANSWER 6: E

---

Problem 7:
Ted's grandfather covered $2$ miles on each of the $3$ days. 
First, find the time he actually spent each day using $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
- **Monday:** $\frac{2\text{ miles}}{5\text{ mph}} = \frac{2}{5}\text{ hour} = \frac{2}{5} \times 60 = 24\text{ minutes}$.
- **Wednesday:** $\frac{2\text{ miles}}{3\text{ mph}} = \frac{2}{3}\text{ hour} = \frac{2}{3} \times 60 = 40\text{ minutes}$.
- **Friday:** $\frac{2\text{ miles}}{4\text{ mph}} = \frac{2}{4}\text{ hour} = \frac{1}{2} \times 60 = 30\text{ minutes}$.

Total actual time spent:
$$24 + 40 + 30 = 94\text{ minutes}$$

If he had walked at $4\text{ mph}$ every day, the time spent each day would be $\frac{2}{4}\text{ hour} = 30\text{ minutes}$, so for all $3$ days:
$$3 \times 30 = 90\text{ minutes}$$

He would have spent:
$$94 - 90 = 4\text{ minutes less}$$

ANSWER 7: D

---

Problem 8:
Let four consecutive odd integers be represented as:
$$n, \quad n+2, \quad n+4, \quad n+6$$
where $n$ is an odd integer.

Their sum is:
$$S = n + (n+2) + (n+4) + (n+6) = 4n + 12 = 4(n+3)$$

Since $n$ is an odd integer, $n+3$ must be an even integer, which can be written as $2k$ for some integer $k$.
Substituting this in gives:
$$S = 4(2k) = 8k$$

Therefore, the sum of four consecutive odd integers must be a multiple of $8$.
Let's check the given options:
- A. $16 = 8 \times 2$ ($1 + 3 + 5 + 7 = 16$)
- B. $40 = 8 \times 5$ ($7 + 9 + 11 + 13 = 40$)
- C. $72 = 8 \times 9$ ($15 + 17 + 19 + 21 = 72$)
- D. $100$: not divisible by $8$ ($100 / 8 = 12.5$)
- E. $200 = 8 \times 25$ ($47 + 49 + 51 + 53 = 200$)

Hence, $100$ cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

---

Problem 9:
Let Blake's scores on the four tests be $B_1, B_2, B_3, B_4$.
Jenny's scores are:
- First test: $B_1 + 10$
- Second test: $B_2 - 10$
- Third test: $B_3 + 20$
- Fourth test: $B_4 + 20$

The sum of Jenny's scores minus the sum of Blake's scores is:
$$(B_1 + 10 + B_2 - 10 + B_3 + 20 + B_4 + 20) - (B_1 + B_2 + B_3 + B_4) = 10 - 10 + 20 + 20 = 40\text{ points}$$

Since there are $4$ tests, the difference between Jenny's average and Blake's aver

# Problem 1:
We need the product of the 9 factors (1 − 1/n) for n = 2, 3, 4, …, 10.

Rewrite each factor: 1 − 1/n = (n−1)/n. So the product becomes

(1/2)(2/3)(3/4)(4/5)(5/6)(6/7)(7/8)(8/9)(9/10)

This **telescopes**: each numerator cancels the denominator before it (2 cancels 2, 3 cancels 3, …, 9 cancels 9), leaving only the first numerator and the last denominator:

Product = 1/10

Ruling out traps: every factor is less than 1, so the product must be *smaller* than any single factor — this eliminates D (10/11) and E (11/2) immediately. Choice C (1/2) would be the answer if only the first factor were used, and B (1/9) would result from miscounting the factors (stopping at 1 − 1/9). Since n runs from 2 to 10 inclusive, there are exactly 9 factors and the answer is 1/10.

ANSWER 1: A

---

# Problem 2:
Factor 2016 into primes by repeated division:

2016 = 2 × 1008 = 2² × 504 = 2³ × 252 = 2⁴ × 126 = 2⁵ × 63, and 63 = 3² × 7.

So 2016 = 2⁵ × 3² × 7.

The **distinct** prime divisors are just 2, 3, and 7 (the exponents don't matter — we count each prime once).

Sum = 2 + 3 + 7 = **12**

Ruling out traps: A (9) would come from forgetting the prime 3; E (63) is the odd part 3² × 7 of 2016, not a sum of primes; adding primes *with repetition* gives 2+2+2+2+2+3+3+7 = 23, which isn't even a choice. The correct distinct-prime sum is 12.

ANSWER 2: B

---

# Problem 3:
Starting at 200 people in 1998, the population triples every 25 years:

- 1998: 200
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16 200

We want about 6000. Check how many triplings are needed: 6000/200 = 30, and 3³ = 27 ≈ 30 (while 3⁴ = 81 is way too big). So about **three triplings ≈ 75 years** after 1998, i.e., around 2073.

Comparing to the choices: at 2073 the population is 5400 (off by only 600), while at 2098 it is 16 200 (off by over 10 000). The closest estimate offered is **2075**.

Ruling out traps: 2050 and 2100 bracket the answer but are farther from the true crossing point than 2075; 2125 and 2150 correspond to populations of tens of thousands.

ANSWER 3: B

---

# Problem 4:
If N people were asked and exactly 74% said yes, then 0.74N must be a whole number of people:

0.74N = (74/100)N = (37/50)N

Since 37 is prime and shares no factor with 50, the fraction 37/50 is fully reduced. For (37/50)N to be an integer, **50 must divide N**. The smallest such N is 50.

Check: with 50 people, 0.74 × 50 = 37 people said yes — exactly 74%. ✓

Ruling out the smaller choices: 10 people → 7.4 yes-votes; 20 → 14.8; 25 → 18.5 — none of these is a whole number of people, so each is impossible. Hence 50 is the fewest.

ANSWER 4: D

---

# Problem 5:
First, confirm the calculator display: entering 075 × 256 (no decimal points) gives 75 × 256 = 19 200. ✓ That matches.

Now place the decimal correctly. The rule for multiplying decimals: the product has as many decimal places as the two factors combined.

- 0.075 has **3** decimal places
- 2.56 has **2** decimal places
- Product has 3 + 2 = **5** decimal places

Start from 19 200 and move the decimal point 5 places to the left:

19 200 → 0.19200 = **0.192**

Ruling out traps: 192 (E) and 19.2 (D) come from shifting the point only 2 or 3 places; 0.0192 (A) from shifting 6 places. A quick sanity check: 0.075 × 2.56 should be a bit less than 0.1 × 2.56 = 0.256, and 0.192 fits perfectly.

ANSWER 5: B

---

# Problem 6:
Convert the time limits to minutes, then to words:

- Minimum time: 1/2 hour = 30 minutes → 150 × 30 = **4500 words**
- Maximum time: 3/4 hour = 45 minutes → 150 × 45 = **6750 words**

So the speech needs between 4500 and 6750 words.

Check each choice: 2250 words = 15 min (too short); 3000 = 20 min (too short); 4200 = 28 min (too short); 4350 = 29 min (just under a half hour — too short); 5650 = 5650/150 ≈ 37.7 minutes, which lies nicely between 30 and 45 minutes. ✓

Only 5650 works.

ANSWER 6: E

---

# Problem 7:
Use time = distance ÷ speed, and convert to minutes (1 hour = 60 min):

- Monday: 2 miles at 5 mph → 2/5 hour = 24 minutes
- Wednesday: 2 miles at 3 mph → 2/3 hour = 40 minutes
- Friday: 2 miles at 4 mph → 2/4 = 1/2 hour = 30 minutes

Actual total: 24 + 40 + 30 = **94 minutes**.

If he had always gone 4 mph: total distance 6 miles at 4 mph → 6/4 = 1.5 hours = 90 minutes.

