🔬 Model Lab

I need to find the number of factors of 7 in the prime factorization of the superfactorial of 51, which is 1! · 2! · 3! · ... · 51!

**Step 1: Understand what we're counting**

I need to count how many times 7 appears as a prime factor in the product:
$$S(51) = 1! \cdot 2! \cdot 3! \cdot ... \cdot 51!$$

This is equivalent to counting the total power of 7 in this product.

**Step 2: Count factors of 7 in n!**

Using Legendre's formula, the exponent of prime p in n! is:
$$\nu_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + ...$$

For p = 7, the exponent of 7 in n! is:
$$\nu_7(n!) = \left\lfloor \frac{n}{7} \right\rfloor + \left\lfloor \frac{n}{49} \right\rfloor + \left\lfloor \frac{n}{343} \right\rfloor + ...$$

**Step 3: Calculate the total**

The total number of factors of 7 in S(51) is:
$$\sum_{k=1}^{51} \nu_7(k!)$$

I need to compute this sum systematically. For each k from 1 to 51, I calculate how many factors of 7 are in k!.

**Step 4: Organize the calculation**

For k = 1 to 6: $\nu_7(k!) = 0$ (no factors of 7)

For k = 7 to 13: $\nu_7(k!) = \lfloor k/7 \rfloor = 1$ (7 terms with 1 factor each)

For k = 14 to 20: $\nu_7(k!) = \lfloor k/7 \rfloor = 2$ (7 terms with 2 factors each)

For k = 21 to 27: $\nu_7(k!) = \lfloor k/7 \rfloor + \lfloor k/49 \rfloor = 3 + 0 = 3$ (7 terms with 3 factors each)

For k = 28 to 34: $\nu_7(k!) = \lfloor k/7 \rfloor = 4$ (7 terms with 4 factors each)

For k = 35 to 41: $\nu_7(k!) = \lfloor k/7 \rfloor = 5$ (7 terms with 5 factors each)

For k = 42 to 48: $\nu_7(k!) = \lfloor k/7 \rfloor = 6$ (7 terms with 6 factors each)

For k = 49: $\nu_7(49!) = \lfloor 49/7 \rfloor + \lfloor 49/49 \rfloor = 7 + 1 = 8$

For k = 50: $\nu_7(50!) = \lfloor 50/7 \rfloor + \lfloor 50/49 \rfloor = 7 + 1 = 8$

For k = 51: $\nu_7(51!) = \lfloor 51/7 \rfloor + \lfloor 51/49 \rfloor = 7 + 1 = 8$

**Step 5: Sum it all up**

$$\text{Total} = 0 \cdot 6 + 1 \cdot 7 + 2 \cdot 7 + 3 \cdot 7 + 4 \cdot 7 + 5 \cdot 7 + 6 \cdot 7 + 8 + 8 + 8$$

$$= 7(1 + 2 + 3 + 4 + 5 + 6) + 24$$

$$= 7 \cdot 21 + 24$$

$$= 147 + 24 = 171$$

**Step 6: Verify**

The sum $1 + 2 + 3 + 4 + 5 + 6 = 21$ ✓

$7 \times 21 = 147$ ✓

$147 + 24 = 171$ ✓

ANSWER: E

1. We need the exponent of 7 in the product 1!·2!·····51!.

2. The exponent of 7 in k! is floor(k/7) + floor(k/49) + ...

3. Total exponent = sum over k=1 to 51 of (exponent of 7 in k!).

4. Alternative: count contribution of each multiple of 7. The number 7m appears in factorial k! for all k≥7m. Actually let me count how many times each factor contributes.

5. Better approach: exponent of 7 in superfactorial = sum_{k=1}^{51} v_7(k!), where v_7 is the 7-adic valuation.

6. Note v_7(k!) = sum_{j=1}^{k} v_7(j). So total = sum_{k=1}^{51} sum_{j=1}^{k} v_7(j) = sum_{j=1}^{51} v_7(j) · (number of k from j to 51) = sum_{j=1}^{51} v_7(j)·(52 - j).

7. Find j with v_7(j)>0 up to 51: multiples of 7: 7,14,21,28,35,42,49. v_7=1 each except 49 has v_7=2.

8. Compute v_7(j)·(52-j):
- j=7: 1·45 = 45
- j=14: 1·38 = 38
- j=21: 1·31 = 31
- j=28: 1·24 = 24
- j=35: 1·17 = 17
- j=42: 1·10 = 10
- j=49: 2·3 = 6

9. Sum: 45+38=83, +31=114, +24=138, +17=155, +10=165, +6=171.

10. Total = 171.

11. Ruling out: 147,150,156,168 are too low; we carefully accounted for the extra factor from 49 (giving +3 beyond a single count) which pushes us to 171.

ANSWER: E

I need to find the number of factors of 7 in the superfactorial of 51, which is 1! · 2! · 3! · ... · 51!

## Step 1: Set Up the Problem

The superfactorial of 51 = 1! · 2! · 3! · ... · 51!

I need to find the total power of 7 in this product, which equals:
$$\sum_{k=1}^{51} v_7(k!)$$

where $v_7(k!)$ is the largest power of 7 dividing k!.

## Step 2: Formula for Power of 7 in k!

By Legendre's formula:
$$v_7(k!) = \left\lfloor \frac{k}{7} \right\rfloor + \left\lfloor \frac{k}{49} \right\rfloor + \left\lfloor \frac{k}{343} \right\rfloor + \cdots$$

Since 343 > 51, only the first two terms matter for k ≤ 51.

## Step 3: Compute the Sum

$$\sum_{k=1}^{51} v_7(k!) = \sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor + \sum_{k=1}^{51} \left\lfloor \frac{k}{49} \right\rfloor$$

**First sum:** $\sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor$

Group by value of $\lfloor k/7 \rfloor$:
- k = 1-6: value 0, appears 6 times → contributes 0
- k = 7-13: value 1, appears 7 times → contributes 7
- k = 14-20: value 2, appears 7 times → contributes 14
- k = 21-27: value 3, appears 7 times → contributes 21
- k = 28-34: value 4, appears 7 times → contributes 28
- k = 35-41: value 5, appears 7 times → contributes 35
- k = 42-48: value 6, appears 7 times → contributes 42
- k = 49-51: value 7, appears 3 times → contributes 21

Total first sum: 0 + 7 + 14 + 21 + 28 + 35 + 42 + 21 = **168**

**Second sum:** $\sum_{k=1}^{51} \left\lfloor \frac{k}{49} \right\rfloor$

- k = 1-48: value 0 → contributes 0
- k = 49-51: value 1, appears 3 times → contributes 3

Total second sum: **3**

## Step 4: Add the Results

Total = 168 + 3 = **171**

## Step 5: Verify by Checking the Logic

Let me double-check the first sum: 7(1+2+3+4+5+6) + 21 = 7(21) + 21 = 147 + 21 = 168 ✓
Second sum: 3 ✓
Total: 171 ✓

## Step 6: Rule Out Wrong Choices

- A (147): This is just $\sum \lfloor k/7 \rfloor$ without the last group adjustment — actually it's 7×21=147, missing the k=49-51 contribution to first sum
- B (150): 147 + 3 = 150, this forgets the k=49-51 portion of first sum
- C (156): Not matching our calculation
- D (168): This is just the first sum without the second sum
- E (171): Matches our complete calculation ✓

ANSWER: E

1. **Restate the question.**  
We want the number of factors of \(7\) in
\[
1!\cdot 2!\cdot 3!\cdots 51!.
\]
So we need the exponent of \(7\) in this product.

2. **Use exponents of 7 in a product.**  
If \(v_7(N)\) means “the exponent of 7 in \(N\),” then
\[
v_7(1!\cdot 2!\cdot \cdots \cdot 51!)=\sum_{k=1}^{51} v_7(k!).
\]

3. **Rewrite by counting how many times each multiple of 7 appears.**  
Each \(k!\) contains all integers from \(1\) to \(k\).  
A factor of \(7\) coming from the integer \(7\) appears in every factorial from \(7!\) up to \(51!\), so it appears \(51-7+1=45\) times.

Similarly:
- the factor from \(14\) appears in \(14!,15!,\dots,51!\): \(51-14+1=38\) times,
- from \(21\): \(31\) times,
- from \(28\): \(24\) times,
- from \(35\): \(17\) times,
- from \(42\): \(10\) times,
- from \(49\): \(3\) times.

So first count all multiples of \(7\):
\[
45+38+31+24+17+10+3.
\]

4. **Be careful about extra 7's from powers like \(49=7^2\).**  
The number \(49\) contributes **two** factors of \(7\), not just one.  
We already counted one of them in the list above, so we need to add the extra one 3 more times (since \(49\) appears in \(49!,50!,51!\)).

Thus total exponent is
\[
(45+38+31+24+17+10+3)+3.
\]

5. **Compute.**
\[
45+38=83,\quad 83+31=114,\quad 114+24=138,
\]
\[
138+17=155,\quad 155+10=165,\quad 165+3=168.
\]
Then add the extra \(3\):
\[
168+3=171.
\]

6. **Check another way to confirm.**  
For each \(k!\),
\[
v_7(k!)=\left\lfloor \frac{k}{7}\right\rfloor+\left\lfloor \frac{k}{49}\right\rfloor.
\]
So
\[
\sum_{k=1}^{51}\left\lfloor \frac{k}{7}\right\rfloor
\]
has blocks:
- \(7\) numbers with value 1,
- \(7\) with value 2,
- ...
- \(7\) with value 7,
- then \(2\) numbers with value 7 more? Let's do it carefully:

For \(k=1\) to \(6\): 0  
\(7\) to \(13\): 1 (7 terms)  
\(14\) to \(20\): 2 (7 terms)  
\(21\) to \(27\): 3  
\(28\) to \(34\): 4  
\(35\) to \(41\): 5  
\(42\) to \(48\): 6  
\(49\) to \(51\): 7 (3 terms)

So
\[
7(1+2+3+4+5+6)+3\cdot 7
=7\cdot 21+21
=147+21
=168.
\]
Also,
\[
\sum_{k=1}^{51}\left\lfloor \frac{k}{49}\right\rfloor = 1+1+1=3.
\]
Total:
\[
168+3=171.
\]
Confirmed.

7. **Match with the choices.**  
\(171\) is choice **E**.

ANSWER: E

1. **What is being asked?**  
   We need the number of factors of \(7\) in
   \[
   1!\cdot 2!\cdot 3!\cdots 51!.
   \]
   In other words, we want the exponent of \(7\) in this product.

2. **Use the exponent-counting idea**  
   The number of factors of \(7\) in \(n!\) is
   \[
   \left\lfloor \frac{n}{7} \right\rfloor + \left\lfloor \frac{n}{49} \right\rfloor + \left\lfloor \frac{n}{343} \right\rfloor + \cdots
   \]
   Since \(51<343\), only the first two terms matter for numbers up to \(51\).

3. **Count the 7s in each factorial from \(1!\) to \(51!\)**  
   We want
   \[
   \sum_{n=1}^{51} v_7(n!).
   \]
   For each \(n\), the exponent of \(7\) in \(n!\) is:
   \[
   \left\lfloor \frac{n}{7} \right\rfloor + \left\lfloor \frac{n}{49} \right\rfloor.
   \]

4. **Sum the first part: \(\sum_{n=1}^{51}\left\lfloor \frac{n}{7} \right\rfloor\)**  
   Group the numbers by intervals:
   - \(1\) through \(6\): contribution \(0\) each
   - \(7\) through \(13\): contribution \(1\) each, 7 numbers
   - \(14\) through \(20\): contribution \(2\) each, 7 numbers
   - \(21\) through \(27\): contribution \(3\) each, 7 numbers
   - \(28\) through \(34\): contribution \(4\) each, 7 numbers
   - \(35\) through \(41\): contribution \(5\) each, 7 numbers
   - \(42\) through \(48\): contribution \(6\) each, 7 numbers
   - \(49\) through \(51\): contribution \(7\) each, 3 numbers

   So:
   \[
   7(1+2+3+4+5+6)+3\cdot 7
   =7\cdot 21+21
   =147+21
   =168.
   \]

5. **Sum the second part: \(\sum_{n=1}^{51}\left\lfloor \frac{n}{49} \right\rfloor\)**  
   This is \(0\) for \(n=1\) through \(48\), and \(1\) for \(n=49,50,51\).  
   So the total is:
   \[
   3.
   \]

6. **Total number of factors of 7**  
   \[
   168+3=171.
   \]

7. **Check against the choices**  
   The only choice matching \(171\) is **E**.

ANSWER: E

To find the number of factors of 7 in the prime factorization of the superfactorial of 51, denoted as $S(51) = 1! \cdot 2! \cdot 3! \cdot \dots \cdot 51!$, we follow these steps:

1. **Understand the structure of the problem:**
   The exponent of a prime $p$ in the prime factorization of $n!$ is given by Legendre's Formula: $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor$.
   The exponent of 7 in $S(51)$ is the sum of the exponents of 7 in each factorial from $1!$ to $51!$:
   $E_7(S(51)) = \sum_{n=1}^{51} E_7(n!)$.

2. **Express the sum using the definition of $E_7(n!)$:**
   $E_7(S(51)) = \sum_{n=1}^{51} \sum_{k=1}^{\infty} \lfloor \frac{n}{7^k} \rfloor$.
   We can swap the summation order:
   $E_7(S(51)) = \sum_{k=1}^{\infty} \sum_{n=1}^{51} \lfloor \frac{n}{7^k} \rfloor$.
   Since $7^2 = 49$ and $7^3 = 343 > 51$, we only need to consider $k=1$ and $k=2$.

3. **Calculate the sum for $k=1$:**
   We need $\sum_{n=1}^{51} \lfloor \frac{n}{7} \rfloor$.
   - For $n=1$ to $6$, $\lfloor n/7 \rfloor = 0$ (6 terms).
   - For $n=7$ to $13$, $\lfloor n/7 \rfloor = 1$ (7 terms).
   - For $n=14$ to $20$, $\lfloor n/7 \rfloor = 2$ (7 terms).
   - For $n=21$ to $27$, $\lfloor n/7 \rfloor = 3$ (7 terms).
   - For $n=28$ to $34$, $\lfloor n/7 \rfloor = 4$ (7 terms).
   - For $n=35$ to $41$, $\lfloor n/7 \rfloor = 5$ (7 terms).
   - For $n=42$ to $48$, $\lfloor n/7 \rfloor = 6$ (7 terms).
   - For $n=49$ to $51$, $\lfloor n/7 \rfloor = 7$ (3 terms).
   Sum $= 7(1+2+3+4+5+6) + 3(7) = 7(21) + 21 = 147 + 21 = 168$.

4. **Calculate the sum for $k=2$:**
   We need $\sum_{n=1}^{51} \lfloor \frac{n}{49} \rfloor$.
   - For $n=1$ to $48$, $\lfloor n/49 \rfloor = 0$.
   - For $n=49$ to $51$, $\lfloor n/49 \rfloor = 1$ (3 terms).
   Sum $= 3(1) = 3$.

5. **Combine the results:**
   $E_7(S(51)) = 168 + 3 = 171$.

6. **Evaluate the choices:**
   - A (147): Incorrect, missed the higher multiples.
   - B (150): Incorrect.
   - C (156): Incorrect.
   - D (168): Incorrect, missed the $k=2$ contribution.
   - E (171): Matches our calculation.

ANSWER: E

To find the number of factors of 7 in the prime factorization of the superfactorial of 51, we need to determine the exponent of the highest power of 7 that divides it. 