Difference: 94 − 90 = **4 minutes less**.

Ruling out traps: note Monday would actually take *longer* at 4 mph (30 vs. 24 min), so you can't just assume each day saves time — Wednesday's big savings (40 → 30) must be combined with Monday's loss (24 → 30). Net saving = 10 − 6 = 4 minutes, matching the totals.

ANSWER 7: D

---

# Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6, where a is odd. Their sum is

a + (a+2) + (a+4) + (a+6) = 4a + 12 = 4(a + 3).

Since a is odd, a + 3 is **even**, so the sum equals 4 × (an even number), i.e., every such sum is **divisible by 8**.

Test divisibility by 8 for each choice:

- 16 ÷ 8 = 2 ✓ (indeed 1 + 3 + 5 + 7 = 16)
- 40 ÷ 8 = 5 ✓ (7 + 9 + 11 + 13 = 40)
- 72 ÷ 8 = 9 ✓ (15 + 17 + 19 + 21 = 72)
- 100 ÷ 8 = 12.5 ✗ — **not divisible by 8**
- 200 ÷ 8 = 25 ✓ (47 + 49 + 51 + 53 = 200)

So 100 cannot be written as such a sum. (Check directly: 100 = 4(a+3) forces a + 3 = 25, so a = 22 — but 22 is even, a contradiction.)

ANSWER 8: D

---

# Problem 9:
Blake's total over four tests: 4 × 78 = 312 points.

Jenny's scores compared to Blake's, test by test: +10, −10, +20, +20.

Total difference: +10 − 10 + 20 + 20 = **+40 points**.

Jenny's total: 312 + 40 = 352, so her average is 352 ÷ 4 = 88.

Jenny's average exceeds Blake's by 88 − 78 = **10 points**.

(Equivalently: the *average* difference is the total difference divided by the number of tests: 40 ÷ 4 = 10.)

Ruling out traps: E (40) is the total point difference — it forgets to divide by the 4 tests; C (20) might come from averaging only the +20's and ignoring the +10/−10 pair; B and D have no valid basis.

ANSWER 9: A

---

# Problem 10:
Compute the numerator and denominator separately:

- (0.2)³ = 

Problem 1:  
The product is  
\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots\left(1-\frac1{10}\right).
\]
Rewrite each factor:
\[
\frac12\cdot \frac23\cdot \frac34\cdot \frac45\cdots \frac9{10}.
\]
Everything cancels except the first numerator and the last denominator:
\[
\frac1{10}.
\]
So the answer is A.

ANSWER 1: A

Problem 2:  
We need the sum of the distinct prime factors of 2016. Factor:
\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot 252=2^4\cdot 126=2^5\cdot 63.
\]
Also,
\[
63=3^2\cdot 7.
\]
Thus
\[
2016=2^5\cdot 3^2\cdot 7.
\]
The distinct prime divisors are 2, 3, and 7. Their sum is
\[
2+3+7=12.
\]

ANSWER 2: B

Problem 3:  
The population triples every 25 years. Starting with 200 in 1998, after \(t\) years the population is approximately
\[
200\cdot 3^{t/25}.
\]
We want this to be about 6000:
\[
200\cdot 3^{t/25}=6000.
\]
Divide by 200:
\[
3^{t/25}=30.
\]
Now,
\[
3^3=27
\]
and
\[
3^4=81.
\]
So \(t/25\) is a little more than 3, meaning \(t\) is a little more than 75 years.  
\[
1998+75=2073,
\]
so the closest choice is 2075.

ANSWER 3: B

Problem 4:  
Exactly 74% answered “yes.” Write 74% as a fraction in lowest terms:
\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]
For the number of “yes” answers to be an integer, the total number of people asked must be a multiple of 50. The smallest such number among the choices is 50.  
Check: \(\frac{37}{50}=74\%\).

ANSWER 4: D

Problem 5:  
Betty forgot the decimal points, so she effectively multiplied
\[
75\cdot 256=19200.
\]
Now restore the decimal places.  
\(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places, so the correct product must have
\[
3+2=5
\]
decimal places. Therefore,
\[
0.075\cdot 2.56=\frac{19200}{100000}=0.192.
\]

ANSWER 5: B

Problem 6:  
The speech must last between one-half hour and three-quarters of an hour, which is between
\[
30\text{ minutes}
\]
and
\[
45\text{ minutes}.
\]
At 150 words per minute, the acceptable word counts are between
\[
150\cdot 30=4500
\]
and
\[
150\cdot 45=6750.
\]
Among the choices, only 5650 lies between 4500 and 6750.

ANSWER 6: E

Problem 7:  
Each day he went 2 miles.

Monday: speed 5 mph, time
\[
\frac{2}{5}\text{ hour}=24\text{ minutes}.
\]
Wednesday: speed 3 mph, time
\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]
Friday: speed 4 mph, time
\[
\frac{2}{4}\text{ hour}=30\text{ minutes}.
\]
Total time:
\[
24+40+30=94\text{ minutes}.
\]
If he had walked all 6 miles at 4 mph, the time would be
\[
\frac{6}{4}\text{ hour}=1.5\text{ hours}=90\text{ minutes}.
\]
The difference is
\[
94-90=4\text{ minutes}.
\]

ANSWER 7: D

Problem 8:  
Let four consecutive odd integers be
\[
2k+1,\;2k+3,\;2k+5,\;2k+7.
\]
Their sum is
\[
(2k+1)+(2k+3)+(2k+5)+(2k+7)=8k+16=8(k+2).
\]
So any sum of four consecutive odd integers must be divisible by 8.

Check the choices:
- 16 is divisible by 8.
- 40 is divisible by 8.
- 72 is divisible by 8.
- 100 is not divisible by 8.
- 200 is divisible by 8.

Therefore, 100 cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

Problem 9:  
Blake’s average was 78 on four tests, so his total score was
\[
4\cdot 78=312.
\]
Jenny’s scores differed from Blake’s by
\[
+10,\;-10,\;+20,\;+20.
\]
The total difference is
\[
10-10+20+20=40.
\]
So Jenny’s total score was 40 points higher than Blake’s. Since there are 4 tests, her average was higher by
\[
\frac{40}{4}=10.
\]

ANSWER 9: A

Problem 10:  
Compute:
\[
(0.2)^3=0.008
\]
and
\[
(0.02)^2=0.0004.
\]
Thus
\[
\frac{(0.2)^3}{(0.02)^2}=\frac{0.008}{0.0004}=20.
\]

ANSWER 10: E

Problem 1:
Each factor simplifies: (1 − 1/2) = 1/2, (1 − 1/3) = 2/3, …, (1 − 1/10) = 9/10.

Product = (1/2)(2/3)(3/4)⋯(9/10).

Everything telescopes: the numerator of each fraction cancels the denominator of the previous one, leaving 1/10.

(Choice B, 1/9, would come from miscounting the last denominator; the product is definitely less than 1/2, ruling out C, D, E.)

ANSWER 1: A

Problem 2:
Factor 2016: 2016 = 2 · 1008 = 2·2·504 = 2·2·2·252 = 2^5 · 63 = 2^5 · 3^2 · 7.

Distinct primes: 2, 3, 7. Sum = 2 + 3 + 7 = 12.

(9 = 2+7 misses 3; 16 = 2·5+... etc.; 49 and 63 come from adding non‑prime factors or all factors with multiplicity.)

ANSWER 2: B

Problem 3:
Start: 1998 → 200 people. Tripling every 25 years:
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16200

The population reaches about 6000 shortly after 2073, so the best estimate among the choices is 2075.

(2050 gives ~1800, 2100 gives ~16000 — both far off.)

ANSWER 3: B

Problem 4:
We need 74% of the number of people to be a whole number: 0.74n = 37n/50 must be an integer.