**1. Understand the Superfactorial and Set Up the Math**
The superfactorial of 51 is the product of the first 51 factorials:
$sf(51) = 1! \cdot 2! \cdot 3! \cdot \ldots \cdot 51!$
Let $E_7(N)$ be the exponent of 7 in the prime factorization of $N$. Because $sf(51)$ is a product, we can add the exponents of 7 from each factorial:
$E_7(sf(51)) = \sum_{k=1}^{51} E_7(k!)$

**2. Apply Legendre's Formula**
Legendre's formula states that the exponent of a prime $p$ dividing $k!$ is given by the sum of the floor functions of $k$ divided by the powers of $p$:
$E_7(k!) = \lfloor \frac{k}{7} \rfloor + \lfloor \frac{k}{49} \rfloor + \lfloor \frac{k}{343} \rfloor + \ldots$
Since our maximum $k$ is 51, any power of 7 greater than 49 will result in a floor of 0. Thus, the formula simplifies to:
$E_7(k!) = \lfloor \frac{k}{7} \rfloor + \lfloor \frac{k}{49} \rfloor$

Substituting this back into our total sum, we get:
$E_7(sf(51)) = \sum_{k=1}^{51} \left( \lfloor \frac{k}{7} \rfloor + \lfloor \frac{k}{49} \rfloor \right) = \sum_{k=1}^{51} \lfloor \frac{k}{7} \rfloor + \sum_{k=1}^{51} \lfloor \frac{k}{49} \rfloor$

**3. Compute the First Sum**
Let's evaluate $\sum_{k=1}^{51} \lfloor \frac{k}{7} \rfloor$ by grouping the terms based on their floor values:
- For $k \in \{1, 2, \dots, 6\}$, $\lfloor \frac{k}{7} \rfloor = 0$ (6 terms)
- For $k \in \{7, 8, \dots, 13\}$, $\lfloor \frac{k}{7} \rfloor = 1$ (7 terms)
- For $k \in \{14, 15, \dots, 20\}$, $\lfloor \frac{k}{7} \rfloor = 2$ (7 terms)
- For $k \in \{21, 22, \dots, 27\}$, $\lfloor \frac{k}{7} \rfloor = 3$ (7 terms)
- For $k \in \{28, 29, \dots, 34\}$, $\lfloor \frac{k}{7} \rfloor = 4$ (7 terms)
- For $k \in \{35, 36, \dots, 41\}$, $\lfloor \frac{k}{7} \rfloor = 5$ (7 terms)
- For $k \in \{42, 43, \dots, 48\}$, $\lfloor \frac{k}{7} \rfloor = 6$ (7 terms)
- For $k \in \{49, 50, 51\}$, $\lfloor \frac{k}{7} \rfloor = 7$ (3 terms)

Adding these up:
Sum $= 7(1 + 2 + 3 + 4 + 5 + 6) + 3(7)$
Sum $= 7(21) + 21 = 147 + 21 = 168$

**4. Compute the Second Sum**
Next, we evaluate $\sum_{k=1}^{51} \lfloor \frac{k}{49} \rfloor$:
- For $k < 49$, $\lfloor \frac{k}{49} \rfloor = 0$ (48 terms)
- For $k \in \{49, 50, 51\}$, $\lfloor \frac{k}{49} \rfloor = 1$ (3 terms)

Adding these up:
Sum $= 3(1) = 3$

**5. Total Exponent and Ruling Out Wrong Choices**
Adding both sums together gives the total number of factors of 7:
Total $= 168 + 3 = 171$

Looking at the incorrect choices:
- *Choice A (147)* is what you would get if you completely forgot to include the $k=49, 50, 51$ terms.
- *Choice B (150)* is what you would get if you added the 3 from the second sum to 147, still missing the 21 from the first sum for $k \ge 49$.
- *Choice D (168)* is what you get if you only calculate the first sum $\sum \lfloor \frac{k}{7} \rfloor$, forgetting that multiples of $7^2 = 49$ contribute an additional factor of 7.

ANSWER: E

An elegant way to solve this problem is by breaking down the contribution of the prime number $7$ to each factorial in the product.

### Step 1: Restate the problem
We want to find the number of factors of $7$ in the prime factorization of the superfactorial of $51$, which is defined as:
$$\text{sf}(51) = 1! \cdot 2! \cdot 3! \cdots 51!$$
The exponent of $7$ in the prime factorization of $\text{sf}(51)$ is the sum of the exponents of $7$ in each of the individual factorials $k!$ for $k = 1, 2, \dots, 51$. Let $E_7(N)$ denote the exponent of $7$ in $N$. Then:
$$E_7(\text{sf}(51)) = \sum_{k=1}^{51} E_7(k!)$$

### Step 2: Set up the math using Legendre's Formula
By Legendre's Formula, the exponent of a prime $p$ in $k!$ is given by:
$$E_p(k!) = \sum_{i=1}^{\infty} \left\lfloor \frac{k}{p^i} \right\rfloor$$
For $p = 7$, since $7^3 = 343 > 51$, we only need to consider powers of $7$ up to $7^2 = 49$. Thus:
$$E_7(k!) = \left\lfloor \frac{k}{7} \right\rfloor + \left\lfloor \frac{k}{49} \right\rfloor$$

We can split the total sum into two parts:
$$E_7(\text{sf}(51)) = \sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor + \sum_{k=1}^{51} \left\lfloor \frac{k}{49} \right\rfloor$$

### Step 3: Compute the first sum $\sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor$
Let's analyze the value of $\left\lfloor \frac{k}{7} \right\rfloor$ as $k$ goes from $1$ to $51$:
* For $k \in [1, 6]$, $\left\lfloor \frac{k}{7} \right\rfloor = 0$ (6 terms)
* For $k \in [7, 13]$, $\left\lfloor \frac{k}{7} \right\rfloor = 1$ (7 terms)
* For $k \in [14, 20]$, $\left\lfloor \frac{k}{7} \right\rfloor = 2$ (7 terms)
* For $k \in [21, 27]$, $\left\lfloor \frac{k}{7} \right\rfloor = 3$ (7 terms)
* For $k \in [28, 34]$, $\left\lfloor \frac{k}{7} \right\rfloor = 4$ (7 terms)
* For $k \in [35, 41]$, $\left\lfloor \frac{k}{7} \right\rfloor = 5$ (7 terms)
* For $k \in [42, 48]$, $\left\lfloor \frac{k}{7} \right\rfloor = 6$ (7 terms)
* For $k \in [49, 51]$, $\left\lfloor \frac{k}{7} \right\rfloor = 7$ (3 terms)

Summing these up:
$$\sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor = 0(6) + 1(7) + 2(7) + 3(7) + 4(7) + 5(7) + 6(7) + 7(3)$$
$$\sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor = 7 \cdot (1 + 2 + 3 + 4 + 5 + 6) + 21$$
$$\sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor = 7 \cdot 21 + 21 = 147 + 21 = 168$$

### Step 4: Compute the second sum $\sum_{k=1}^{51} \left\lfloor \frac{k}{49} \right\rfloor$
Now we analyze the value of $\left\lfloor \frac{k}{49} \right\rfloor$ as $k$ goes from $1$ to $51$:
* For $k \in [1, 48]$, $\left\lfloor \frac{k}{49} \right\rfloor = 0$ (48 terms)
* For $k \in [49, 51]$, $\left\lfloor \frac{k}{49} \right\rfloor = 1$ (3 terms)

Summing these up:
$$\sum_{k=1}^{51} \left\lfloor \frac{k}{49} \right\rfloor = 0(48) + 1(3) = 3$$

### Step 5: Add the two sums
The total number of factors of $7$ is:
$$E_7(\text{sf}(51)) = 168 + 3 = 171$$

### Step 6: Explain and rule out the wrong choices
* **A. 147** is incorrect; this is only $7 \times 21$, which misses the contribution of the last few terms and the multiples of 49.
* **B. 150** is incorrect; it is too small.
* **C. 156** is incorrect; it is too small.
* **D. 168** is incorrect; this is only the first sum $\sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor$, which forgets to account for the extra factor of $7$ contributed by $49!$, $50!$, and $51!$ (since $49 = 7^2$).
* **E. 171** is the correct total.

ANSWER: E

The superfactorial of 51 is the product
\[
SF(51) = 1! \cdot 2! \cdots 51!.
\]
The exponent of the prime 7 in the prime factorization of \(SF(51)\) is therefore the sum of the exponents of 7 in each factorial \(k!\) for \(k = 1\) to \(51\):
\[
E = \sum_{k=1}^{51} v_7(k!).
\]
The exponent of 7 in a single factorial is given by the formula
\[
v_7(k!) = \sum_{m \ge 1} \lfloor k/7^m \rfloor.
\]
Hence
\[
E = \sum_{k=1}^{51} \sum_{m \ge 1} \lfloor k/7^m \rfloor.
\]
Interchanging the order of summation yields
\[
E = \sum_{m \ge 1} \sum_{k=1}^{51} \lfloor k/7^m \rfloor.
\]
Because \(k \le 51 < 343 = 7^3\), only the terms \(m=1\) (\(7^1=7\)) and \(m=2\) (\(7^2=49\)) can be nonzero. Consequently
\[
E = \sum_{k=1}^{51} \lfloor k/7 \rfloor + \sum_{k=1}^{51} \lfloor k/49 \rfloor.
\]

For the second sum, \(\lfloor k/49 \rfloor = 0\) when \(1 \le k \le 48\) and equals 1 when \(49 \le k \le 51\). Thus the sum equals 3.

For the first sum, write
\[
\sum_{k=1}^{51} \lfloor k/7 \rfloor = 1\cdot7 + 2\cdot7 + 3\cdot7 + 4\cdot7 + 5\cdot7 + 6\cdot7 + 7\cdot3,
\]
which evaluates to
\[
7(1+2+3+4+5+6) + 21 = 147 + 21 = 168.
\]
Adding the two contributions therefore gives
\[
E = 168 + 3 = 171.
\]
(The choices 147, 150, 156 and 168 are each obtained by omitting one or both of the contributions coming from the higher powers of 7 or from the incomplete final block of multiples of 7.)

ANSWER: E

1. **Understanding the problem**: We need to find the number of factors of 7 in the prime factorization of the superfactorial of 51. The superfactorial of 51 is the product of the factorials from 1! to 51!.

2. **Breaking down the superfactorial**: To find the number of factors of 7, we first need to understand how many times 7 appears in the prime factorization of each factorial from 1! to 51!, and then sum those counts.

3. **Counting factors of 7 in n!**: The number of factors of 7 in n! can be found by counting the multiples of 7, then adding the multiples of 7^2, and so on, until 7^k > n. For n!, the count is given by $\left\lfloor \frac{n}{7} \right\rfloor + \left\lfloor \frac{n}{7^2} \right\rfloor + \left\lfloor \frac{n}{7^3} \right\rfloor + \cdots$ until the quotient becomes 0.

4. **Calculating the count for each n!**: We need to calculate $\left\lfloor \frac{n}{7} \right\rfloor + \left\lfloor \frac{n}{49} \right\rfloor$ for n = 1 to 51 because $7^3 = 343 > 51$, so we don't need to consider $7^3$ or higher powers.

5. **Summing the counts**: The total count of factors of 7 in the superfactorial of 51 is the sum of the counts for each n! from 1! to 51!. So, we need to compute $\sum_{n=1}^{51} \left( \left\lfloor \frac{n}{7} \right\rfloor + \left\lfloor \frac{n}{49} \right\rfloor \right)$.

6. **Computing the sum**: 
- For $\left\lfloor \frac{n}{7} \right\rfloor$, the values are 0 (for n = 1 to 6), 1 (for n = 7 to 13), and so on, increasing by 1 for every 7 numbers. So, the sum can be calculated by grouping: $1 \cdot 7 + 2 \cdot 7 + \cdots + 7 \cdot 7 + (8 \cdot 3)$, because we have 7 numbers for each quotient from 1 to 7, and then 3 more numbers (49, 50, 51) that give a quotient of 8 (though 8 is not reached for n=51 for the first term, it is for the second term for n=49).
- The sum of the first part is $7(1+2+3+4+5+6+7) + 8 \cdot 3 = 7 \cdot \frac{7 \cdot 8}{2} + 24 = 7 \cdot 28 + 24 = 196 + 24 = 220$.
- For $\left\lfloor \frac{n}{49} \right\rfloor$, the values are 0 (for n = 1 to 48) and 1 (for n = 49, 50, 51). So, the sum is $1 \cdot 3 = 3$.
- The total count is $220 + 3 = 223$ but we made a mistake in detailed step-by-step count for the first part. Let's directly simplify the summation by using the formula for the sum of floor functions.

7. **Simplifying the summation**: 
The correct way to simplify is to directly calculate $\sum_{n=1}^{51} \left\lfloor \frac{n}{7} \right\rfloor$ and $\sum_{n=1}^{51} \left\lfloor \frac{n}{49} \right\rfloor$.
- $\sum_{n=1}^{51} \left\lfloor \frac{n}{7} \right\rfloor = 0 + 0 + 0 + 0 + 0 + 0 + 1 + 1 + \cdots + 7 + 7 + 7 = 7(1+2+3+4+5+6+7) + 3(8) = 196 + 24 = 220$ is an overcomplication. Instead, directly: $\sum_{n=1}^{51} \left\lfloor \frac{n}{7} \right\rfloor = \sum_{k=1}^{7} k \cdot 7 + 3 \cdot 8 = 7 \cdot \frac{7(7+1)}{2} + 24 = 196 + 24 = 220$.
- $\sum_{n=1}^{51} \left\lfloor \frac{n}{49} \right\rfloor = 0 + \cdots + 0 + 1 + 1 + 1 = 3$.

8. **Correcting and simplifying step 6 and 7**: Let's directly count:
- For $\left\lfloor \frac{n}{7} \right\rfloor$, we have: $0 \cdot 7 + 1 \cdot 7 + 2 \cdot 7 + 3 \cdot 7 + 4 \cdot 7 + 5 \cdot 7 + 6 \cdot 7 + 7 \cdot 3 = 7(0+1+2+3+4+5+6) + 7*3 = 7*21 + 21 = 168$.
- For $\left\lfloor \frac{n}{49} \right\rfloor$, we have: $0 \cdot 48 + 1 \cdot 3 = 3$.
- Total = $168 + 3 = 171$.

9. **Ruling out wrong choices**: 
- A. 147 is too low.
- B. 150 is too low.
- C. 156 is too low.
- D. 168 is close but doesn't account for the additional factors from $\left\lfloor \frac{n}{49} \right\rfloor$.
- E. 171 matches our calculation.

ANSWER: E

The exponent of prime 7 in the superfactorial of 51 is found by summing the exponent of 7 in each factorial from 1! to 51!. This sum equals the total number of factors of 7 in the product.

1. For each integer \(k\) (1 ≤ k ≤ 51), the exponent of 7 in \(k!\) is  
   \[
   v_7(k!) = \left\lfloor \frac{k}{7} \right\rfloor + \left\lfloor \frac{k}{49} \right\rfloor,
   \]
   since \(7^3 = 343 > 51\), so higher powers do not contribute.

2. Compute \(\displaystyle\sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor\).  
   The values of \(\lfloor k/7 \rfloor\) increase by 1 every 7 numbers. With \(51 = 7\cdot7 + 2\), we have:
   - For \(j = 1,2,3,4,5,6\): each appears for 7 values of \(k\).
   - For \(j = 7\): appears for \(k = 49,50,51\) (3 values).  
   Sum = \(7(1+2+3+4+5+6) + 7\cdot3 = 7\cdot21 + 21 = 147 + 21 = 168\).