Since 37/50 is in lowest terms (37 is prime), n must be a multiple of 50. The smallest is n = 50 (then 37 people said yes).

Check the smaller options: 74% of 10 = 7.4, of 20 = 14.8, of 25 = 18.5 — none are whole numbers.

ANSWER 4: D

Problem 5:
Ignoring decimals: 75 × 256 = 19200, matching the display.

Count decimal places: 0.075 has 3, 2.56 has 2, total 5 decimal places.

So 0.075 × 2.56 = 19200 with the decimal point moved 5 places left = 0.19200 = 0.192.

Sanity check: 0.075 ≈ 0.08 and 2.56 ≈ 2.5, product ≈ 0.2. So 0.192 is right (not 0.0192 or 1.92).

ANSWER 5: B

Problem 6:
Time allowed: 30 minutes to 45 minutes.

At 150 words per minute:
- Minimum words = 150 × 30 = 4500
- Maximum words = 150 × 45 = 6750

We need a count between 4500 and 6750. Only 5650 qualifies.

(2250 would take 15 min, 3000 → 20 min, 4200 → 28 min, 4350 → 29 min — all too short.)

ANSWER 6: E

Problem 7:
Time = distance ÷ speed, with 2 miles each day.

- Monday at 5 mph: 2/5 h = 24 min
- Wednesday at 3 mph: 2/3 h = 40 min
- Friday at 4 mph: 2/4 h = 30 min

Total actual = 24 + 40 + 30 = 94 minutes.

If always 4 mph: 30 min each day → 3 × 30 = 90 minutes.

Difference: 94 − 90 = 4 minutes less.

ANSWER 7: D

Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6 with a odd.

Sum = 4a + 12 = 4(a + 3). Since a is odd, a + 3 is even, say a + 3 = 2k, so the sum = 8k — a multiple of 8.

Check the choices for divisibility by 8:
- 16 = 8·2 ✓ (1+3+5+7)
- 40 = 8·5 ✓ (7+9+11+13)
- 72 = 8·9 ✓ (15+17+19+21)
- 100 ÷ 8 = 12.5 ✗
- 200 = 8·25 ✓ (47+49+51+53)

So 100 cannot be written that way.

ANSWER 8: D

Problem 9:
Jenny's total differs from Blake's total by:
+10 − 10 + 20 + 20 = +40 points.

Average difference = 40 ÷ 4 = 10 points.

(40 is the total difference, not the average — that's the trap in choice E. Blake's 78 average is not even needed, though Jenny averages 88, which is under 100, so it's consistent.)

ANSWER 9: A

Problem 10:
(0.2)³ = 0.008 and (0.02)² = 0.0004.

0.008 ÷ 0.0004 = 80/4 = 20.

(Alternatively: (2×10⁻¹)³ / (2×10⁻²)² = 8×10⁻³ / (4×10⁻⁴) = 2 × 10¹ = 20.)

ANSWER 10: E

Problem 1:

We need to compute

\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots \left(1-\frac1{10}\right).
\]

Rewrite each factor:

\[
1-\frac12=\frac12,\qquad
1-\frac13=\frac23,\qquad
1-\frac14=\frac34,
\]

and so on. Thus the product is

\[
\frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdots\frac9{10}.
\]

Nearly everything cancels:

\[
\frac{\cancel1}{2}\cdot\frac{\cancel2}{3}\cdot\frac{\cancel3}{4}\cdots
\frac{\cancel9}{10}
=\frac1{10}.
\]

Choice B, \(\frac19\), might seem plausible if one thinks there are 9 factors, but the cancellation leaves the first numerator \(1\) and final denominator \(10\).

\[
\boxed{\frac1{10}}
\]

ANSWER 1: A

Problem 2:

We need the sum of the distinct prime divisors of \(2016\).

Factor \(2016\):

\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot252=2^4\cdot126=2^5\cdot63.
\]

Since

\[
63=3^2\cdot7,
\]

we have

\[
2016=2^5\cdot3^2\cdot7.
\]

The distinct prime divisors are

\[
2,\ 3,\ 7.
\]

Their sum is

\[
2+3+7=12.
\]

Do not add repeated prime factors such as all five 2s and both 3s; the question asks for distinct primes only.

\[
\boxed{12}
\]

ANSWER 2: B

Problem 3:

The population starts at \(200\) in 1998 and triples every 25 years.

We want it to be about \(6000\). Compare:

\[
\frac{6000}{200}=30.
\]

So the population must grow by a factor of about \(30\).

Powers of \(3\) are:

\[
3^1=3,\qquad 3^2=9,\qquad 3^3=27,\qquad 3^4=81.
\]

A factor of \(27\) is close to \(30\), so we need about 3 periods of 25 years:

\[
3\cdot25=75\text{ years}.
\]

Therefore the estimated year is

\[
1998+75=2073.
\]

The nearest choice is \(2075\).

Indeed, after 75 years the estimated population is

\[
200\cdot 27=5400,
\]

which is reasonably close to \(6000\). Waiting 100 years would give \(200\cdot81=16200\), much too large.

\[
\boxed{2075}
\]

ANSWER 3: B

Problem 4:

Exactly \(74\%\) of the people answered yes.

Convert \(74\%\) to a fraction:

\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]

For the number of yes answers to be a whole number, the total number surveyed must be a multiple of \(50\).

The smallest possible total is therefore

\[
50.
\]

For example,

\[
74\%\text{ of }50=\frac{74}{100}\cdot 50=37,
\]

which is an integer number of people.

Smaller choices such as 10, 20, and 25 do not work, since \(74\%\) of them is not a whole number.

\[
\boxed{50}
\]

ANSWER 4: D

Problem 5:

Without decimal points, Betty entered

\[
75\times256=19200.
\]

Now place the decimal points correctly.

The number \(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places. Therefore the product must have

\[
3+2=5
\]

decimal places.

So move the decimal point in \(19200\) five places left:

\[
19200\longrightarrow 0.19200=0.192.
\]

Thus

\[
0.075\times2.56=0.192.
\]

Choice A, \(0.0192\), has one too many decimal places, while \(1.92\) and larger answers are too large because \(0.075\) is less than \(0.1\).

\[
\boxed{0.192}
\]

ANSWER 5: B

Problem 6:

Julie’s speech must last between one-half hour and three-quarters of an hour.

Convert these times to minutes:

\[
\frac12\text{ hour}=30\text{ minutes},
\qquad
\frac34\text{ hour}=45\text{ minutes}.
\]

At \(150\) words per minute, the minimum appropriate word count is

\[
30\cdot150=4500.
\]

The maximum appropriate word count is

\[
45\cdot150=6750.
\]

So her speech should contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) and \(3000\) are too few.
- \(4200\) and \(4350\) are still below \(4500\).
- \(5650\) is between \(4500\) and \(6750\).

Therefore the appropriate word count is

\[
\boxed{5650}.
\]

ANSWER 6: E

Problem 7:

Grandfather traveled 2 miles on each of 3 days.

Find his actual treadmill time.

On Monday, he jogged at 5 mph:

\[
\frac{2}{5}\text{ hour}=0.4\text{ hour}=24\text{ minutes}.
\]

On Wednesday, he walked at 3 mph:

\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]

On Friday, he walked at 4 mph:

\[
\frac24\text{ hour}=\frac12\text{ hour}=30\text{ minutes}.
\]

So his actual total time was

\[
24+40+30=94\text{ minutes}.
\]

If he had walked at 4 mph on all three days, each 2-mile trip would take 30 minutes:

\[
3\cdot30=90\text{ minutes}.
\]

Thus he would have spent

\[
94-90=4
\]

fewer minutes on the treadmill.