3. Compute \(\displaystyle\sum_{k=1}^{51} \left\lfloor \frac{k}{49} \right\rfloor\).  
   Only \(k = 49,50,51\) give quotient 1; all others give 0. Sum = \(1+1+1 = 3\).

4. Total exponent = \(168 + 3 = 171\).

Thus the prime factorization of the superfactorial of 51 contains 171 factors of 7.

ANSWER: E

**Step 1: Understand the problem**  
We need the exponent of the prime 7 in the *superfactorial* of 51.  
The superfactorial of \(n\) is defined as \(1! \cdot 2! \cdot 3! \cdots n!\).  
So we want the exponent of 7 in \(S = \prod_{i=1}^{51} i!\).

**Step 2: Set up the exponent calculation**  
The exponent of a prime \(p\) in \(n!\) is \(\sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor\) (Legendre’s formula).  
The exponent in the superfactorial is the sum of the exponents in each \(i!\) for \(i = 1\) to \(51\):
\[
v_7(S) = \sum_{i=1}^{51} v_7(i!) = \sum_{i=1}^{51} \sum_{k=1}^{\infty} \left\lfloor \frac{i}{7^k} \right\rfloor.
\]

**Step 3: Swap the order of summation**  
\[
v_7(S) = \sum_{k=1}^{\infty} \sum_{i=1}^{51} \left\lfloor \frac{i}{7^k} \right\rfloor.
\]
Since \(51 < 7^3 = 343\), only \(k = 1\) and \(k = 2\) contribute.

**Step 4: Compute for \(k = 1\)**  
We need \(A = \sum_{i=1}^{51} \left\lfloor \frac{i}{7} \right\rfloor\).  
Values of \(\lfloor i/7 \rfloor\):
- 0 for \(i = 1\) to \(6\) (6 numbers): \(0 \times 6 = 0\)
- 1 for \(i = 7\) to \(13\) (7 numbers): \(1 \times 7 = 7\)
- 2 for \(i = 14\) to \(20\) (7 numbers): \(2 \times 7 = 14\)
- 3 for \(i = 21\) to \(27\) (7 numbers): \(3 \times 7 = 21\)
- 4 for \(i = 28\) to \(34\) (7 numbers): \(4 \times 7 = 28\)
- 5 for \(i = 35\) to \(41\) (7 numbers): \(5 \times 7 = 35\)
- 6 for \(i = 42\) to \(48\) (7 numbers): \(6 \times 7 = 42\)
- 7 for \(i = 49\) to \(51\) (3 numbers): \(7 \times 3 = 21\)

Sum: \(0 + 7 + 14 + 21 + 28 + 35 + 42 + 21 = 168\).  
So \(A = 168\).

**Step 5: Compute for \(k = 2\)**  
We need \(B = \sum_{i=1}^{51} \left\lfloor \frac{i}{49} \right\rfloor\).  
Only \(i = 49, 50, 51\) give \(\lfloor i/49 \rfloor = 1\); all others give 0.  
So \(B = 3\).

**Step 6: Add the contributions**  
Total exponent \(v_7(S) = A + B = 168 + 3 = 171\).

**Step 7: Match with choices**  
The options are 147, 150, 156, 168, 171. Our computed value 171 corresponds to choice E.

ANSWER: E

sum** of the exponents of 7 in each $ k! $, for $ k = 1 $ to $ 51 $.

So, define $ v_7(m) $ as the exponent of 7 in the prime factorization of $ m $. Then:

$$
v_7\left( \prod_{k=1}^{51} k! \right) = \sum_{k=1}^{51} v_7(k!)
$$

We need to compute this sum.

---

### Step 2: Legendre’s formula for $ v_7(k!) $

Legendre’s formula gives:

$$
v_7(k!) = \left\lfloor \frac{k}{7} \right\rfloor + \left\lfloor \frac{k}{7^2} \right\rfloor + \left\lfloor \frac{k}{7^3} \right\rfloor + \cdots
$$

Since $ 7^3 = 343 > 51 $, we only need terms up to $ 7^2 = 49 $. So:

$$
v_7(k!) = \left\lfloor \frac{k}{7} \right\rfloor + \left\lfloor \frac{k}{49} \right\rfloor
$$

Therefore, the total exponent is:

$$
\sum_{k=1}^{51} \left( \left\lfloor \frac{k}{7} \right\rfloor + \left\lfloor \frac{k}{49} \right\rfloor \right)
= \sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor + \sum_{k=1}^{51} \left\lfloor \frac{k}{49} \right\rfloor
$$

We’ll compute these two sums separately.

---

### Step 3: Compute $ \sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor $

We can think of this as: for each integer $ q \geq 0 $, how many $ k \in [1, 51] $ satisfy $ \left\lfloor \frac{k}{7} \right\rfloor = q $? Then multiply $ q $ by the count and sum.

Note: $ \left\lfloor \frac{k}{7} \right\rfloor = q $ when $ 7q \leq k \leq 7q + 6 $. So each $ q $ appears 7 times, except possibly the last one.

Maximum $ \left\lfloor \frac{51}{7} \right\rfloor = 7 $, since $ 7 \cdot 7 = 49 $, and $ 7 \cdot 8 = 56 > 51 $.

So values of $ q = 0, 1, 2, ..., 7 $

- For $ q = 0 $: $ k = 1 $ to $ 6 $ → actually $ k = 1 $ to $ 6 $, but note $ \left\lfloor \frac{0}{7} \right\rfloor = 0 $, but we start at $ k=1 $. However, $ \left\lfloor \frac{k}{7} \right\rfloor = 0 $ for $ k = 1 $ to $ 6 $, and also $ k=0 $, but we don’t include 0. So 6 values.
- Wait! Actually, $ \left\lfloor \frac{k}{7} \right\rfloor = 0 $ for $ k = 1,2,3,4,5,6 $ → 6 values.
- But $ k=7 $: $ \left\lfloor 7/7 \right\rfloor = 1 $

But better approach: just compute the sum directly using known formula:

$$
\sum_{k=1}^{n} \left\lfloor \frac{k}{d} \right\rfloor = \sum_{q=1}^{\left\lfloor n/d \right\rfloor} q \cdot (\text{number of } k \text{ with } \left\lfloor k/d \right\rfloor = q)
$$

But maybe even easier: note that

$$
\sum_{k=1}^{n} \left\lfloor \frac{k}{d} \right\rfloor = \sum_{m=1}^{n} \sum_{\substack{k=1 \\ d \mid k}}^{m} 1 = \sum_{j=1}^{\left\lfloor n/d \right\rfloor} (n - dj + 1)
$$

Wait, another known identity:

$$
\sum_{k=1}^{n} \left\lfloor \frac{k}{d} \right\rfloor = \frac{(n+1)(q) - d \cdot q(q+1)/2}{}, \text{ where } q = \left\lfloor \frac{n}{d} \right\rfloor
$$

But maybe just do it manually since numbers are small.

Let’s list $ \left\lfloor \frac{k}{7} \right\rfloor $ for $ k = 1 $ to $ 51 $:

- $ k = 1 $ to $ 6 $: floor = 0 → 6 terms
- $ k = 7 $ to $ 13 $: floor = 1 → 7 terms
- $ k = 14 $ to $ 20 $: floor = 2 → 7 terms
- $ k = 21 $ to $ 27 $: floor = 3 → 7 terms
- $ k = 28 $ to $ 34 $: floor = 4 → 7 terms
- $ k = 35 $ to $ 41 $: floor = 5 → 7 terms
- $ k = 42 $ to $ 48 $: floor = 6 → 7 terms
- $ k = 49 $ to $ 51 $: floor = 7 → 3 terms

Now compute sum:

$$
0 \cdot 6 + 1 \cdot 7 + 2 \cdot 7 + 3 \cdot 7 + 4 \cdot 7 + 5 \cdot 7 + 6 \cdot 7 + 7 \cdot 3
$$

Compute:

- $ 1+2+3+4+5+6 = 21 $, so $ 21 \cdot 7 = 147 $
- Plus $ 7 \cdot 3 = 21 $

Total: $ 147 + 21 = 168 $

So:

$$
\sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor = 168
$$

---

### Step 4: Compute $ \sum_{k=1}^{51} \left\lfloor \frac{k}{49} \right\rfloor $

Since $ 49 = 7^2 $, and $ 2 \cdot 49 = 98 > 51 $, we have:

- For $ k = 1 $ to $ 48 $: $ \left\lfloor \frac{k}{49} \right\rfloor = 0 $
- For $ k = 49, 50, 51 $: $ \left\lfloor \frac{k}{49} \right\rfloor = 1 $

So sum = $ 0 \cdot 48 + 1 \cdot 3 = 3 $

---

### Step 5: Total exponent of 7

Add both parts:

$$
168 + 3 = 171
$$

---

### Step 6: Check answer choices

Choices:

A. 147  
B. 150  
C. 156  
D. 168  
E. 171  

Our result: **171** → Choice **E**

---

### Final Answer:

ANSWER: E

I need to find the maximum number of students that can be seated if no two students sit next to each other in the same row.

**Step 1: Restate the problem**
- 20 rows total
- Row 1 has 10 seats
- Each successive row has one more seat (Row 2 has 11, Row 3 has 12, etc.)
- No two students can sit next to each other in the same row
- Find the maximum number of students

**Step 2: Determine seating pattern for each row**

If no two students can sit next to each other in a row, the maximum number of students in a row follows a pattern. For a row with n seats, the maximum number of non-adjacent students is ⌈n/2⌉ (ceiling of n/2).

This is because we can use a pattern like: Student, Empty, Student, Empty, etc.

- Row with n seats: maximum students = ⌈n/2⌉

**Step 3: Calculate for each row**

- Row 1: 10 seats → ⌈10/2⌉ = 5 students
- Row 2: 11 seats → ⌈11/2⌉ = 6 students
- Row 3: 12 seats → ⌈12/2⌉ = 6 students
- Row 4: 13 seats → ⌈13/2⌉ = 7 students
- Row 5: 14 seats → ⌈14/2⌉ = 7 students
- ...
- Row 20: 29 seats → ⌈29/2⌉ = 15 students

**Step 4: Find the pattern**

For row k (where k goes from 1 to 20):
- Number of seats = 10 + (k - 1) = 9 + k
- Maximum students = ⌈(9+k)/2⌉

For even values of k: (9+k) is odd, so ⌈(9+k)/2⌉ = (9+k+1)/2 = (10+k)/2 = 5 + k/2
For odd values of k: (9+k) is even, so ⌈(9+k)/2⌉ = (9+k)/2 = 4.5 + k/2

Let me recalculate more carefully:
- k=1: 10 seats → 5 students
- k=2: 11 seats → 6 students
- k=3: 12 seats → 6 students
- k=4: 13 seats → 7 students
- k=5: 14 seats → 7 students
- k=6: 15 seats → 8 students
- ...
- k=19: 28 seats → 14 students
- k=20: 29 seats → 15 students

**Step 5: Sum all students**

The pattern is: 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15

This can be grouped as pairs (except we need to be careful):
- (5) + (6+6) + (7+7) + (8+8) + (9+9) + (10+10) + (11+11) + (12+12) + (13+13) + (14+14) + (15)
- = 5 + 2(6) + 2(7) + 2(8) + 2(9) + 2(10) + 2(11) + 2(12) + 2(13) + 2(14) + 15
- = 5 + 15 + 2(6+7+8+9+10+11+12+13+14)
- = 20 + 2(90)
- = 20 + 180
- = 200

**Step 6: Verify the sum**
Sum of 6 through 14 = (6+14)×9/2 = 20×9/2 = 90 ✓

Total = 5 + 180 + 15 = 200

ANSWER: C

1. **Restate the problem:** There are 20 rows. Row 1 has 10 seats, each next row has one more. Students can't sit next to each other in a row. Find max students.

2. **Seats per row:** Row n has 10 + (n-1) seats, so rows have 10, 11, 12, ..., 29 seats.

3. **Max students in a row with k seats (no two adjacent):** This is ⌈k/2⌉ = (k+1)/2 for odd k, k/2 for even k.

4. **Compute per row:**
   - 10 seats → 5
   - 11 → 6
   - 12 → 6
   - 13 → 7
   - 14 → 7
   - 15 → 8
   - 16 → 8
   - 17 → 9
   - 18 → 9
   - 19 → 10
   - 20 → 10
   - 21 → 11
   - 22 → 11
   - 23 → 12
   - 24 → 12
   - 25 → 13
   - 26 → 13
   - 27 → 14
   - 28 → 14
   - 29 → 15

5. **Sum:** Pair them up:
   - 5+6 = 11
   - 6+7 = 13
   - 7+8 = 15
   - 8+9 = 17
   - 9+10 = 19
   - 10+11 = 21
   - 11+12 = 23
   - 12+13 = 25
   - 13+14 = 27
   - 14+15 = 29

   Sum these: 11+13+15+17+19+21+23+25+27+29.
   These are 10 terms, arithmetic, average (11+29)/2 = 20, times 10 = 200.

6. **Rule out others:** 
   - A (150): too low.
   - B (180): too low.
   - D (400): would require ~all seats; total seats = sum 10..29 = 390, so 400 impossible.
   - E (460): exceeds total seats, impossible.
   - C (200): matches.

ANSWER: C

## Step-by-Step Solution

**Step 1: Understand what's being asked.**
We need to find the maximum number of students that can be seated in an auditorium where no two students sit next to each other in the same row.

**Step 2: Determine the number of seats in each row.**
- Row 1: 10 seats
- Row 2: 11 seats
- Row 3: 12 seats
- ...
- Row n: (9 + n) seats
- Row 20: 29 seats

**Step 3: Figure out how many students can sit in a row with k seats, with no two adjacent.**

If a row has k seats, the maximum number of non-adjacent students is ⌈k/2⌉ (ceiling of k/2).
- If k is even: k/2 students
- If k is odd: (k+1)/2 students

**Step 4: Calculate for each row.**

| Row | Seats (k) | Max students ⌈k/2⌉ |
|-----|-----------|---------------------|
| 1 | 10 | 5 |
| 2 | 11 | 6 |
| 3 | 12 | 6 |
| 4 | 13 | 7 |
| 5 | 14 | 7 |
| 6 | 15 | 8 |
| 7 | 16 | 8 |
| 8 | 17 | 9 |
| 9 | 18 | 9 |
| 10 | 19 | 10 |
| 11 | 20 | 10 |
| 12 | 21 | 11 |
| 13 | 22 | 11 |
| 14 | 23 | 12 |
| 15 | 24 | 12 |
| 16 | 25 | 13 |
| 17 | 26 | 13 |
| 18 | 27 | 14 |
| 19 | 28 | 14 |
| 20 | 29 | 15 |

**Step 5: Sum all the maximums.**

Group the pairs:
- Rows 1-2: 5 + 6 = 11
- Rows 3-4: 6 + 7 = 13
- Rows 5-6: 7 + 8 = 15
- Rows 7-8: 8 + 9 = 17
- Rows 9-10: 9 + 10 = 19
- Rows 11-12: 10 + 11 = 21
- Rows 13-14: 11 + 12 = 23
- Rows 15-16: 12 + 13 = 25
- Rows 17-18: 13 + 14 = 27
- Rows 19-20: 14 + 15 = 29

**Total = 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29**

This is an arithmetic sequence with first term 11, last term 29, and 10 terms.

Sum = 10 × (11 + 29)/2 = 10 × 20 = **200**

**Step 6: Check the answer choices.**
- 150: Too low
- 180: Too low  
- **200: ✓ Matches our calculation**
- 400: Would require nearly all seats filled
- 460: More than total seats

ANSWER: C

1. **Restate the problem.**  
   There are 20 rows in an auditorium.  
   - Row 1 has 10 seats.  
   - Each next row has 1 more seat than the row before it.  

   So the rows have:
   \[
   10,11,12,\dots,29
   \]
   seats.

   Students may sit in any row, but **no two students in the same row can sit next to each other**. We want the **maximum** number of students that can be seated.

2. **Find the maximum students in one row.**  
   If a row has \(n\) seats, the most students you can seat with no two adjacent is:
   \[
   \left\lceil \frac{n}{2} \right\rceil
   \]
   because you can alternate occupied and empty seats.

   For example:
   - 10 seats \(\to 5\) students
   - 11 seats \(\to 6\) students

3. **Apply this to all 20 rows.**  
   The seat counts are:
   \[
   10,11,12,13,\dots,28,29
   \]

   Pair them:
   \[
   (10,11), (12,13), (14,15), \dots, (28,29)
   \]

   For each pair:
   - \(10 \to 5\), \(11 \to 6\), total \(11\)
   - \(12 \to 6\), \(13 \to 7\), total \(13\)
   - \(14 \to 7\), \(15 \to 8\), total \(15\)

   In general, each pair \((2k,2k+1)\) contributes:
   \[
   k + (k+1) = 2k+1
   \]

4. **Add the totals.**  
   The pair totals are:
   \[
   11,13,15,17,19,21,23,25,27,29
   \]

   This is an arithmetic sequence with 10 terms. Its sum is:
   \[
   \frac{10(11+29)}{2} = \frac{10\cdot 40}{2}=200
   \]

5. **Check by another method.**  
   Since there are 10 even-numbered seat rows and 10 odd-numbered seat rows:
   - Even rows: \(10,12,\dots,28\) give \(5,6,\dots,14\), sum
     \[
     5+6+\cdots+14 = \frac{10(5+14)}{2}=95
     \]
   - Odd rows: \(11,13,\dots,29\) give \(6,7,\dots,15\), sum
     \[
     6+7+\cdots+15 = \frac{10(6+15)}{2}=105
     \]
   Total:
   \[
   95+105=200
   \]

6. **Rule out the choices.**
   - **A. 150**: too small; we found a legal seating of 200.
   - **B. 180**: too small.
   - **C. 200**: matches our calculation.
   - **D. 400**: impossible, since total seats are
     \[
     \frac{20(10+29)}{2}=390,
     \]
     so 400 students cannot fit.
   - **E. 460**: also impossible for the same reason.

ANSWER: C

1. **Restate the problem.**  
   There are **20 rows**. The first row has **10 seats**, and each row after that has **1 more seat** than the row before it.  
   Students may sit in any row, but **no two students can sit next to each other in the same row**.  
   We want the **maximum** number of students that can be seated.

2. **Find the number of seats in each row.**  
   The rows have:
   \[
   10, 11, 12, \dots, 29
   \]
   because there are 20 rows total, starting at 10 and increasing by 1 each time.

3. **Figure out the maximum students per row.**  
   If a row has \(n\) seats, and no two students can sit next to each other, the most students you can place is:
   \[
   \left\lceil \frac{n}{2} \right\rceil
   \]
   because you can seat them in alternating seats.

4. **Apply this to each row.**  
   For the rows from 10 to 29 seats:

   - 10 seats \(\to 5\) students
   - 11 seats \(\to 6\) students
   - 12 seats \(\to 6\) students
   - 13 seats \(\to 7\) students
   - 14 seats \(\to 7\) students
   - etc.

   So each pair of consecutive rows contributes:
   \[
   5+6=11,\quad 6+6=12,\quad 6+7=13,\ \text{etc.}
   \]
   But it is easier to add all the row maxima directly.

5. **Count the total.**  
   There are 10 even-numbered rows and 10 odd-numbered rows.

   - Even seat counts: \(10,12,14,\dots,28\)  
     Their maxima are \(5,6,7,\dots,14\), which sum to:
     \[
     5+6+\cdots+14 = \frac{(5+14)\cdot 10}{2} = 95
     \]

   - Odd seat counts: \(11,13,15,\dots,29\)  
     Their maxima are \(6,7,8,\dots,15\), which sum to:
     \[
     6+7+\cdots+15 = \frac{(6+15)\cdot 10}{2} = 105
     \]

   Total:
   \[
   95+105=200
   \]

6. **Check the choices.**  
   The total is **200**, which matches choice **C**.

ANSWER: C

To find the maximum number of students that can be seated in the auditorium under the given constraints, we follow these steps:

1. **Understand the seating arrangement:**
   The auditorium has 20 rows.
   - Row 1 has 10 seats.
   - Row 2 has 11 seats.
   - Row 3 has 12 seats.
   - ...
   - Row $n$ has $10 + (n - 1) = n + 9$ seats.
   - Row 20 has $20 + 9 = 29$ seats.

2. **Determine the maximum capacity per row:**
   The rule is that no student can sit next to another student in the same row. In a row with $k$ seats, the maximum number of students that can be seated is $\lceil k/2 \rceil$ (the ceiling of $k/2$).
   - If $k$ is even, we can seat $k/2$ students (e.g., seats 1, 3, 5, ..., $k-1$).
   - If $k$ is odd, we can seat $(k+1)/2$ students (e.g., seats 1, 3, 5, ..., $k$).

3. **Calculate the number of students for each row:**
   - Row 1 ($k=10$): $\lceil 10/2 \rceil = 5$
   - Row 2 ($k=11$): $\lceil 11/2 \rceil = 6$
   - Row 3 ($k=12$): $\lceil 12/2 \rceil = 6$
   - Row 4 ($k=13$): $\lceil 13/2 \rceil = 7$
   - Row 5 ($k=14$): $\lceil 14/2 \rceil = 7$
   - ...
   - Row 19 ($k=28$): $\lceil 28/2 \rceil = 14$
   - Row 20 ($k=29$): $\lceil 29/2 \rceil = 15$

4. **Sum the capacities:**
   The sequence of capacities is: 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15.
   We can group these:
   Sum = $5 + (6+6) + (7+7) + (8+8) + (9+9) + (10+10) + (11+11) + (12+12) + (13+13) + (14+14) + 15$
   Sum = $5 + 2(6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14) + 15$
   The sum of the arithmetic series $6 + 7 + ... + 14$ is $\frac{9}{2}(6 + 14) = \frac{9}{2}(20) = 9 \times 10 = 90$.
   Sum = $5 + 2(90) + 15 = 5 + 180 + 15 = 200$.

5. **Evaluate the choices:**
   - A (150), B (180), D (400), and E (460) are incorrect based on the calculation of the series sum. The calculation confirms the total capacity is 200.

ANSWER: C

To find the maximum number of students that can be seated in the auditorium under the given conditions, we can break the problem down into logical steps:

**1. Restate what is being asked:**
We need to find the maximum number of students that can sit in a 20-row auditorium where the first row has 10 seats, each subsequent row has 1 additional seat, and no two students are allowed to sit next to each other in the same row. 

**2. Set up the math:**
First, let's determine the number of seats in each of the 20 rows. 
The first row has 10 seats. Since each successive row has one more seat, the number of seats forms an arithmetic sequence: 10, 11, 12, 13, ..., up to the 20th row.
The 20th row will have $10 + (20 - 1) = 29$ seats.

To maximize the number of students in a single row of $n$ seats without any two sitting together, students must sit in every other seat (e.g., seats 1, 3, 5, etc.). 
- If a row has an even number of seats (like 10), the maximum number of students is exactly half: $10 / 2 = 5$.
- If a row has an odd number of seats (like 11), the maximum number of students is half rounded up: $\lceil 11 / 2 \rceil = 6$. 
In general, the maximum number of students a row of $n$ seats can hold is $\lceil n/2 \rceil$, which equals $n/2$ for even rows and $n/2 + 0.5$ for odd rows.

**3. Do the computation:**
Let's find the total number of seats in the auditorium by summing the arithmetic sequence from 10 to 29:
$$\text{Total Seats} = \frac{\text{Number of rows}}{2} \times (\text{First row} + \text{Last row})$$
$$\text{Total Seats} = \frac{20}{2} \times (10 + 29) = 10 \times 39 = 390 \text{ seats}$$

Next, we count how many rows have an odd number of seats. The odd-numbered seat counts are 11, 13, 15, 17, 19, 21, 23, 25, 27, and 29. There are exactly 10 rows with an odd number of seats.

Using our logic from Step 2, the maximum number of students is the total number of seats divided by 2, plus an extra $0.5$ for every row that has an odd number of seats:
$$\text{Maximum Students} = \frac{390}{2} + (10 \times 0.5)$$
$$\text{Maximum Students} = 195 + 5 = 200$$

**4. Explain how to rule the wrong choices out:**
- **Choices D (400) and E (460):** The absolute maximum capacity of the entire auditorium is only 390 seats. It is physically impossible to seat 400 or 460 students, so these can be immediately ruled out.
- **Choices A (150) and B (180):** Even if the auditorium had an even number of seats in every single row, we would be able to fill exactly half of the total seats ($390 / 2 = 195$). Because we can seat at least 195 students (and slightly more due to the odd-length rows), 150 and 180 are far too low.
- **Choice C (200):** This perfectly matches our calculated maximum.

ANSWER: C

An elegant way to solve this problem is to break it down into clear, logical steps.

### Step 1: Restate the Problem
We are asked to find the maximum number of students that can be seated in an auditorium with 20 rows of seats. 
* The first row has 10 seats.
* Each subsequent row has 1 more seat than the previous row.
* Students cannot sit next to each other in any row (i.e., no two students can occupy adjacent seats in the same row).

### Step 2: Determine the Number of Seats in Each Row
Let $r$ represent the row number, where $1 \le r \le 20$. 
The number of seats in row $r$ is given by:
$$\text{Seats}(r) = 10 + (r - 1)$$

* Row 1 has $10$ seats.
* Row 2 has $11$ seats.
* Row 3 has $12$ seats.
* ...
* Row 20 has $10 + 19 = 29$ seats.

### Step 3: Determine the Seating Capacity per Row
To maximize the number of students in a row of $N$ seats such that no two students are adjacent, we should place students in alternating seats starting from the first seat (e.g., seats 1, 3, 5, ...). 

The maximum number of students that can sit in a row of $N$ seats is given by the ceiling function:
$$\text{Students}(N) = \lceil N/2 \rceil$$

Let's test this formula:
* For an even number of seats, say $N = 10$: $\lceil 10/2 \rceil = 5$ students (seated at positions 1, 3, 5, 7, 9).
* For an odd number of seats, say $N = 11$: $\lceil 11/2 \rceil = 6$ students (seated at positions 1, 3, 5, 7, 9, 11).

### Step 4: Compute the Capacity for Each of the 20 Rows
Using our formula $\lceil N/2 \rceil$, we calculate the maximum number of students for each row:
* Row 1 (10 seats): $\lceil 10/2 \rceil = 5$
* Row 2 (11 seats): $\lceil 11/2 \rceil = 6$
* Row 3 (12 seats): $\lceil 12/2 \rceil = 6$
* Row 4 (13 seats): $\lceil 13/2 \rceil = 7$
* Row 5 (14 seats): $\lceil 14/2 \rceil = 7$
* ...
* Row 18 (27 seats): $\lceil 27/2 \rceil = 14$
* Row 19 (28 seats): $\lceil 28/2 \rceil = 14$
* Row 20 (29 seats): $\lceil 29/2 \rceil = 15$

### Step 5: Sum the Capacities
The sequence of student capacities for the 20 rows is:
$$5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15$$

We can group these terms to make the summation easier:
$$\text{Total} = 5 + (6 + 6) + (7 + 7) + \dots + (14 + 14) + 15$$
$$\text{Total} = 5 + 15 + 2 \times (6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14)$$

First, find the sum of the arithmetic series $6 + 7 + \dots + 14$:
$$\text{Sum} = \frac{9}{2} \times (6 + 14) = \frac{9}{2} \times 20 = 90$$

Now, substitute this back into our total:
$$\text{Total} = 20 + 2 \times 90 = 200$$

### Step 6: Rule Out Wrong Choices
* **A. 150** is too small (it assumes fewer than 8 students per row on average).
* **B. 180** is incorrect (it misses the extra capacity gained from the odd-numbered rows).
* **D. 400** and **E. 460** are too large (they exceed the maximum possible capacity even if students were allowed to sit next to each other in some rows).

Thus, the maximum number of students that can be seated is **200**.

ANSWER: C

The auditorium has 20 rows, with the number of seats in row \(k\) (for \(k = 1\) to \(20\)) given by the arithmetic sequence starting at 10 and ending at 29. That is, row \(k\) has \(m_k = 9 + k\) seats.

Students may sit in any combination of rows, but no two students may occupy adjacent seats in the same row. (There are no adjacency restrictions between rows.) Thus, the problem reduces to independently maximizing the number of students per row and then summing over all rows.

For a row with \(m\) seats in a line, the maximum number of non-adjacent seats is \(\lceil m/2 \rceil\). This quantity is realized by placing students in every other seat (starting from either end).

The required total is therefore
\[
\sum_{k=1}^{20} \lceil m_k/2 \rceil = \sum_{m=10}^{29} \lceil m/2 \rceil.
\]
The 20 summands may be grouped into the 10 consecutive pairs
\[
(10,11),\ (12,13),\ \dots,\ (28,29).
\]
The corresponding pair sums are
\[
5+6=11,\ 6+7=13,\ \dots,\ 14+15=29.
\]
These 10 values themselves form an arithmetic series with first term 11, last term 29, and common difference 2. Their sum is
\[
\frac{10}{2} \times (11+29) = 5 \times 40 = 200.
\]
Hence the maximum number of students that can be seated is 200.

This value matches choice C and exceeds the smaller options A and B. It is strictly less than the total number of seats (which is 390) and far less than the unrealistic figures D and E, both of which would require violating the non-adjacency rule inside rows.

Final answer: C

(The computation above already shows that every admissible seating yields at most 200 students, so no larger choice is feasible.)  
ANSWER: C

1. **Understanding the Problem**: The problem asks us to find the maximum number of students that can be seated in an auditorium with specific seating arrangements, under the condition that no two students can sit next to each other in the same row.

2. **Setting Up the Seating Arrangement**: The auditorium has 20 rows. The first row has 10 seats, and each successive row has one more seat than the previous row. So, the number of seats in each row forms an arithmetic sequence starting at 10 and increasing by 1 for each subsequent row.

3. **Calculating the Number of Seats in Each Row**: To understand the seating capacity, we first calculate the number of seats in each row. The number of seats in the rows are: 10, 11, 12, ..., 29. This is an arithmetic sequence where the first term, $a_1 = 10$, the last term, $a_{20} = 29$, and the common difference, $d = 1$.

4. **Determining the Maximum Number of Students per Row**: To maximize the number of students without having them sit next to each other, we can seat them in every other seat. For a row with $n$ seats, the maximum number of students that can be seated is $\lceil \frac{n}{2} \rceil$ if $n$ is odd or $\frac{n}{2}$ if $n$ is even, because we are alternating seats.

5. **Calculating the Maximum Number of Students for Each Row**:
   - For rows with an even number of seats (10, 12, 14, ..., 28), the maximum number of students is half the number of seats.
   - For rows with an odd number of seats (11, 13, 15, ..., 29), the maximum number of students is $\frac{n+1}{2}$.

6. **Explicit Calculation for Even and Odd Rows**:
   - Even rows: 10, 12, 14, 16, 18, 20, 22, 24, 26, 28. Their respective maximum students are: 5, 6, 7, 8, 9, 10, 11, 12, 13, 14.
   - Odd rows: 11, 13, 15, 17, 19, 21, 23, 25, 27, 29. Their respective maximum students are: 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.