The 5-mph jogging day saves time, but the 3-mph walking day costs more time; together the actual time is 4 minutes longer than always walking at 4 mph.

\[
\boxed{4}
\]

ANSWER 7: D

Problem 8:

Let the four consecutive odd integers be

\[
n,\ n+2,\ n+4,\ n+6,
\]

where \(n\) is odd.

Their sum is

\[
n+(n+2)+(n+4)+(n+6)=4n+12.
\]

Factor:

\[
4n+12=4(n+3).
\]

Since \(n\) is odd, \(n+3\) is even. Therefore \(4(n+3)\) must be divisible by \(8\).

So a number can be written as the sum of four consecutive odd integers only if it is divisible by \(8\).

Check the choices:

\[
16\div8=2,
\quad
40\div8=5,
\quad
72\div8=9,
\quad
200\div8=25,
\]

so all of these are divisible by 8.

But

\[
100\div8=12.5,
\]

so \(100\) is not divisible by 8. Therefore it cannot be the sum of four consecutive odd integers.

For example, \(16=1+3+5+7\), showing that the condition is possible for the other type of choices.

\[
\boxed{100}
\]

ANSWER 8: D

Problem 9:

Blake’s average is not actually needed, because we can compare Jenny’s scores directly to Blake’s.

Jenny’s score changes relative to Blake’s are:

- first test: \(+10\),
- second test: \(-10\),
- third test: \(+20\),
- fourth test: \(+20\).

Her total score exceeds Blake’s by

\[
10-10+20+20=40.
\]

Since there are four tests, her average exceeds Blake’s by

\[
\frac{40}{4}=10.
\]

The \(40\) is the difference in total points, not the difference in average points.

\[
\boxed{10}
\]

ANSWER 9: A

Problem 10:

Compute

\[
\frac{(0.2)^3}{(0.02)^2}.
\]

First,

\[
(0.2)^3=0.2\cdot0.2\cdot0.2=0.008.
\]

Next,

\[
(0.02)^2=0.02\cdot0.02=0.0004.
\]

Therefore,

\[
\frac{0.008}{0.0004}=20.
\]

Another way is to use powers of 10:

\[
0.2=2\cdot10^{-1},\qquad 0.02=2\cdot10^{-2}.
\]

Then

\[
\fra

Problem 1:
We are asked for the value of $(1-1/2)(1-1/3)\cdots (1-1/10)$.

Write each factor as a single fraction:
$1-1/2 = 1/2$, $1-1/3 = 2/3$, $1-1/4 = 3/4$, ..., $1-1/10 = 9/10$.

So the product is:
$$\frac12 \cdot \frac23 \cdot \frac34 \cdots \frac89 \cdot \frac9{10}.$$

This telescopes: the numerator 2 cancels denominator 2, numerator 3 cancels denominator 3, etc. All cancel except the first denominator 2? Let's see:
$$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10} = \frac{1}{10}.$$
Formally numerator $=1\cdot2\cdots9$, denominator $=2\cdot3\cdots10$, ratio $=1/10$.

Choices B $1/9$, C $1/2$, D $10/11$, E $11/2$ do not result from cancellation.
ANSWER 1: A

Problem 2:
We need the sum of the distinct prime divisors of 2016.

Factor 2016. It is even:
$2016 = 2\times1008 = 2\times2\times504 = 2^2\times504$.
$504 = 2\times252 = 2^2\times252$? Continue systematically:
$2016 /2 =1008$, $/2=504$, $/2=252$, $/2=126$, $/2=63$.
So $2016 = 2^5 \times 63$.
$63 = 7\times9 = 3^2\times7$.

Hence $2016 = 2^5 \cdot 3^2 \cdot 7^1$.
The distinct primes occurring are $2,3,7$.

Their sum is $2+3+7 = 12$.

This rules out A 9, C 16, D 49, E 63 which would come from missing a prime or including powers.
ANSWER 2: B

Problem 3:
In 1998 population is 200, triples every 25 years. When is it about 6000?

Ratio wanted: $6000/200 = 30$. So we need about 30 times growth.

Each 25 years multiply by 3:
After 25 yr: $200\times3=600$.
After 50 yr: $600\times3=1800$.
After 75 yr: $1800\times3=5400$.

5400 is about 6000. So about 3 triplings, i.e. about $3\times25=75$ years after 1998.

$1998+75 = 2073$, closest to 2075 among choices.

2050 is only ~2 triplings, 2100+ is too many.
ANSWER 3: B

Problem 4:
Some $n$ people asked, exactly $74\%$ said yes. Find smallest possible $n$.

Let $k$ be number saying yes. Then $k/n = 74\% = 74/100 = 37/50$.

So $k = 37n/50$ must be an integer. Since 37 and 50 are coprime (gcd=1), $n$ must be a multiple of 50 for $37n/50$ to be integer.

Smallest positive multiple is $n=50$, giving $k=37$.

$n=10,20,25$ give $7.4,14.8,18.5$ yes-votes, impossible. $n=100$ works but is not fewest.
ANSWER 4: D

Problem 5:
True product is $0.075 \times 2.56$. Entering $75 \times 256$ gave $19200$. What is correct with decimals?

$0.075$ has 3 decimal places: $0.075 = 75/1000$.
$2.56$ has 2 decimal places: $2.56 = 256/100$.

So product $= 75\times256 / (1000\times100) = 19200 / 100000$.

$19200/100000 = 0.192$.

In general move decimal 3+2=5 places left. .0192 would be 6 places, 1.92 would be 4 places, etc.
ANSWER 5: B

Problem 6:
Speech must be $1/2$ hour to $3/4$ hour at 150 words per minute. Which word count fits?

Convert hours to minutes: $1/2$ hr $=30$ min, $3/4$ hr $=45$ min.

At 150 wpm:
minimum words $=30\times150 = 4500$.
maximum words $=45\times150 = 6750$.

We need word count between 4500 and 6750 inclusive.

A 2250 is 15 min, B 3000 is 20 min, C 4200 is 28 min, D 4350 is 29 min — all too short. Only E 5650 satisfies $4500 \le 5650 \le 6750$.
ANSWER 6: E

Problem 7:
2 miles each day for 3 days. Speeds: Monday 5 mph, Wednesday 3 mph, Friday 4 mph. Compare total time to time if all 3 days at 4 mph. How many minutes saved?

Time = distance / speed.

Actual:
Monday: $2/5$ hr $=0.4$ hr $=0.4\times60=24$ min.
Wednesday: $2/3$ hr $= (2/3)\times60 =40$ min.
Friday: $2/4=1/2$ hr $=30$ min.
Total $=24+40+30=94$ min.

If always 4 mph: each day $30$ min, total $3\times30=90$ min.

Saving $=94-90=4$ minutes.
ANSWER 7: D

Problem 8:
Which integer cannot be sum of four consecutive odd integers?

Let the four be $n, n+2, n+4, n+6$ with $n$ odd.
Sum $=4n+12 =4(n+3)$.

Since $n$ is odd, $n+3$ is even, say $n+3=2m$. Then sum $=4\cdot2m=8m$, a multiple of 8. So any such sum must be divisible by 8.

Test:
16: $16/8=2$ possible; indeed $1+3+5+7=16$.
40: $40/8=5$ possible; indeed $7+9+11+13=40$.
72: $72/8=9$ possible; e.g. $15+17+19+21=72$.
200: $200/8=25$ possible; e.g. $47+49+51+53=200$.
100: $100/8=12.5$ not integer, so impossible.

Thus 100 cannot be written that way.
ANSWER 8: D

Problem 9:
Blake averaged 78 over four tests. Jenny vs Blake: +10, -10, +20, +20. By how much does Jenny's average exceed Blake's?

Only differences matter. Total points difference over 4 tests:
$+10-10+20+20 = +40$.