7. **Summing the Maximum Number of Students**:
   - Sum for even rows: $5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 = 95$.
   - Sum for odd rows: $6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 105$.
   - Total maximum number of students = $95 + 105 = 200$.

8. **Ruling Out Wrong Choices**:
   - A. 150: Too low because our calculation yields a higher number.
   - B. 180: Still too low compared to our total.
   - C. 200: Matches our calculation.
   - D. 400: Too high because it exceeds the total number of seats (which is the sum of an arithmetic series: $S = \frac{n}{2}(a_1 + a_n) = \frac{20}{2}(10 + 29) = 10 \times 39 = 390$), and seating every other student cannot exceed half of this total.
   - E. 460: Also too high for the same reason as D.

ANSWER: C

The maximum number of students that can be seated is the sum over all rows of the maximum number of students that can sit in that row without adjacency. The first row has 10 seats, and each successive row has one more seat, so row \(r\) (with \(r = 1,2,\dots,20\)) has \(9 + r\) seats, giving seats from 10 to 29. For a row with \(n\) seats, the maximum number of students without two adjacent is \(\lceil n/2 \rceil\).

Split the rows into evens and odds. Even numbers: 10,12,…,28 correspond to \(n = 2m\) with \(m = 5,6,\dots,14\); their contribution is \(\sum_{m=5}^{14} m = 95\). Odd numbers: 11,13,…,29 correspond to \(n = 2m+1\) with \(m = 5,6,\dots,14\); their contribution is \(\sum_{m=5}^{14} (m+1) = 95 + 10 = 105\). Total = \(95 + 105 = 200\).

Other choices are too small (150, 180) or too large (400, 460) because they exceed the total seats (390) or ignore the adjacency restriction.

ANSWER: C

**Step 1: Restate the problem**  
We have an auditorium with 20 rows. The first row has 10 seats, and each following row has one more seat than the previous row. Students may sit in any row, but no two students may sit next to each other in the same row. We need the maximum number of students that can be seated.

**Step 2: Determine the number of seats in each row**  
Row 1: 10 seats  
Row 2: 11 seats  
Row 3: 12 seats  
...  
Row 20: 10 + 19 = 29 seats.

**Step 3: Maximum students per row**  
In a row with \(n\) seats, the maximum number of students that can sit without sitting next to each other is \(\lceil n/2 \rceil\). (For example, if a row has 10 seats, you can seat 5 students by using every other seat; for 11 seats, you can seat 6 students.)

**Step 4: Compute the maximum for rows 10 through 29**  
We have rows with seat counts \(n = 10, 11, 12, \dots, 29\).

- For even \(n\): \(\lceil n/2 \rceil = n/2\)  
- For odd \(n\): \(\lceil n/2 \rceil = (n+1)/2\)

List the maximum students per row:
- 10 → 5
- 11 → 6
- 12 → 6
- 13 → 7
- 14 → 7
- 15 → 8
- 16 → 8
- 17 → 9
- 18 → 9
- 19 → 10
- 20 → 10
- 21 → 11
- 22 → 11
- 23 → 12
- 24 → 12
- 25 → 13
- 26 → 13
- 27 → 14
- 28 → 14
- 29 → 15

**Step 5: Sum the maximum students**  
We can group the rows into 10 pairs of consecutive seat counts:
(10,11): 5+6=11  
(12,13): 6+7=13  
(14,15): 7+8=15  
(16,17): 8+9=17  
(18,19): 9+10=19  
(20,21): 10+11=21  
(22,23): 11+12=23  
(24,25): 12+13=25  
(26,27): 13+14=27  
(28,29): 14+15=29

These sums form an arithmetic sequence: 11, 13, 15, ..., 29 (10 terms).
Sum = (number of terms) × (first + last) / 2 = 10 × (11 + 29) / 2 = 10 × 40 / 2 = 10 × 20 = 200.

(Alternatively, sum the even-row students: 5+6+7+...+14 = 95, and odd-row students: 6+7+8+...+15 = 105; total = 95 + 105 = 200.)

**Step 6: Eliminate wrong choices**  
- A. 150: too low; this might be obtained if you mistakenly used a smaller number of rows or a different seating rule.  
- B. 180: also too low.  
- C. 200: matches our calculation.  
- D. 400 and E. 460: far too high; possibly from forgetting the "not next to another student" rule (sum of all seats = 10+11+...+29 = 390) or misapplying it.

ANSWER: C

We are given:

- An auditorium has **20 rows**.
- **Row 1** has **10 seats**.
- Each next row has **1 more seat** than the previous → arithmetic sequence.
- Students **cannot sit next to another student in the same row** → at most **every other seat** can be occupied.
- Goal: Find the **maximum number of students** that can be seated under this rule.

---

### Step 1: Determine number of seats in each row

This is an arithmetic sequence:

- First term $ a = 10 $
- Common difference $ d = 1 $
- Number of terms $ n = 20 $

So, row $ k $ (from 1 to 20) has:
$$
s_k = 10 + (k - 1) = k + 9 \text{ seats}
$$

So row 1: 10 seats, row 2: 11 seats, ..., row 20: $10 + 19 = 29$ seats.

---

### Step 2: Determine maximum students per row given "no adjacent students"

In a row with $ s $ seats in a **line**, the maximum number of people you can seat **without any two adjacent** is:

$$
\left\lceil \frac{s}{2} \right\rceil
$$

**Why?** Because you can occupy every other seat: seats 1, 3, 5, ..., and if $ s $ is odd, you get $(s+1)/2$; if even, $s/2$. That’s the ceiling of $ s/2 $.

Examples:
- 10 seats → max 5 students
- 11 seats → max 6 students
- 12 seats → 6 students
- 13 seats → 7 students, etc.

So for row $ k $, with $ s_k = k + 9 $, max students = $ \left\lceil \frac{k + 9}{2} \right\rceil $

---

### Step 3: Compute total maximum students over 20 rows

We need:
$$
\sum_{k=1}^{20} \left\lceil \frac{k + 9}{2} \right\rceil
$$

Let’s compute this carefully.

Let’s list $ s_k = k + 9 $, and then $ m_k = \left\lceil \frac{s_k}{2} \right\rceil $

| Row k | Seats s_k = k+9 | Max students m_k = ceil(s_k/2) |
|-------|------------------|-------------------------------|
| 1     | 10               | 5                             |
| 2     | 11               | 6                             |
| 3     | 12               | 6                             |
| 4     | 13               | 7                             |
| 5     | 14               | 7                             |
| 6     | 15               | 8                             |
| 7     | 16               | 8                             |
| 8     | 17               | 9                             |
| 9     | 18               | 9                             |
|10     | 19               |10                             |
|11     | 20               |10                             |
|12     | 21               |11                             |
|13     | 22               |11                             |
|14     | 23               |12                             |
|15     | 24               |12                             |
|16     | 25               |13                             |
|17     | 26               |13                             |
|18     | 27               |14                             |
|19     | 28               |14                             |
|20     | 29               |15                             |

Now sum the "Max students" column.

Group them in pairs to simplify:

- (Row 1 & 2): 5 + 6 = 11  
- (3 & 4): 6 + 7 = 13  
- (5 & 6): 7 + 8 = 15  
- (7 & 8): 8 + 9 = 17  
- (9 &10): 9 +10 = 19  
- (11&12):10 +11 = 21  
- (13&14):11 +12 = 23  
- (15&16):12 +13 = 25  
- (17&18):13 +14 = 27  
- (19&20):14 +15 = 29  

Now sum these 10 pair totals:

11 + 13 = 24  
24 + 15 = 39  
39 + 17 = 56  
56 + 19 = 75  
75 + 21 = 96  
96 + 23 = 119  
119 + 25 = 144  
144 + 27 = 171  
171 + 29 = **200**

So total maximum students = **200**

---

### Step 4: Check answer choices

Choices:
A. 150  
B. 180  
C. **200** ← matches  
D. 400  
E. 460

So correct answer is **C**

---

### Step 5: Double-check logic

- Seating rule: not next to another **in that row** → we handled per row independently. Students in different rows can be adjacent — that’s fine.
- Max per row = ceiling(seats/2) → correct for linear row (not circular).
- Row seat counts increase by 1 → correct.
- Sum computed carefully → verified via pairing, total 200.

All good.

ANSWER: C

I need to find the probability that three specific friends (Al, Bob, and Carol) are all assigned to the same lunch group.

**Step 1: Restate the problem**
- There are 3 lunch groups
- Each student is randomly assigned to one of the 3 groups with equal probability (1/3 each)
- I need to find P(Al, Bob, and Carol all in the same group)

**Step 2: Set up the approach**
Since the assignments are random and independent, I can think about this sequentially:
- Al gets assigned to some group
- Bob must be assigned to the same group as Al
- Carol must be assigned to the same group as Al and Bob

**Step 3: Calculate the probability**

Method 1 (Sequential thinking):
- Al is assigned to some group (say, Group 1). This happens with probability 1 (he has to go somewhere).
- Bob must be assigned to the same group as Al: P(Bob in same group as Al) = 1/3
- Carol must be assigned to the same group as Al and Bob: P(Carol in same group) = 1/3

Since these events are independent:
P(all three in same group) = 1 × (1/3) × (1/3) = 1/9

Method 2 (Counting outcomes):
- Total possible assignments for three students: 3 × 3 × 3 = 27
- Favorable outcomes (all in same group):
  - All in Group 1: 1 way
  - All in Group 2: 1 way
  - All in Group 3: 1 way
  - Total: 3 ways

P(all in same group) = 3/27 = 1/9

**Step 4: Check against the choices**
- A. 1/27 - This would be if they all had to be in a specific group
- B. 1/9 - This matches our calculation ✓
- C. 1/8 - No clear reason for this
- D. 1/6 - Too large
- E. 1/3 - Too large (this is just the probability for one person)

The answer is **B. 1/9**

ANSWER: B

1. We want the probability all three friends are in the same group of three equal groups.

2. Fix Al's group. Bob has 1/3 chance to match Al, Carol has 1/3 chance to match Al.

3. Probability = (1/3)(1/3) = 1/9.

4. Ruling out: 1/27 would be all three independently landing in one specific group (too small); 1/3 is just one person matching; others don't fit.

ANSWER: B

## Step-by-Step Solution

**Step 1: Understand what is being asked.**
We need to find the probability that Al, Bob, and Carol are all assigned to the same lunch group (any one of the three groups).

**Step 2: Set up the probability calculation.**
Each student is randomly assigned to one of three groups. We want all three friends in the same group.

**Step 3: Calculate the probability.**
- Al can be assigned to any group (probability = 1, this is certain).
- Bob must be assigned to the **same** group as Al: probability = 1/3
- Carol must be assigned to the **same** group as Al and Bob: probability = 1/3

**Step 4: Multiply the probabilities.**
P(all three in same group) = 1 × (1/3) × (1/3) = **1/9**

**Step 5: Verify by counting all outcomes.**
- Total ways to assign 3 people to 3 groups: 3³ = 27
- Favorable outcomes: all three in Group 1, all in Group 2, or all in Group 3 = **3 outcomes**
- Probability = 3/27 = **1/9** ✓

**Step 6: Rule out wrong choices.**
- **A (1/27):** This would be the probability of all three being assigned to one *specific* group (say, Group 1 only). It ignores that any of the 3 groups works.
- **C (1/8):** Would apply if there were only 2 groups (½ × ½ = 1/4... actually not even this). Doesn't apply here.
- **D (1/6):** No clear basis for this answer.
- **E (1/3):** Would be the probability that just two specific people match, not all three.

The correct answer is **1/9**.

ANSWER: B

1. **Restate the question.**  
We want the probability that **all three friends** — Al, Bob, and Carol — end up in the **same** lunch group.

2. **Understand the setup.**  
There are 3 lunch groups, and each student is assigned randomly to one of the 3 groups.

3. **Assign the students one at a time.**  
- Al can be put in **any** of the 3 groups. No restriction yet.
- For Bob to be in the same group as Al, the probability is  
  \[
  \frac{1}{3}
  \]
- For Carol also to be in that same group, the probability is  
  \[
  \frac{1}{3}
  \]

4. **Multiply the probabilities.**  
Since both Bob and Carol must match Al’s group,
\[
\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}
\]

5. **Check against the answer choices.**  
\[
\frac{1}{9}
\]
matches choice **B**.

6. **Why not the others?**  
- **A. \(1/27\)** would be the probability of all three going to one particular fixed lunch group, not just any same group.
- **C, D, E** are all too large.

ANSWER: B

1. **What is being asked?**  
   We want the probability that **Al, Bob, and Carol all end up in the same one of the 3 lunch groups**.

2. **Understand the setup**  
   Since the 600 students are divided into 3 equal groups, each student is equally likely to be placed in any of the 3 groups.  
   So for each friend, the chance of being assigned to a particular group is \( \frac13 \).

3. **Compute the probability step by step**  
   - Put Al in any group. This happens with probability 1, since we are just using Al’s group as the reference.  
   - Bob must be in the **same** group as Al: probability \( \frac13 \).  
   - Carol must also be in that same group: probability \( \frac13 \).

   Therefore,
   \[
   P(\text{all three in same group}) = 1 \cdot \frac13 \cdot \frac13 = \frac19
   \]

4. **Check against the choices**  
   - A. \( \frac{1}{27} \) — too small  
   - **B. \( \frac{1}{9} \) — matches**  
   - C. \( \frac{1}{8} \) — not equal  
   - D. \( \frac{1}{6} \) — not equal  
   - E. \( \frac{1}{3} \) — too large  

So the probability is approximately \( \frac19 \).

ANSWER: B

To solve this problem, we follow these steps:

1. **Restate the problem:** We have 600 students divided into three lunch groups of equal size (200 students per group). Three specific students (Al, Bob, and Carol) are assigned to these groups randomly. We need to find the probability that all three are assigned to the same group.

2. **Set up the math:**
   - First, consider Al. Al is assigned to one of the three groups. It does not matter which group he is in; let's assume he is in Group 1.
   - Now, consider Bob. Since there are 600 total spots and 200 spots per group, once Al is placed, there are 599 spots remaining. In Group 1, there are 199 spots remaining. The probability that Bob is assigned to the same group as Al is:
     $P(\text{Bob in Al's group}) = \frac{199}{599}$
   - Next, consider Carol. If both Al and Bob are in Group 1, there are 598 spots remaining in total, and 198 spots remaining in Group 1. The probability that Carol is assigned to the same group as Al and Bob is:
     $P(\text{Carol in Al/Bob's group}) = \frac{198}{598}$

3. **Perform the computation:**
   - The probability that all three are in the same group is the product of these two probabilities:
     $P = \frac{199}{599} \times \frac{198}{598}$
   - Since 600 is a large number, we can approximate these fractions:
     $\frac{199}{599} \approx \frac{200}{600} = \frac{1}{3}$
     $\frac{198}{598} \approx \frac{200}{600} = \frac{1}{3}$
   - Multiplying these approximations:
     $P \approx \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

4. **Rule out wrong choices:**
   - A (1/27) would be the probability if the students were assigned to groups independently with replacement (like rolling a 3-sided die), but the groups have fixed sizes.
   - C (1/8) is a common distractor based on $1/2^3$, which is incorrect here.
   - D (1/6) and E (1/3) do not match the calculated probability of approximately 1/9.
   - The exact value is $\frac{199 \times 198}{599 \times 598} \approx 0.1100$, while $1/9 \approx 0.1111$. This is very close, confirming B is the correct choice.

ANSWER: B

To find the approximate probability that Al, Bob, and Carol are assigned to the same lunch group, we can break the problem down into logical steps.

**Step 1: Restate what is being asked**
We need to find the approximate probability that three specific students out of 600 are placed into the exact same lunch group, given that the students are divided into three groups of equal size.

**Step 2: Set up the math**
Since there are 600 students in total and they are divided into 3 equal-sized groups, each lunch group will have exactly:
600 ÷ 3 = 200 students.

We can determine the probability by placing the three friends into a group one by one.

**Step 3: Do the computation**
- **Al's placement:** Al will be assigned to one of the three groups. The probability that he is placed in *some* group is simply **1** (or 100%). 
- **Bob's placement:** Once Al is placed in a group, there are 599 total spots left for the remaining students. In Al's specific group, there are 199 spots left. Therefore, the probability that Bob is randomly assigned to the same group as Al is **199/599**.
- **Carol's placement:** If Al and Bob are now in the same group, there are 598 total spots left in the school, and 198 spots left in their group. The probability that Carol is also assigned to this group is **198/598**.

To find the probability of all these events happening together, we multiply them:
Probability = 1 × (199 / 599) × (198 / 598)

Because the problem asks for an *approximate* probability, we can round these fractions:
199 / 599 ≈ 200 / 600 = 1/3
198 / 598 ≈ 200 / 600 = 1/3

Multiplying these approximations gives:
1 × (1/3) × (1/3) = **1/9**

**Step 4: Explain how to rule out the wrong choices**
- **A. 1/27:** This would be the probability if the question asked for the three friends to be placed in one *specific* pre-chosen group (e.g., "What is the probability they are all in the 11:00 AM lunch group?"). Since they can be together in *any* of the three groups, we multiply 1/27 by 3 to get 1/9.
- **C. 1/8 & D. 1/6:** These fractions do not logically follow from the base-3 nature of the group divisions (3 groups, 1/3 chance roughly per person). 
- **E. 1/3:** This is the approximate probability that only *two* friends (for example, just Al and Bob) end up in the same group. It forgets to factor in the third friend, Carol. 

ANSWER: B

An elegant way to solve this problem is by using probability and combinations, then simplifying using an approximation.

### Step 1: Restate what is being asked
We need to find the approximate probability that three specific students (Al, Bob, and Carol) are assigned to the same lunch group when 600 students are randomly divided into three equal groups of 200.

### Step 2: Set up the exact probability
There are 600 total slots for students, divided into 3 groups of 200 slots each.
The total number of ways to assign slots to Al, Bob, and Carol is the number of ways to choose 3 slots out of 600:
$$\text{Total ways} = \binom{600}{3}$$

For the three friends to be in the same lunch group, they must all be assigned to slots within the same group. There are 3 groups, and for each group, the number of ways to choose 3 slots is $\binom{200}{3}$. Thus:
$$\text{Favorable ways} = 3 \times \binom{200}{3}$$

The exact probability $P$ is:
$$P = \frac{3 \times \binom{200}{3}}{\binom{600}{3}}$$

### Step 3: Simplify the computation
Using the formula for combinations, we write:
$$P = 3 \times \frac{\frac{200 \times 199 \times 198}{3 \times 2 \times 1}}{\frac{600 \times 599 \times 598}{3 \times 2 \times 1}}$$

Cancel out the denominators:
$$P = 3 \times \frac{200 \times 199 \times 198}{600 \times 599 \times 598}$$

Group the terms:
$$P = \left(3 \times \frac{200}{600}\right) \times \frac{199}{599} \times \frac{198}{598}$$
$$P = 1 \times \frac{199}{599} \times \frac{198}{598}$$

### Step 4: Approximate the fractions
Since 600 is a very large number, we can approximate the fractions:
$$\frac{199}{599} \approx \frac{200}{600} = \frac{1}{3}$$
$$\frac{198}{598} \approx \frac{200}{600} = \frac{1}{3}$$

Multiplying these approximations gives:
$$P \approx 1 \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$

*(Note: The exact probability is $\approx 0.110$, which is extremely close to $\frac{1}{9} \approx 0.111$.)*

### Step 5: Rule out the wrong choices
* **A. $1/27$** is too small (this would be the probability if we also specified which *particular* group they all had to join, e.g., all in Group 1).
* **C. $1/8$**, **D. $1/6$**, and **E. $1/3$** are all significantly larger than $1/9 \approx 0.111$.

Thus, the correct choice is B.

ANSWER: B

The problem requires finding the probability that Al, Bob, and Carol are assigned to the same lunch group of exactly 200 students each (out of 600 total students). Because the group sizes are fixed, the assignments are not fully independent, but the large population size of 600 makes an approximation appropriate.

The exact probability that all three students are placed in one specific group (say, group 1) equals
\[
\frac{\binom{597}{197}}{\binom{600}{200}} = \frac{200 \times 199 \times 198}{600 \times 599 \times 598}.
\]
There are three possible groups, so the probability that the three students occupy the same group is
\[
3 \times \frac{200 \times 199 \times 198}{600 \times 599 \times 598} = \frac{199}{599} \times \frac{198}{598}.
\]
Evaluating the numerical value gives
\[
\frac{199}{599} \approx 0.3322, \qquad \frac{198}{598} \approx 0.3311,
\]
and their product is approximately \(0.1099\). This is indistinguishable from \(1/9 \approx 0.1111\) at the level of precision needed for the given choices.

(The same numerical result is obtained by treating the assignments as independent and uniform over the three groups: the probability they all land in any one fixed group is \((1/3)^3\), and multiplying by three groups again yields exactly \(1/9\). The fixed-size constraint merely perturbs this value by less than \(0.002\).)

Thus the probability is approximately \(1/9\).

- Choice A (\(1/27\)) is the probability that the three students receive three completely independent uniform assignments that happen to coincide, which undercounts by a factor of three.
- Choice C (\(1/8\)) would arise only from an erroneous two-group model.
- Choice D (\(1/6\)) has no combinatorial interpretation for three labeled groups.
- Choice E (\(1/3\)) equals the probability that Bob matches Al, ignoring Carol entirely.

Hence the correct selection is B.

ANSWER: B

1. The problem asks us to find the probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group out of three groups when 600 students are randomly divided into these groups.

2. To simplify, we first understand that the total number of students and the exact size of the groups are not directly relevant to calculating the probability for Al, Bob, and Carol being in the same group, as the probability depends on the random assignment process.

3. Since there are three lunch groups, the probability that Al is assigned to any particular group is 1/3.

4. After Al is assigned, the probability that Bob is assigned to the same group as Al is also 1/3, because there are three groups and the assignment is random.

5. Similarly, after Al and Bob are assigned to a particular group, the probability that Carol is assigned to the same group is 1/3.

6. To find the probability that all three (Al, Bob, and Carol) are in the same group, we need to consider the probability of Bob and Carol being in the same group as Al. So, we multiply the probabilities found in steps 3, 4, and 5: (1) * (1/3) * (1/3) = 1/9. The first "(1)" is because Al's assignment doesn't restrict the others; we're just using his group as a reference.

7. Now, let's examine the choices:
   - A. 1/27: This would be the result if we multiplied the probabilities of all three being assigned to a specific group (1/3 * 1/3 * 1/3), not accounting for the fact that we're looking for them to be in "the same group," not a "specific group."
   - B. 1/9: This matches our calculation.
   - C. 1/8: There's no basis for this fraction given the division into three groups.
   - D. 1/6: This doesn't align with our simple probability calculation.
   - E. 1/3: This is too high; it's the probability of two people being in the same group as a third, not all three being in the same group.

8. Our computation directly supports choice B, and we've ruled out the other choices based on the logic of the probability calculation.

ANSWER: B

We are asked: 600 students divided into three equal groups (200 each). Three friends are randomly assigned. We want the probability that all three end up in the same lunch group.

**Step 1: Interpret the problem**  
Since assignments are random and independent (each student equally likely to be in any of the 3 groups), the probability that any particular student goes to a specific group is \( \frac{1}{3} \).

**Step 2: Compute probability for a specific group**  
The probability that Al, Bob, and Carol all go to, say, Group A is:  
\[
\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}.
\]

**Step 3: Account for any of the three groups**  
They could all be together in Group A, Group B, or Group C. These are three mutually exclusive events. So total probability is:  
\[
3 \times \frac{1}{27} = \frac{3}{27} = \frac{1}{9}.
\]

**Step 4: Check if group sizes matter**  
With 600 students and groups of 200, assignments are not independent in a strict sense because groups have fixed sizes. However, for three specific students, the probability they are all in the same group is still exactly \(\frac{1}{9}\) by symmetry: the first friend can go anywhere, the second has probability \(\frac{199}{599}\) to join the same group (since 199 spots left in that group out of 599 remaining), and the third has \(\frac{198}{598}\). This product simplifies to \(\frac{1}{9}\) as well.

**Step 5: Rule out wrong choices**  
- A: \(1/27\) is the probability for one specific group, not all three.  
- C, D, E: These are larger than \(1/9\) and do not match the calculation.  
- B: \(1/9\) matches exactly.

ANSWER: B

1. **Restate the problem**: There are 600 students divided into three equal lunch groups (200 each) at different times. A computer randomly assigns each student to a group. We need the approximate probability that three specific friends (Al, Bob, Carol) all end up in the same lunch group.

2. **Interpret the randomness**: The assignment could be either independent random assignment (each student equally likely to any group) or a random partition into exactly equal groups. Since the problem asks for an approximate probability and 600 is large, both yield about the same answer. We'll check both to be safe.

3. **Independent assignment method**: Each student is independently assigned to one of three groups with probability \( \frac{1}{3} \).  
   - Al can be in any group.  
   - Probability Bob is in the same group as Al: \( \frac{1}{3} \).  
   - Probability Carol is also in that same group: \( \frac{1}{3} \).  
   - Combined probability: \( 1 \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \).

4. **Exact partition method**: The groups are distinct (different lunch times) and exactly 200 each.  
   - Total ways to assign students: \( \frac{600!}{200!\,200!\,200!} \).  
   - Favorable ways (all three in the same specific group): Choose the other 197 members for that group from the remaining 597, then split the rest: \( \binom{597}{197} \times \binom{400}{200} \).  
   - Multiply by 3 (three possible groups).  
   - Probability: \( \frac{3 \times \binom{597}{197} \times \binom{400}{200}}{\binom{600}{200} \times \binom{400}{200}} = 3 \times \frac{\binom{597}{197}}{\binom{600}{200}} \).  
   - Simplify ratio: \( \frac{\binom{597}{197}}{\binom{600}{200}} = \frac{597!}{197!\,400!} \times \frac{200!\,400!}{600!} = \frac{200 \times 199 \times 198}{600 \times 599 \times 598} \).  
   - Probability = \( 3 \times \frac{200 \times 199 \times 198}{600 \times 599 \times 598} = \frac{200}{600} \times 3 \times \frac{199 \times 198}{599 \times 598} = 1 \times \frac{199 \times 198}{599 \times 598} \).  
   - Compute approximate: \( \frac{199 \times 198}{599 \times 598} \approx \frac{39402}{358202} \approx 0.1100 \), while \( \frac{1}{9} \approx 0.1111 \). So it's approximately \( \frac{1}{9} \).

5. **Rule out other choices**: \( \frac{1}{27} \) would be if groups were assigned in order and all had to be in one specific group; \( \frac{1}{8} \), \( \frac{1}{6} \), and \( \frac{1}{3} \) are too large or not matching the computed value.

**Final answer:**
ANSWER: B

nstead, the assignment is **random among all assignments that result in exactly 200 students per group**.

However, for the purpose of computing the probability that **three specific students** end up in the same group, **the difference between these two models is negligible when the total number is large** (600 is large), and in fact, **the answer is the same** in both models! Let's justify that.

But to be precise, let's consider the **exact model**: random assignment **with exactly 200 in each group**.

---

### Step 2: Fix Al’s group, then compute probability Bob and Carol are in the same group

Because the groups are symmetric, we can fix Al’s group without loss of generality.

Suppose Al is assigned to Group 1. (This happens with probability 1/3, but since we’re looking for all three in **any** same group, we can fix Al’s group and just compute the chance Bob and Carol join him.)

Now, **given** that Al is in Group 1, there are **599 remaining students**, and **199 remaining spots** in Group 1 (since Al took one of the 200), and 200 spots each in Groups 2 and 3.

We now compute the probability that **both Bob and Carol are also assigned to Group 1**, given this setup.

So:

- Probability Bob is assigned to Group 1 = $ \frac{199}{599} $
- Then, **given** Bob is in Group 1, there are now 198 spots left in Group 1, and 598 students remaining (excluding Al and Bob), so:
- Probability Carol is assigned to Group 1 = $ \frac{198}{598} $

So probability all three are in Group 1 is:

$$
P_1 = \frac{199}{599} \cdot \frac{198}{598}
$$

But they could also all be in Group 2 or Group 3. However, **by symmetry**, the total probability that all three are in the **same** group is **3 times** the probability they are all in Group 1 **given that Al is in Group 1**? Wait — careful!

Actually, **Al is in some group** — say Group X. The chance that Bob and Carol are in **that same group** is what we need. We don’t multiply by 3 again, because Al is already in one of the groups, and we’re just checking if Bob and Carol join him.

So **total probability** = probability Bob and Carol are assigned to **Al’s group**, given random assignment with equal group sizes.

So the correct probability is:

$$
P = \frac{199}{599} \cdot \frac{198}{598}
$$

Let’s compute this approximately.

---

### Step 3: Approximate the probability

Note that 199/599 ≈ 1/3 (since 600/3 = 200, so 199/599 ≈ 0.3322)

Similarly, 198/598 ≈ 1/3 (since 198/598 ≈ 0.3311)

So the product is approximately (1/3) × (1/3) = **1/9**

Let’s compute more precisely:

- 199 ÷ 599 ≈  
  599 × 0.332 = 198.868 → so 199/599 ≈ 0.3322

- 198 ÷ 598 ≈  
  598 × 0.3311 = 198.0 → so ≈ 0.3311

Multiply: 0.3322 × 0.3311 ≈

Use approximation: (1/3 - ε)² ≈ 1/9 - (2ε)/3

But easier:  
0.3322 × 0.3311 ≈  
= (0.33 + 0.0022)(0.33 + 0.0011)  
≈ 0.33² + 0.33(0.0022 + 0.0011) + (0.0022)(0.0011)  
= 0.1089 + 0.33×0.0033 + negligible  
= 0.1089 + 0.001089 ≈ **0.11**

Now, 1/9 ≈ 0.1111...

So yes, very close to **1/9**.

Thus, the probability is **approximately 1/9**.

---

### Step 4: Consider alternative (independent assignment) model

If each student were independently assigned to one of 3 groups with equal probability (1/3 each), then:

- Probability all 3 in same group =  
  P(all in group 1) + P(all in group 2) + P(all in group 3)  
  = 3 × (1/3)³ = 3 × (1/27) = **1/9**

So **same answer!**

So regardless of whether assignment is independent or constrained to equal group sizes (especially with large total number), the probability is **very close to 1/9**, and in the independent case exactly 1/9.

Since the problem says “approximately,” and gives 1/9 as a choice, that’s the answer.

---

### Step 5: Eliminate other choices

- 1/27 is the probability all three are in a **specific** group (e.g., Group 1), not any same group.
- 1/8, 1/6, 1/3 are all significantly larger than our computed ~0.11.
  - 1/8 = 0.125 (a bit higher)
  - 1/6 ≈ 0.1667
  - 1/3 ≈ 0.333

So **1/9 ≈ 0.111** is the best match.

---

ANSWER: B

I need to find the possible values of $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$.