Average is total/4, so Jenny's average is $40/4=10$ points higher.

Blake's 78 is irrelevant to the difference. Choices 15,20,25,40 confuse total with average.
ANSWER 9: A

Problem 10:
Compute $(.2)^3/(.02)^2$.

$(.2)^3 = 0.2\times0.2\times0.2 = 0.04\times0.2 = 0.008$.
$(.02)^2 = 0.02\times0.02 = 0.0004$.

Ratio $=0.008/0.0004 = 8/0.4 = 80/4 =20$.

Alternatively with powers of 10: $(2\times10^{-1})^3=8\times10^{-3}$, $(2\times10^{-2})^2=4\times10^{-4}$, ratio $=2\times10^{1}=20$.

So not .2, 2, 10, or 15.
ANSWER 10: E

Problem 1:
We are asked to evaluate the product of the 9 factors:
$$\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{10}\right)$$

First, simplify each factor:
$$1 - \frac{1}{2} = \frac{1}{2}$$
$$1 - \frac{1}{3} = \frac{2}{3}$$
$$1 - \frac{1}{4} = \frac{3}{4}$$
$$\vdots$$
$$1 - \frac{1}{10} = \frac{9}{10}$$

Now, multiply the fractions together:
$$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{9}{10}$$

Notice that this is a telescoping product where the denominator of each fraction cancels with the numerator of the following fraction:
$$\frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \cdots \times \frac{\cancel{9}}{10} = \frac{1}{10}$$

ANSWER 1: A

---

Problem 2:
We are asked to find the sum of the distinct prime integer divisors of $2016$.

First, find the prime factorization of $2016$:
- $2016 = 2 \times 1008$
- $1008 = 2 \times 504$
- $504 = 2 \times 252$
- $252 = 2 \times 126$
- $126 = 2 \times 63$
- $63 = 9 \times 7 = 3^2 \times 7$

Thus, the prime factorization of $2016$ is:
$$2016 = 2^5 \times 3^2 \times 7$$

The distinct prime divisors are $2$, $3$, and $7$. 
Their sum is:
$$2 + 3 + 7 = 12$$

ANSWER 2: B

---

Problem 3:
We are given that the population of Nisos Isles is $200$ in the year $1998$, and it triples every $25$ years. We want to find the year in which the population reaches approximately $6000$.

Let $n$ be the number of $25$-year periods that pass after $1998$. The population can be represented as:
$$P = 200 \times 3^n$$

We want $P \approx 6000$:
$$200 \times 3^n \approx 6000 \implies 3^n \approx \frac{6000}{200} = 30$$

Let's test powers of $3$:
- For $n = 1$: $3^1 = 3 \implies P = 600$
- For $n = 2$: $3^2 = 9 \implies P = 1800$
- For $n = 3$: $3^3 = 27 \implies P = 5400$
- For $n = 4$: $3^4 = 81 \implies P = 16200$

$5400$ is very close to $6000$, so about $n = 3$ periods of $25$ years are needed:
$$3 \times 25 = 75\text{ years}$$

Adding $75$ years to $1998$:
$$1998 + 75 = 2073 \approx 2075$$

ANSWER 3: B

---

Problem 4:
We are given that exactly $74\%$ of the people surveyed answered "yes." We need to find the minimum number of people who could have been surveyed.

Let $N$ be the total number of people surveyed. The number of people who answered "yes" is:
$$74\% \times N = \frac{74}{100} N = \frac{37}{50} N$$

Since the number of people who answered "yes" must be a whole number, $\frac{37}{50} N$ must be an integer. 
Since $37$ and $50$ share no common factors other than $1$ (they are coprime), $N$ must be a multiple of $50$. 

The smallest positive integer multiple of $50$ is $50$. 
With $N = 50$, exactly $\frac{37}{50} \times 50 = 37$ people answered "yes."

ANSWER 4: D

---

Problem 5:
Betty is multiplying $0.075$ by $2.56$.
She calculated $75 \times 256 = 19200$.

Count the total number of digits to the right of the decimal point in the original numbers:
- $0.075$ has $3$ decimal places.
- $2.56$ has $2$ decimal places.

The product must therefore have $3 + 2 = 5$ decimal places:
$$19200 \times 10^{-5} = 0.19200 = 0.192$$

ANSWER 5: B

---

Problem 6:
Julie's speech must last between one-half hour ($30$ minutes) and three-quarters of an hour ($45$ minutes). Her speaking rate is $150$ words per minute.

Calculate the minimum and maximum acceptable word counts:
- Minimum words: $30\text{ minutes} \times 150\text{ words/minute} = 4500\text{ words}$
- Maximum words: $45\text{ minutes} \times 150\text{ words/minute} = 6750\text{ words}$

An appropriate word count must lie in the range $[4500, 6750]$. 
Looking at the choices:
- A. $2250$ (too low)
- B. $3000$ (too low)
- C. $4200$ (too low)
- D. $4350$ (too low)
- E. $5650$ is between $4500$ and $6750$ ($\frac{5650}{150} \approx 37.7$ minutes).

ANSWER 6: E

---

Problem 7:
Ted's grandfather covered $2$ miles on each of the $3$ days. 
First, find the time he actually spent each day using $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
- **Monday:** $\frac{2\text{ miles}}{5\text{ mph}} = \frac{2}{5}\text{ hour} = \frac{2}{5} \times 60 = 24\text{ minutes}$.
- **Wednesday:** $\frac{2\text{ miles}}{3\text{ mph}} = \frac{2}{3}\text{ hour} = \frac{2}{3} \times 60 = 40\text{ minutes}$.
- **Friday:** $\frac{2\text{ miles}}{4\text{ mph}} = \frac{2}{4}\text{ hour} = \frac{1}{2} \times 60 = 30\text{ minutes}$.

Total actual time spent:
$$24 + 40 + 30 = 94\text{ minutes}$$

If he had walked at $4\text{ mph}$ every day, the time spent each day would be $\frac{2}{4}\text{ hour} = 30\text{ minutes}$, so for all $3$ days:
$$3 \times 30 = 90\text{ minutes}$$

He would have spent:
$$94 - 90 = 4\text{ minutes less}$$

ANSWER 7: D

---

Problem 8:
Let four consecutive odd integers be represented as:
$$n, \quad n+2, \quad n+4, \quad n+6$$
where $n$ is an odd integer.

Their sum is:
$$S = n + (n+2) + (n+4) + (n+6) = 4n + 12 = 4(n+3)$$

Since $n$ is an odd integer, $n+3$ must be an even integer, which can be written as $2k$ for some integer $k$.
Substituting this in gives:
$$S = 4(2k) = 8k$$

Therefore, the sum of four consecutive odd integers must be a multiple of $8$.
Let's check the given options:
- A. $16 = 8 \times 2$ ($1 + 3 + 5 + 7 = 16$)
- B. $40 = 8 \times 5$ ($7 + 9 + 11 + 13 = 40$)
- C. $72 = 8 \times 9$ ($15 + 17 + 19 + 21 = 72$)
- D. $100$: not divisible by $8$ ($100 / 8 = 12.5$)
- E. $200 = 8 \times 25$ ($47 + 49 + 51 + 53 = 200$)

Hence, $100$ cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

---

Problem 9:
Let Blake's scores on the four tests be $B_1, B_2, B_3, B_4$.
Jenny's scores are:
- First test: $B_1 + 10$
- Second test: $B_2 - 10$
- Third test: $B_3 + 20$
- Fourth test: $B_4 + 20$

The sum of Jenny's scores minus the sum of Blake's scores is:
$$(B_1 + 10 + B_2 - 10 + B_3 + 20 + B_4 + 20) - (B_1 + B_2 + B_3 + B_4) = 10 - 10 + 20 + 20 = 40\text{ points}$$

Since there are $4$ tests, the difference between Jenny's average and Blake's aver

# Problem 1:
We need the product of the 9 factors (1 − 1/n) for n = 2, 3, 4, …, 10.

Rewrite each factor: 1 − 1/n = (n−1)/n. So the product becomes

(1/2)(2/3)(3/4)(4/5)(5/6)(6/7)(7/8)(8/9)(9/10)

This **telescopes**: each numerator cancels the denominator before it (2 cancels 2, 3 cancels 3, …, 9 cancels 9), leaving only the first numerator and the last denominator:

Product = 1/10

Ruling out traps: every factor is less than 1, so the product must be *smaller* than any single factor — this eliminates D (10/11) and E (11/2) immediately. Choice C (1/2) would be the answer if only the first factor were used, and B (1/9) would result from miscounting the factors (stopping at 1 − 1/9). Since n runs from 2 to 10 inclusive, there are exactly 9 factors and the answer is 1/10.