**Step 1: Understand what each term means**

For any nonzero real number $x$:
- $\frac{x}{|x|} = 1$ if $x > 0$
- $\frac{x}{|x|} = -1$ if $x < 0$

So each term equals either 1 or -1 depending on the sign of the variable.

**Step 2: Use the constraint $a + b + c = 0$**

Since $a + b + c = 0$ and all are nonzero, at least one must be positive and at least one must be negative.

**Step 3: Analyze possible sign distributions**

Let me denote the sign pattern. If $k$ variables are positive and $(3-k)$ are negative:
- The sum of signs (from $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|}$) = $k - (3-k) = 2k - 3$

**Case 1: One positive, two negative** ($k=1$)
- Sum of first three terms: $2(1) - 3 = -1$
- Example: $a = 2, b = -1, c = -1$ ✓ (sum = 0)
- Then $abc = 2(-1)(-1) = 2 > 0$, so $\frac{abc}{|abc|} = 1$
- Total: $-1 + 1 = 0$

**Case 2: Two positive, one negative** ($k=2$)
- Sum of first three terms: $2(2) - 3 = 1$
- Example: $a = 1, b = 1, c = -2$ ✓ (sum = 0)
- Then $abc = 1(1)(-2) = -2 < 0$, so $\frac{abc}{|abc|} = -1$
- Total: $1 + (-1) = 0$

Wait, let me reconsider. Can we have all three positive or all three negative?

**Case 3: All positive** ($k=3$)
- This would give $a + b + c > 0$, contradicting $a + b + c = 0$ ✗

**Case 4: All negative** ($k=0$)
- This would give $a + b + c < 0$, contradicting $a + b + c = 0$ ✗

So only Cases 1 and 2 are possible.

**Step 4: Check the sign of $abc$ more carefully**

In **Case 1** (one positive, two negative):
- $abc = (+)(-)(-) = (+)$
- So $\frac{abc}{|abc|} = +1$

In **Case 2** (two positive, one negative):
- $abc = (+)(+)(-) = (-)$
- So $\frac{abc}{|abc|} = -1$

**Step 5: Calculate total for each case**

**Case 1:** $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|} = -1 + 1 = 0$

**Case 2:** $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|} = 1 + (-1) = 0$

**Step 6: Verify with concrete examples**

Example 1: $a = 2, b = -1, c = -1$
- $\frac{2}{2} + \frac{-1}{1} + \frac{-1}{1} + \frac{2}{2} = 1 - 1 - 1 + 1 = 0$ ✓

Example 2: $a = 1, b = 1, c = -2$
- $\frac{1}{1} + \frac{1}{1} + \frac{-2}{2} + \frac{-2}{2} = 1 + 1 - 1 - 1 = 0$ ✓

**Conclusion**

The only possible value is **0**.

ANSWER: A

1. We need possible values of a/|a| + b/|b| + c/|c| + abc/|abc|, where each fraction is the sign (±1).

2. Since a+b+c=0, they can't all be same sign. So either two positive one negative, or two negative one positive.

3. Case: two positive, one negative. Sum of signs = +1+1−1 = +1. Product abc = (+)(+)(−) = negative, so sign = −1. Total = 1 + (−1) = 0.

4. Case: two negative, one positive. Sum of signs = −1−1+1 = −1. Product = (−)(−)(+) = positive, sign = +1. Total = −1 + 1 = 0.

5. Both cases give 0.

6. Rule out others: only 0 occurs, eliminating B, C, D, E.

ANSWER: A

I need to find all possible values of $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$ given that $a + b + c = 0$ and $a, b, c$ are nonzero reals.

## Step 1: Understand the components

Each term $\frac{x}{|x|}$ is the **sign function**: it equals +1 if x > 0, and -1 if x < 0.

So let $s_a = \text{sgn}(a)$, $s_b = \text{sgn}(b)$, $s_c = \text{sgn}(c)$, each equal to ±1.

Note that $\frac{abc}{|abc|} = \text{sgn}(abc) = s_a \cdot s_b \cdot s_c$.

## Step 2: Determine possible sign combinations given a + b + c = 0

Since $a + b + c = 0$ with all nonzero, the numbers can't all be the same sign (all positive would give positive sum, all negative would give negative sum).

**Case 1: Two positives, one negative**
Say $a > 0, b > 0, c < 0$: Then $s_a + s_b + s_c = 1 + 1 - 1 = 1$.
And $s_a \cdot s_b \cdot s_c = (1)(1)(-1) = -1$.
Expression = $1 + (-1) = 0$.

This is achievable: e.g., $a = 1, b = 1, c = -2$: sum = 0 ✓

**Case 2: Two negatives, one positive**
Say $a < 0, b < 0, c > 0$: Then $s_a + s_b + s_c = -1 - 1 + 1 = -1$.
And $s_a \cdot s_b \cdot s_c = (-1)(-1)(1) = 1$.
Expression = $-1 + 1 = 0$.

This is achievable: e.g., $a = -1, b = -1, c = 2$: sum = 0 ✓

## Step 3: Check if other sign patterns are possible

- **All positive**: $a + b + c > 0 \neq 0$. Impossible.
- **All negative**: $a + b + c < 0 \neq 0$. Impossible.
- **One positive, two negative** or **two positive, one negative**: These are the only cases (covered above).

## Step 4: Compute the expression in all cases

**Case 1 (two positives, one negative):** 
$$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|} = 1 + 1 + (-1) + (-1) = 0$$

**Case 2 (two negatives, one positive):**
$$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|} = (-1) + (-1) + 1 + 1 = 0$$

## Step 5: Conclusion

In all possible cases, the expression equals **0**.

The answer is choice **A**.

ANSWER: A

1. **Restate the problem.**  
We are given nonzero real numbers \(a,b,c\) such that
\[
a+b+c=0.
\]
We want the possible value(s) of
\[
a|a|+b|b|+c|c|+abc|abc|.
\]

We must decide which answer choice is correct.

---

2. **Use the condition \(a+b+c=0\).**  
Since \(a+b+c=0\), we can write
\[
c=-(a+b).
\]

But a more useful fact is this: because \(a,b,c\) are nonzero and add to \(0\), they cannot all have the same sign. So exactly one is positive and two are negative, or two are positive and one is negative.

Also recall:
- if \(x>0\), then \(x|x|=x^2\),
- if \(x<0\), then \(x|x|=-x^2\).

And similarly, \(abc|abc|\) equals \((abc)^2\) if \(abc>0\), and \(- (abc)^2\) if \(abc<0\).

---

3. **Case 1: two positives and one negative.**  
Let the positive numbers be \(a,b\), and let the negative number be \(c\).  
Since \(a+b+c=0\), we have
\[
c=-(a+b).
\]

Now compute:
\[
a|a|=a^2,\qquad b|b|=b^2,\qquad c|c|=-c^2.
\]
So
\[
a|a|+b|b|+c|c|=a^2+b^2-c^2.
\]
Substitute \(c=-(a+b)\), so \(c^2=(a+b)^2\):
\[
a^2+b^2-(a+b)^2
= a^2+b^2-(a^2+2ab+b^2)
= -2ab.
\]

Now look at \(abc\). Since \(a>0\), \(b>0\), \(c<0\), we have
\[
abc<0.
\]
Thus
\[
abc|abc|=-(abc)^2.
\]

So the full expression becomes
\[
-2ab-(abc)^2.
\]
This is definitely **negative**, but it is not obviously one of the listed constants. So let us check whether we interpreted the expression correctly.

The expression in the problem is intended as
\[
a|a|+b|b|+c|c|+abc|abc|,
\]
and the answer choices are just small fixed numbers. That suggests the quantity must simplify much more dramatically. So let us instead factor using signs carefully.

---

4. **Try a simpler substitution.**  
Let
\[
x=|a|,\quad y=|b|,\quad z=|c|.
\]
Because \(a+b+c=0\), one of the three has magnitude equal to the sum of the other two.

### Subcase 1a: \(a,b>0\), \(c<0\)
Then \(c=-(a+b)\), so
\[
|c|=a+b.
\]
Thus
\[
a|a|+b|b|+c|c|
= a^2+b^2-(a+b)^2
= -2ab.
\]
Also
\[
abc = ab(-(a+b))=-ab(a+b),
\]
so
\[
|abc|=ab(a+b),
\]
and therefore
\[
abc|abc| = -[ab(a+b)]^2.
\]
So the whole expression is
\[
-2ab-[ab(a+b)]^2,
\]
which varies with \(a,b\), not matching the answer choices.

Therefore, the only way the problem makes sense is that the intended expression must be
\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|},
\]
since that gives values like \(0,\pm2,\pm1\). The choices strongly confirm this.

So we solve that intended sign-expression.

---

5. **Evaluate the sign expression.**  
For any nonzero real \(x\),
\[
\frac{x}{|x|}=
\begin{cases}
1,&x>0\\
-1,&x<0
\end{cases}
\]
which is just the sign of \(x\).

So we need the possible values of
\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}.
\]

Because \(a+b+c=0\) and none are zero, the signs must be either:
- two positive, one negative, or
- two negative, one positive.

---

6. **Case 1: two positive, one negative.**  
Then the first three sign terms add to
\[
1+1-1=1.
\]
Also, the product \(abc\) has sign
\[
(+)(+)(-)=-,
\]
so
\[
\frac{abc}{|abc|}=-1.
\]
Total:
\[
1+(-1)=0.
\]

---

7. **Case 2: two negative, one positive.**  
Then the first three sign terms add to
\[
-1-1+1=-1.
\]
Also, the product \(abc\) has sign
\[
(-)(-)(+)=+,
\]
so
\[
\frac{abc}{|abc|}=1.
\]
Total:
\[
-1+1=0.
\]

---

8. **So the only possible value is \(0\).**  
Thus the correct choice is **A**.

---

9. **Rule out the other choices.**
- **B** says \(1\) and \(-1\), but we always got \(0\).
- **C** says \(2\) and \(-2\), impossible.
- **D** says \(0,2,-2\), but only \(0\) occurs.
- **E** says \(0,1,-1\), but \(1\) and \(-1\) do not occur.

Therefore the correct answer choice is **A**.

ANSWER: A

1. We are asked to find all possible values of  
   \[
   a|a|+b|b|+c|c|+abc|abc|
   \]
   given that \(a,b,c\) are nonzero real numbers and
   \[
   a+b+c=0.
   \]

2. Since \(a+b+c=0\), the three numbers cannot all have the same sign. So there must be either:
   - two positive and one negative, or
   - two negative and one positive.

3. Let us use the condition \(a+b+c=0\) to rewrite one variable in terms of the others.  
   The expression involves absolute values, so it is helpful to examine sign cases.

---

### Case 1: \(a\) and \(b\) are positive, \(c\) is negative
Then \(c=-(a+b)\), so \(c<0\).

- \(a|a|=a^2\)
- \(b|b|=b^2\)
- \(c|c|=-c^2=-(a+b)^2\)

So
\[
a|a|+b|b|+c|c|=a^2+b^2-(a+b)^2=-2ab.
\]

Now look at \(abc\). Since \(a,b>0\) and \(c<0\), we have \(abc<0\), so
\[
abc|abc|=-(abc)^2.
\]
But \(abc=a b (-(a+b))=-ab(a+b)\), so
\[
abc|abc|=-a^2b^2(a+b)^2.
\]

That makes the whole expression
\[
-2ab-a^2b^2(a+b)^2,
\]
which is always negative. This suggests we should look for a simpler global approach.

---

### Better approach: use \(c=-(a+b)\) and factor by signs
Because the expression has \(x|x|\), it equals:
- \(x^2\) if \(x>0\),
- \(-x^2\) if \(x<0\).

So the first three terms depend on the signs. The key observation is that because \(a+b+c=0\), the expression is forced into one of a few constant values.

Let us test the sign patterns.

---

### Case 2: \(a>0,\ b>0,\ c<0\)
Then \(c=-(a+b)\), and
\[
a|a|+b|b|+c|c|=a^2+b^2-(a+b)^2=-2ab.
\]
Also \(abc<0\), so
\[
abc|abc|=-(abc)^2.
\]
But because \(c=-(a+b)\),
\[
abc=-ab(a+b),
\]
so
\[
abc|abc|=-a^2b^2(a+b)^2.
\]
The total is not a fixed simple value, so let us try another way.

---

### Key algebraic simplification
Notice that if we let
\[
x=a|a|,\quad y=b|b|,\quad z=c|c|,
\]
then \(x,y,z\) have the same signs as \(a,b,c\), and the condition \(a+b+c=0\) implies the values of \(a,b,c\) are in a special ratio within each sign pattern.

A cleaner route is to use an example to identify possible values.

If \(a=1,\ b=1,\ c=-2\), then:
\[
a|a|+b|b|+c|c| = 1+1-4=-2,
\]
and
\[
abc=-2,\quad abc|abc|=-4.
\]
Total \(=-6\), which is not among the choices, so that cannot be the intended interpretation.

So the expression must be read carefully as
\[
a|a|+b|b|+c|c|+abc\,|abc|
\]
and the multiple-choice answers suggest the total simplifies to only a small set of values.

---

### Observe the sign of the whole expression
Let
\[
S=a|a|+b|b|+c|c|.
\]
If \(a,b>0\) and \(c<0\), then \(S=a^2+b^2-c^2\). Since \(c=-(a+b)\),
\[
S=a^2+b^2-(a+b)^2=-2ab.
\]
Now \(abc<0\), so \(abc|abc|=-(abc)^2\), which is also negative. Thus the whole expression is negative.

Likewise, if \(a,b<0\) and \(c>0\), the whole expression is positive by symmetry.

This suggests the only possible magnitudes are determined by a normalized form. Let us test the standard equal-magnitude choice:
\[
a=1,\ b=1,\ c=-2.
\]
Then total \(=-2-4=-6\), not a choice.

Try \(a=2,\ b=-1,\ c=-1\):
\[
a|a|=4,\quad b|b|=-1,\quad c|c|=-1,
\]
sum of first three \(=2\).
Also \(abc=2\), so \(abc|abc|=4\).
Total \(=6\), again not a choice.

This indicates the intended expression is likely meant as
\[
a|a|+b|b|+c|c|+abc\,|abc|
\]
and the only choice matching a sign-invariant simplification is \(0\).

4. Therefore the possible value is \(0\).

5. The correct choice is **A**.

ANSWER: A

To find the possible values of the expression $E = \frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$, we analyze the signs of the variables $a, b,$ and $c$.

1. **Restate the expression:**
   Let $s_a = \frac{a}{|a|}$, $s_b = \frac{b}{|b|}$, and $s_c = \frac{c}{|c|}$. These variables can only take the values $1$ (if the variable is positive) or $-1$ (if the variable is negative). The expression becomes $E = s_a + s_b + s_c + (s_a s_b s_c)$.

2. **Analyze the constraint $a + b + c = 0$:**
   Since $a, b, c$ are nonzero and sum to zero, they cannot all have the same sign.
   * Case 1: Two variables are positive and one is negative.
     Suppose $a > 0, b > 0, c < 0$. Then $s_a = 1, s_b = 1, s_c = -1$.
     The product $s_a s_b s_c = (1)(1)(-1) = -1$.
     $E = 1 + 1 - 1 + (-1) = 0$.
   * Case 2: One variable is positive and two are negative.
     Suppose $a > 0, b < 0, c < 0$. Then $s_a = 1, s_b = -1, s_c = -1$.
     The product $s_a s_b s_c = (1)(-1)(-1) = 1$.
     $E = 1 - 1 - 1 + (1) = 0$.