ANSWER 1: A

---

# Problem 2:
Factor 2016 into primes by repeated division:

2016 = 2 × 1008 = 2² × 504 = 2³ × 252 = 2⁴ × 126 = 2⁵ × 63, and 63 = 3² × 7.

So 2016 = 2⁵ × 3² × 7.

The **distinct** prime divisors are just 2, 3, and 7 (the exponents don't matter — we count each prime once).

Sum = 2 + 3 + 7 = **12**

Ruling out traps: A (9) would come from forgetting the prime 3; E (63) is the odd part 3² × 7 of 2016, not a sum of primes; adding primes *with repetition* gives 2+2+2+2+2+3+3+7 = 23, which isn't even a choice. The correct distinct-prime sum is 12.

ANSWER 2: B

---

# Problem 3:
Starting at 200 people in 1998, the population triples every 25 years:

- 1998: 200
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16 200

We want about 6000. Check how many triplings are needed: 6000/200 = 30, and 3³ = 27 ≈ 30 (while 3⁴ = 81 is way too big). So about **three triplings ≈ 75 years** after 1998, i.e., around 2073.

Comparing to the choices: at 2073 the population is 5400 (off by only 600), while at 2098 it is 16 200 (off by over 10 000). The closest estimate offered is **2075**.

Ruling out traps: 2050 and 2100 bracket the answer but are farther from the true crossing point than 2075; 2125 and 2150 correspond to populations of tens of thousands.

ANSWER 3: B

---

# Problem 4:
If N people were asked and exactly 74% said yes, then 0.74N must be a whole number of people:

0.74N = (74/100)N = (37/50)N

Since 37 is prime and shares no factor with 50, the fraction 37/50 is fully reduced. For (37/50)N to be an integer, **50 must divide N**. The smallest such N is 50.

Check: with 50 people, 0.74 × 50 = 37 people said yes — exactly 74%. ✓

Ruling out the smaller choices: 10 people → 7.4 yes-votes; 20 → 14.8; 25 → 18.5 — none of these is a whole number of people, so each is impossible. Hence 50 is the fewest.

ANSWER 4: D

---

# Problem 5:
First, confirm the calculator display: entering 075 × 256 (no decimal points) gives 75 × 256 = 19 200. ✓ That matches.

Now place the decimal correctly. The rule for multiplying decimals: the product has as many decimal places as the two factors combined.

- 0.075 has **3** decimal places
- 2.56 has **2** decimal places
- Product has 3 + 2 = **5** decimal places

Start from 19 200 and move the decimal point 5 places to the left:

19 200 → 0.19200 = **0.192**

Ruling out traps: 192 (E) and 19.2 (D) come from shifting the point only 2 or 3 places; 0.0192 (A) from shifting 6 places. A quick sanity check: 0.075 × 2.56 should be a bit less than 0.1 × 2.56 = 0.256, and 0.192 fits perfectly.

ANSWER 5: B

---

# Problem 6:
Convert the time limits to minutes, then to words:

- Minimum time: 1/2 hour = 30 minutes → 150 × 30 = **4500 words**
- Maximum time: 3/4 hour = 45 minutes → 150 × 45 = **6750 words**

So the speech needs between 4500 and 6750 words.

Check each choice: 2250 words = 15 min (too short); 3000 = 20 min (too short); 4200 = 28 min (too short); 4350 = 29 min (just under a half hour — too short); 5650 = 5650/150 ≈ 37.7 minutes, which lies nicely between 30 and 45 minutes. ✓

Only 5650 works.

ANSWER 6: E

---

# Problem 7:
Use time = distance ÷ speed, and convert to minutes (1 hour = 60 min):

- Monday: 2 miles at 5 mph → 2/5 hour = 24 minutes
- Wednesday: 2 miles at 3 mph → 2/3 hour = 40 minutes
- Friday: 2 miles at 4 mph → 2/4 = 1/2 hour = 30 minutes

Actual total: 24 + 40 + 30 = **94 minutes**.

If he had always gone 4 mph: total distance 6 miles at 4 mph → 6/4 = 1.5 hours = 90 minutes.

Difference: 94 − 90 = **4 minutes less**.

Ruling out traps: note Monday would actually take *longer* at 4 mph (30 vs. 24 min), so you can't just assume each day saves time — Wednesday's big savings (40 → 30) must be combined with Monday's loss (24 → 30). Net saving = 10 − 6 = 4 minutes, matching the totals.

ANSWER 7: D

---

# Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6, where a is odd. Their sum is

a + (a+2) + (a+4) + (a+6) = 4a + 12 = 4(a + 3).

Since a is odd, a + 3 is **even**, so the sum equals 4 × (an even number), i.e., every such sum is **divisible by 8**.

Test divisibility by 8 for each choice:

- 16 ÷ 8 = 2 ✓ (indeed 1 + 3 + 5 + 7 = 16)
- 40 ÷ 8 = 5 ✓ (7 + 9 + 11 + 13 = 40)
- 72 ÷ 8 = 9 ✓ (15 + 17 + 19 + 21 = 72)
- 100 ÷ 8 = 12.5 ✗ — **not divisible by 8**
- 200 ÷ 8 = 25 ✓ (47 + 49 + 51 + 53 = 200)

So 100 cannot be written as such a sum. (Check directly: 100 = 4(a+3) forces a + 3 = 25, so a = 22 — but 22 is even, a contradiction.)

ANSWER 8: D

---

# Problem 9:
Blake's total over four tests: 4 × 78 = 312 points.

Jenny's scores compared to Blake's, test by test: +10, −10, +20, +20.

Total difference: +10 − 10 + 20 + 20 = **+40 points**.

Jenny's total: 312 + 40 = 352, so her average is 352 ÷ 4 = 88.

Jenny's average exceeds Blake's by 88 − 78 = **10 points**.

(Equivalently: the *average* difference is the total difference divided by the number of tests: 40 ÷ 4 = 10.)

Ruling out traps: E (40) is the total point difference — it forgets to divide by the 4 tests; C (20) might come from averaging only the +20's and ignoring the +10/−10 pair; B and D have no valid basis.

ANSWER 9: A

---

# Problem 10:
Compute the numerator and denominator separately:

- (0.2)³ = 

Problem 1:  
The product is  
\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots\left(1-\frac1{10}\right).
\]
Rewrite each factor:
\[
\frac12\cdot \frac23\cdot \frac34\cdot \frac45\cdots \frac9{10}.
\]
Everything cancels except the first numerator and the last denominator:
\[
\frac1{10}.
\]
So the answer is A.

ANSWER 1: A

Problem 2:  
We need the sum of the distinct prime factors of 2016. Factor:
\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot 252=2^4\cdot 126=2^5\cdot 63.
\]
Also,
\[
63=3^2\cdot 7.
\]
Thus
\[
2016=2^5\cdot 3^2\cdot 7.
\]
The distinct prime divisors are 2, 3, and 7. Their sum is
\[
2+3+7=12.
\]

ANSWER 2: B

Problem 3:  
The population triples every 25 years. Starting with 200 in 1998, after \(t\) years the population is approximately
\[
200\cdot 3^{t/25}.
\]
We want this to be about 6000:
\[
200\cdot 3^{t/25}=6000.
\]
Divide by 200:
\[
3^{t/25}=30.
\]
Now,
\[
3^3=27
\]
and
\[
3^4=81.
\]
So \(t/25\) is a little more than 3, meaning \(t\) is a little more than 75 years.  
\[
1998+75=2073,
\]
so the closest choice is 2075.

ANSWER 3: B

Problem 4:  
Exactly 74% answered “yes.” Write 74% as a fraction in lowest terms:
\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]
For the number of “yes” answers to be an integer, the total number of people asked must be a multiple of 50. The smallest such number among the choices is 50.  
Check: \(\frac{37}{50}=74\%\).

ANSWER 4: D

Problem 5:  
Betty forgot the decimal points, so she effectively multiplied
\[
75\cdot 256=19200.
\]
Now restore the decimal places.  
\(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places, so the correct product must have
\[
3+2=5
\]
decimal places. Therefore,
\[
0.075\cdot 2.56=\frac{19200}{100000}=0.192.
\]

ANSWER 5: B

Problem 6:  
The speech must last between one-half hour and three-quarters of an hour, which is between
\[
30\text{ minutes}
\]
and
\[
45\text{ minutes}.
\]
At 150 words per minute, the acceptable word counts are between
\[
150\cdot 30=4500
\]
and
\[
150\cdot 45=6750.
\]
Among the choices, only 5650 lies between 4500 and 6750.

ANSWER 6: E

Problem 7:  
Each day he went 2 miles.

Monday: speed 5 mph, time
\[
\frac{2}{5}\text{ hour}=24\text{ minutes}.
\]
Wednesday: speed 3 mph, time
\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]
Friday: speed 4 mph, time
\[
\frac{2}{4}\text{ hour}=30\text{ minutes}.
\]
Total time:
\[
24+40+30=94\text{ minutes}.
\]
If he had walked all 6 miles at 4 mph, the time would be
\[
\frac{6}{4}\text{ hour}=1.5\text{ hours}=90\text{ minutes}.
\]
The difference is
\[
94-90=4\text{ minutes}.
\]

ANSWER 7: D

Problem 8:  
Let four consecutive odd integers be
\[
2k+1,\;2k+3,\;2k+5,\;2k+7.
\]
Their sum is
\[
(2k+1)+(2k+3)+(2k+5)+(2k+7)=8k+16=8(k+2).
\]
So any sum of four consecutive odd integers must be divisible by 8.

Check the choices:
- 16 is divisible by 8.
- 40 is divisible by 8.
- 72 is divisible by 8.
- 100 is not divisible by 8.
- 200 is divisible by 8.

Therefore, 100 cannot be written as the sum of four consecutive odd integers.

ANSWER 8: D

Problem 9:  
Blake’s average was 78 on four tests, so his total score was
\[
4\cdot 78=312.
\]
Jenny’s scores differed from Blake’s by
\[
+10,\;-10,\;+20,\;+20.
\]
The total difference is
\[
10-10+20+20=40.
\]
So Jenny’s total score was 40 points higher than Blake’s. Since there are 4 tests, her average was higher by
\[
\frac{40}{4}=10.
\]

ANSWER 9: A

Problem 10:  
Compute:
\[
(0.2)^3=0.008
\]
and
\[
(0.02)^2=0.0004.
\]
Thus
\[
\frac{(0.2)^3}{(0.02)^2}=\frac{0.008}{0.0004}=20.
\]

ANSWER 10: E

Problem 1:
Each factor simplifies: (1 − 1/2) = 1/2, (1 − 1/3) = 2/3, …, (1 − 1/10) = 9/10.

Product = (1/2)(2/3)(3/4)⋯(9/10).

Everything telescopes: the numerator of each fraction cancels the denominator of the previous one, leaving 1/10.

(Choice B, 1/9, would come from miscounting the last denominator; the product is definitely less than 1/2, ruling out C, D, E.)

ANSWER 1: A

Problem 2:
Factor 2016: 2016 = 2 · 1008 = 2·2·504 = 2·2·2·252 = 2^5 · 63 = 2^5 · 3^2 · 7.

Distinct primes: 2, 3, 7. Sum = 2 + 3 + 7 = 12.

(9 = 2+7 misses 3; 16 = 2·5+... etc.; 49 and 63 come from adding non‑prime factors or all factors with multiplicity.)

ANSWER 2: B

Problem 3:
Start: 1998 → 200 people. Tripling every 25 years:
- 2023: 600
- 2048: 1800
- 2073: 5400
- 2098: 16200

The population reaches about 6000 shortly after 2073, so the best estimate among the choices is 2075.

(2050 gives ~1800, 2100 gives ~16000 — both far off.)

ANSWER 3: B

Problem 4:
We need 74% of the number of people to be a whole number: 0.74n = 37n/50 must be an integer.

Since 37/50 is in lowest terms (37 is prime), n must be a multiple of 50. The smallest is n = 50 (then 37 people said yes).

Check the smaller options: 74% of 10 = 7.4, of 20 = 14.8, of 25 = 18.5 — none are whole numbers.

ANSWER 4: D

Problem 5:
Ignoring decimals: 75 × 256 = 19200, matching the display.

Count decimal places: 0.075 has 3, 2.56 has 2, total 5 decimal places.

So 0.075 × 2.56 = 19200 with the decimal point moved 5 places left = 0.19200 = 0.192.

Sanity check: 0.075 ≈ 0.08 and 2.56 ≈ 2.5, product ≈ 0.2. So 0.192 is right (not 0.0192 or 1.92).

ANSWER 5: B

Problem 6:
Time allowed: 30 minutes to 45 minutes.

At 150 words per minute:
- Minimum words = 150 × 30 = 4500
- Maximum words = 150 × 45 = 6750

We need a count between 4500 and 6750. Only 5650 qualifies.

(2250 would take 15 min, 3000 → 20 min, 4200 → 28 min, 4350 → 29 min — all too short.)

ANSWER 6: E

Problem 7:
Time = distance ÷ speed, with 2 miles each day.

- Monday at 5 mph: 2/5 h = 24 min
- Wednesday at 3 mph: 2/3 h = 40 min
- Friday at 4 mph: 2/4 h = 30 min

Total actual = 24 + 40 + 30 = 94 minutes.

If always 4 mph: 30 min each day → 3 × 30 = 90 minutes.

Difference: 94 − 90 = 4 minutes less.

ANSWER 7: D

Problem 8:
Let the four consecutive odd integers be a, a+2, a+4, a+6 with a odd.

Sum = 4a + 12 = 4(a + 3). Since a is odd, a + 3 is even, say a + 3 = 2k, so the sum = 8k — a multiple of 8.

Check the choices for divisibility by 8:
- 16 = 8·2 ✓ (1+3+5+7)
- 40 = 8·5 ✓ (7+9+11+13)
- 72 = 8·9 ✓ (15+17+19+21)
- 100 ÷ 8 = 12.5 ✗
- 200 = 8·25 ✓ (47+49+51+53)

So 100 cannot be written that way.

ANSWER 8: D

Problem 9:
Jenny's total differs from Blake's total by:
+10 − 10 + 20 + 20 = +40 points.

Average difference = 40 ÷ 4 = 10 points.

(40 is the total difference, not the average — that's the trap in choice E. Blake's 78 average is not even needed, though Jenny averages 88, which is under 100, so it's consistent.)

ANSWER 9: A

Problem 10:
(0.2)³ = 0.008 and (0.02)² = 0.0004.

0.008 ÷ 0.0004 = 80/4 = 20.

(Alternatively: (2×10⁻¹)³ / (2×10⁻²)² = 8×10⁻³ / (4×10⁻⁴) = 2 × 10¹ = 20.)

ANSWER 10: E

Problem 1:

We need to compute

\[
\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdots \left(1-\frac1{10}\right).
\]

Rewrite each factor:

\[
1-\frac12=\frac12,\qquad
1-\frac13=\frac23,\qquad
1-\frac14=\frac34,
\]

and so on. Thus the product is

\[
\frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdots\frac9{10}.
\]

Nearly everything cancels:

\[
\frac{\cancel1}{2}\cdot\frac{\cancel2}{3}\cdot\frac{\cancel3}{4}\cdots
\frac{\cancel9}{10}
=\frac1{10}.
\]

Choice B, \(\frac19\), might seem plausible if one thinks there are 9 factors, but the cancellation leaves the first numerator \(1\) and final denominator \(10\).

\[
\boxed{\frac1{10}}
\]

ANSWER 1: A

Problem 2:

We need the sum of the distinct prime divisors of \(2016\).

Factor \(2016\):

\[
2016=2\cdot 1008=2^2\cdot 504=2^3\cdot252=2^4\cdot126=2^5\cdot63.
\]

Since

\[
63=3^2\cdot7,
\]

we have

\[
2016=2^5\cdot3^2\cdot7.
\]

The distinct prime divisors are

\[
2,\ 3,\ 7.
\]

Their sum is

\[
2+3+7=12.
\]

Do not add repeated prime factors such as all five 2s and both 3s; the question asks for distinct primes only.

\[
\boxed{12}
\]

ANSWER 2: B

Problem 3:

The population starts at \(200\) in 1998 and triples every 25 years.

We want it to be about \(6000\). Compare:

\[
\frac{6000}{200}=30.
\]

So the population must grow by a factor of about \(30\).

Powers of \(3\) are:

\[
3^1=3,\qquad 3^2=9,\qquad 3^3=27,\qquad 3^4=81.
\]

A factor of \(27\) is close to \(30\), so we need about 3 periods of 25 years:

\[
3\cdot25=75\text{ years}.
\]

Therefore the estimated year is

\[
1998+75=2073.
\]

The nearest choice is \(2075\).

Indeed, after 75 years the estimated population is

\[
200\cdot 27=5400,
\]

which is reasonably close to \(6000\). Waiting 100 years would give \(200\cdot81=16200\), much too large.

\[
\boxed{2075}
\]

ANSWER 3: B

Problem 4:

Exactly \(74\%\) of the people answered yes.

Convert \(74\%\) to a fraction:

\[
74\%=\frac{74}{100}=\frac{37}{50}.
\]

For the number of yes answers to be a whole number, the total number surveyed must be a multiple of \(50\).

The smallest possible total is therefore

\[
50.
\]

For example,

\[
74\%\text{ of }50=\frac{74}{100}\cdot 50=37,
\]

which is an integer number of people.

Smaller choices such as 10, 20, and 25 do not work, since \(74\%\) of them is not a whole number.

\[
\boxed{50}
\]

ANSWER 4: D

Problem 5:

Without decimal points, Betty entered

\[
75\times256=19200.
\]

Now place the decimal points correctly.

The number \(0.075\) has 3 decimal places, and \(2.56\) has 2 decimal places. Therefore the product must have

\[
3+2=5
\]

decimal places.

So move the decimal point in \(19200\) five places left:

\[
19200\longrightarrow 0.19200=0.192.
\]

Thus

\[
0.075\times2.56=0.192.
\]

Choice A, \(0.0192\), has one too many decimal places, while \(1.92\) and larger answers are too large because \(0.075\) is less than \(0.1\).

\[
\boxed{0.192}
\]

ANSWER 5: B

Problem 6:

Julie’s speech must last between one-half hour and three-quarters of an hour.

Convert these times to minutes:

\[
\frac12\text{ hour}=30\text{ minutes},
\qquad
\frac34\text{ hour}=45\text{ minutes}.
\]

At \(150\) words per minute, the minimum appropriate word count is

\[
30\cdot150=4500.
\]

The maximum appropriate word count is

\[
45\cdot150=6750.
\]

So her speech should contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) and \(3000\) are too few.
- \(4200\) and \(4350\) are still below \(4500\).
- \(5650\) is between \(4500\) and \(6750\).

Therefore the appropriate word count is

\[
\boxed{5650}.
\]

ANSWER 6: E

Problem 7:

Grandfather traveled 2 miles on each of 3 days.

Find his actual treadmill time.

On Monday, he jogged at 5 mph:

\[
\frac{2}{5}\text{ hour}=0.4\text{ hour}=24\text{ minutes}.
\]

On Wednesday, he walked at 3 mph:

\[
\frac{2}{3}\text{ hour}=40\text{ minutes}.
\]

On Friday, he walked at 4 mph:

\[
\frac24\text{ hour}=\frac12\text{ hour}=30\text{ minutes}.
\]

So his actual total time was

\[
24+40+30=94\text{ minutes}.
\]

If he had walked at 4 mph on all three days, each 2-mile trip would take 30 minutes:

\[
3\cdot30=90\text{ minutes}.
\]

Thus he would have spent

\[
94-90=4
\]

fewer minutes on the treadmill.

The 5-mph jogging day saves time, but the 3-mph walking day costs more time; together the actual time is 4 minutes longer than always walking at 4 mph.

\[
\boxed{4}
\]

ANSWER 7: D

Problem 8:

Let the four consecutive odd integers be

\[
n,\ n+2,\ n+4,\ n+6,
\]

where \(n\) is odd.

Their sum is

\[
n+(n+2)+(n+4)+(n+6)=4n+12.
\]

Factor:

\[
4n+12=4(n+3).
\]

Since \(n\) is odd, \(n+3\) is even. Therefore \(4(n+3)\) must be divisible by \(8\).

So a number can be written as the sum of four consecutive odd integers only if it is divisible by \(8\).

Check the choices:

\[
16\div8=2,
\quad
40\div8=5,
\quad
72\div8=9,
\quad
200\div8=25,
\]

so all of these are divisible by 8.

But

\[
100\div8=12.5,
\]

so \(100\) is not divisible by 8. Therefore it cannot be the sum of four consecutive odd integers.

For example, \(16=1+3+5+7\), showing that the condition is possible for the other type of choices.

\[
\boxed{100}
\]

ANSWER 8: D

Problem 9:

Blake’s average is not actually needed, because we can compare Jenny’s scores directly to Blake’s.

Jenny’s score changes relative to Blake’s are:

- first test: \(+10\),
- second test: \(-10\),
- third test: \(+20\),
- fourth test: \(+20\).

Her total score exceeds Blake’s by

\[
10-10+20+20=40.
\]

Since there are four tests, her average exceeds Blake’s by

\[
\frac{40}{4}=10.
\]

The \(40\) is the difference in total points, not the difference in average points.

\[
\boxed{10}
\]

ANSWER 9: A

Problem 10:

Compute

\[
\frac{(0.2)^3}{(0.02)^2}.
\]

First,

\[
(0.2)^3=0.2\cdot0.2\cdot0.2=0.008.
\]

Next,

\[
(0.02)^2=0.02\cdot0.02=0.0004.
\]

Therefore,

\[
\frac{0.008}{0.0004}=20.
\]

Another way is to use powers of 10:

\[
0.2=2\cdot10^{-1},\qquad 0.02=2\cdot10^{-2}.
\]

Then

\[
\fra