3. **Check if other values are possible:**
   Are there any other combinations of signs?
   * If $s_a=1, s_b=1, s_c=1$, then $a, b, c > 0$, so $a+b+c > 0$. This contradicts $a+b+c=0$.
   * If $s_a=-1, s_b=-1, s_c=-1$, then $a, b, c < 0$, so $a+b+c < 0$. This contradicts $a+b+c=0$.
   Thus, we must have exactly two positive and one negative, or one positive and two negative.

4. **Evaluate the expression for the valid cases:**
   * In Case 1 ($s_a=1, s_b=1, s_c=-1$): $E = 1 + 1 - 1 + (1 \cdot 1 \cdot -1) = 1 - 1 = 0$.
   * In Case 2 ($s_a=1, s_b=-1, s_c=-1$): $E = 1 - 1 - 1 + (1 \cdot -1 \cdot -1) = -1 + 1 = 0$.
   In all possible configurations allowed by $a+b+c=0$, the sum is always 0.

5. **Rule out other choices:**
   * Choice B, C, D, and E suggest values other than 0 (such as 1, -1, 2, or -2). Since our derivation shows the expression is invariant and equal to 0 for all valid $a, b, c$, these choices are incorrect.

ANSWER: A

To find the correct answer, we will break down the problem step-by-step:

1. **Restate what is being asked:** We are given three nonzero real numbers $a$, $b$, and $c$ such that their sum is $0$. We need to find all possible values of the expression given as `a | a | + b | b | + c | c | + abc | abc |`. Based on the multiple-choice options provided, this is a common typographical (OCR) error for the expression $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$. (If it were not a fraction, the value would depend entirely on the unconstrained magnitudes of $a, b,$ and $c$ and could be any real number, which doesn't match the small integer choices).

2. **Set up the math:** For any nonzero real number $x$, the expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$, and $-1$ if $x < 0$. Essentially, this function extracts the "sign" of the number. We need to find the sum of the signs of $a$, $b$, $c$, and their product $abc$.

3. **Analyze the constraints:** We are given that $a + b + c = 0$. Since $a$, $b$, and $c$ are nonzero, they cannot all be positive (otherwise their sum would be strictly greater than $0$) and they cannot all be negative (otherwise their sum would be strictly less than $0$). Therefore, there are only two possible scenarios for the signs of $a$, $b$, and $c$:
   - Two numbers are positive and one is negative.
   - Two numbers are negative and one is positive.

4. **Do the computation for Case 1 (Two positive, one negative):**
   Without loss of generality, assume $a > 0$, $b > 0$, and $c < 0$.
   - The sign values are $\frac{a}{|a|} = 1$, $\frac{b}{|b|} = 1$, and $\frac{c}{|c|} = -1$.
   - The product $abc$ will be $(+) \times (+) \times (-) = (-)$, meaning the product is negative. Thus, $\frac{abc}{|abc|} = -1$.
   - Substituting these into the expression gives: $1 + 1 + (-1) + (-1) = 0$.

5. **Do the computation for Case 2 (Two negative, one positive):**
   Without loss of generality, assume $a < 0$, $b < 0$, and $c > 0$.
   - The sign values are $\frac{a}{|a|} = -1$, $\frac{b}{|b|} = -1$, and $\frac{c}{|c|} = 1$.
   - The product $abc$ will be $(-) \times (-) \times (+) = (+)$, meaning the product is positive. Thus, $\frac{abc}{|abc|} = 1$.
   - Substituting these into the expression gives: $(-1) + (-1) + 1 + 1 = 0$.

6. **Rule out the wrong choices:** In every possible valid scenario under the constraint $a + b + c = 0$, the expression evaluates exactly to $0$. It can never evaluate to $1, -1, 2,$ or $-2$. 
   - Choice B (1 and -1) is incorrect because the sum is always 0.
   - Choice C (2 and -2) is incorrect for the same reason.
   - Choice D (0, 2, and -2) is incorrect because 2 and -2 are impossible under the given constraints.
   - Choice E (0, 1, and -1) is incorrect because 1 and -1 are impossible.
   This leaves only choice A as the correct answer.

ANSWER: A

To find the possible value(s) of the expression, we first clarify the notation. The expression in the problem is a standard competition math problem where the division slashes were omitted:
$$E = \frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$$

We are given that $a, b, c$ are nonzero real numbers and $a + b + c = 0$.

### Step 1: Analyze the signs of $a, b, c$
Since $a, b, c$ are nonzero and their sum is $0$, they cannot all have the same sign (they cannot all be positive, nor can they all be negative). Thus, there are only two possible cases for their signs:
1. **Case 1:** Two of the numbers are positive and one is negative.
2. **Case 2:** Two of the numbers are negative and one is positive.

Let $s(x) = \frac{x}{|x|}$ denote the sign of $x$, which is $+1$ if $x > 0$ and $-1$ if $x < 0$. Note that $s(abc) = s(a)s(b)s(c)$.

---

### Step 2: Evaluate Case 1 (Two positive, one negative)
Without loss of generality, let $a > 0$, $b > 0$, and $c < 0$.
* $s(a) = 1$
* $s(b) = 1$
* $s(c) = -1$

The product $abc$ is the product of two positive numbers and one negative number, so $abc < 0$, which means $s(abc) = -1$.

Substituting these into the expression:
$$E = s(a) + s(b) + s(c) + s(abc) = 1 + 1 + (-1) + (-1) = 0$$

---

### Step 3: Evaluate Case 2 (Two negative, one positive)
Without loss of generality, let $a < 0$, $b < 0$, and $c > 0$.
* $s(a) = -1$
* $s(b) = -1$
* $s(c) = 1$

The product $abc$ is the product of two negative numbers and one positive number, so $abc > 0$, which means $s(abc) = 1$.

Substituting these into the expression:
$$E = s(a) + s(b) + s(c) + s(abc) = (-1) + (-1) + 1 + 1 = 0$$

---

### Step 4: Conclusion
In both possible cases, the value of the expression is $0$. Therefore, the only possible value is $0$.

* **A** is correct because the only possible value is $0$.
* **B, C, D, E** are incorrect because they include other values like $1, -1, 2, -2$ which are impossible under the given conditions.

ANSWER: A

The expression in question is interpreted as the sum of the signs of the variables:
\[
\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}.
\]
(The given notation is presumed to intend division by the absolute values, as this is the only interpretation consistent with the discrete values appearing among the answer choices.)

Since \(a\), \(b\), and \(c\) are nonzero, each ratio equals \(\pm 1\) according to the sign of the corresponding variable. The final ratio equals the product of the three individual signs. The condition \(a + b + c = 0\) forces the three numbers to have mixed signs: either exactly two are positive and one is negative, or exactly two are negative and one is positive.

- If exactly two are positive, the sum of the three signs equals \(+1\) while the product of the signs equals \(-1\), so the full expression equals \(1 + (-1) = 0\).
- If exactly two are negative, the sum of the three signs equals \(-1\) while the product of the signs equals \(+1\), so the full expression equals \(-1 + 1 = 0\).

In every admissible case the expression therefore evaluates to \(0\). No other value is possible.

Thus the only possible value among the listed choices is \(0\).

ANSWER: A

## Step 1: Understand the given condition
We are given that $a + b + c = 0$ and $a, b, c$ are nonzero real numbers. We need to find the possible values of $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$.

## Step 2: Analyze the expression $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|}$
The term $\frac{x}{|x|}$ equals $1$ if $x > 0$ and $-1$ if $x < 0$. So, $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|}$ depends on the signs of $a, b,$ and $c$. Since $a + b + c = 0$, not all three can be positive or all three negative.

## Step 3: Examine the cases based on the signs of $a, b,$ and $c$
Given $a + b + c = 0$, either one of $a, b, c$ is positive and the other two are negative, or one is negative and the other two are positive.

## Step 4: Calculate $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|}$ for the case when one is positive and two are negative
Without loss of generality, let $a > 0, b < 0, c < 0$. Then, $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} = 1 - 1 - 1 = -1$. Since $a > 0, b < 0, c < 0$, $abc > 0$, so $\frac{abc}{|abc|} = 1$. Thus, $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|} = -1 + 1 = 0$.

## Step 5: Calculate $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|}$ for the case when one is negative and two are positive
Let $a < 0, b > 0, c > 0$. Then, $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} = -1 + 1 + 1 = 1$. Since $a < 0, b > 0, c > 0$, $abc < 0$, so $\frac{abc}{|abc|} = -1$. Thus, $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|} = 1 - 1 = 0$.

## Step 6: Consider all possible scenarios
From steps 4 and 5, we see that regardless of the signs of $a, b,$ and $c$, the expression $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$ equals $0$.

## Step 7: Rule out the wrong choices
- Choice B (1 and −1): Not possible because we found the expression to be 0 in both cases.
- Choice C (2 and −2): Not possible as our calculations did not yield these values.
- Choice D (0, 2, and −2): Not possible because 2 and -2 were not obtained.
- Choice E (0, 1, and −1): Not possible because while 0 is correct, 1 and -1 are not the only other possibilities considered, and our results did not support 1 or -1 as valid outcomes.

## Step 8: Confirm the correct choice
Since $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|} = 0$ in both scenarios, the only possible value is 0.

The final answer is: 
ANSWER: A

The expression is interpreted as the sum of signs: \( \frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|} \). For nonzero real numbers, each term is either \(1\) or \(-1\). The condition \(a+b+c=0\) forces the signs to be mixed: either two positives and one negative, or one positive and two negatives. In both cases, the sum of the three signs is \(\pm1\) and the product sign is the opposite, so the total sum is \(0\). No other values are possible.

ANSWER: A

**Step 1: Restate the problem**  
We are given three nonzero real numbers \(a, b, c\) such that \(a + b + c = 0\). We need to find all possible values of the expression  
\[
\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}.
\]
(The original text lost the division slashes, but the choices clearly match this classic sign-function problem.)

**Step 2: Understand the terms**  
For any nonzero real number \(x\), \(\frac{x}{|x|}\) equals \(1\) if \(x > 0\) and \(-1\) if \(x < 0\). So each fraction is simply the sign of the variable. Let  
\[
\text{sgn}(x) = \frac{x}{|x|}.
\]
The expression becomes \(\text{sgn}(a) + \text{sgn}(b) + \text{sgn}(c) + \text{sgn}(abc)\).

**Step 3: Deduce the signs using \(a+b+c=0\)**  
Since the numbers are nonzero, none of them is zero. If all three were positive, their sum would be positive. If all three were negative, their sum would be negative. Because the sum is exactly zero, the numbers must have mixed signs. Only two cases are possible:
- **Case 1:** Two numbers are positive, one is negative.  
  Sum of signs: \(1 + 1 + (-1) = 1\).  
  Product \(abc\): positive × positive × negative = negative, so \(\text{sgn}(abc) = -1\).  
  Total = \(1 + (-1) = 0\).
- **Case 2:** Two numbers are negative, one is positive.  
  Sum of signs: \((-1) + (-1) + 1 = -1\).  
  Product \(abc\): negative × negative × positive = positive, so \(\text{sgn}(abc) = 1\).  
  Total = \(-1 + 1 = 0\).

**Step 4: Evaluate the choices**  
- A: \(0\) — matches our finding.  
- B: \(1\) and \(-1\) — never occurs.  
- C: \(2\) and \(-2\) — never occurs.  
- D: \(0, 2, -2\) — \(2\) and \(-2\) are impossible.  
- E: \(0, 1, -1\) — \(1\) and \(-1\) are impossible.

Thus the only possible value is \(0\).

**ANSWER: A**

itive, one negative, and the third...** Wait, we covered all.

Is there a case where, say, $ a = 1, b = -1, c = 0 $? No, c must be nonzero.

So only two sign configurations.

Hence, only possible value is **0**.

Thus, answer is **A**.

But now I'm conflicted: the problem as written doesn't match this, but the answer choices do.

Maybe there's **another interpretation**: perhaps the expression is:

$$
\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}
$$

But written in the problem as:

> a/|a| + b/|b| + c/|c| + abc/|abc|

But the user wrote: "**a | a |**"—maybe they meant **a divided by |a|**, but used spaces and bars.

Alternatively, maybe it's:

$$
\frac{|a|}{a} + \frac{|b|}{b} + \frac{|c|}{c} + \frac{|abc|}{abc}
$$

But that's same as above.

Given all this, I think the **intended problem** is the sign version.

But wait—let me check official sources.

This problem is actually a known AMC 8 / AMC 10 problem. Let me recall:

There is a classic problem:

> Suppose $ a, b, c $ are nonzero real numbers such that $ a + b + c = 0 $. What are the possible values of  
> $$
> \frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}?
> $$

Answer: **0**

Yes, that’s a known problem.

Therefore, despite the possibly misleading notation in the user’s message, the **correct interpretation** is the **sign function version**, not the quadratic version.

But wait—the user explicitly wrote: "**a | a |**", not "**a / |a|**". Is it possible that in their plain text, the slash didn't render, and it's actually a over |a|?

Maybe they typed "a/|a|" but the slash was lost? Unlikely.

Alternatively, maybe the original problem uses **absolute value notation with a vertical bar**, and they wrote "a|a|" meaning **a over |a|**? No, that doesn't make sense.

But here's another thought: maybe the problem is:

> $ \frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|} $

And the user, not knowing LaTeX, wrote it as: a | a | + b | b | + ..., meaning a divided by |a|, using the vertical bar as a separator? That would be odd.

Alternatively, maybe the problem is from a source where the expression is:

> $ \frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|} $

And when copying, the "/" got lost.

Given that, and that the answer choices only make sense for the **sign version**, I will proceed under that assumption.

But to be thorough, let’s consider if there's **any way** the original expression **as written** could yield only values in the choices.

Suppose we **restrict to a, b, c ∈ {−1, 1}**, but then a + b + c = 0 is impossible, since sum of three ±1 is odd: −3, −1, 1, 3. So can't be 0.

So to satisfy a + b + c = 0 with nonzero reals, magnitudes must differ.

Thus, the expression $ a|a| + b|b| + c|c| + abc|abc| $ is **not bounded** to small integers.

Therefore, the problem **must** be about signs.

Hence, the expression is:

$$
\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}
$$

= sign(a) + sign(b) + sign(c) + sign(abc)

We already analyzed: always 0.

So answer is **A. 0**

But wait—look again at the choices. Option D is “0, 2, and −2”. Could there be a case where the value is 2 or −2?

Suppose someone mistakenly thought that all three could be positive—but that violates a + b + c = 0.

So no.

Thus, only 0.

Final answer: **A**

But hold on! Let me double-check one more time with a different example for the **sign version**.

Example: $ a = 5, b = -3, c = -2 $ → sum = 0

Signs: +1, -1, -1 → sum = -1

Product: 5 * (-3) * (-2) = 30 → sign = +1

Total = -1 + 1 = 0

Another: $ a = -4, b = 1, c = 3 $

Signs: -1, +1, +1 → sum = 1

Product: (-4)(1)(3) = -12 → sign = -1

Total = 1 -1 = 0

Always 0.

Thus, confirmed.

---

### Final Decision

Given the answer choices and standard contest problems, the intended expression is the **sum of signs plus sign of product**, which always equals **0** under the condition $ a + b + c = 0 $ with nonzero reals.

Therefore, the correct answer is:

ANSWER: A

I need to find the number of teachers at King Middle School.

**Step 1: Restate what is being asked**
Find the total number of teachers needed given the constraints about students, classes, and class size.

**Step 2: Set up the relationship using class size**
Each class has:
- 30 students
- 1 teacher

**Step 3: Find the total number of classes needed**
Total number of student-class enrollments = (number of students) × (classes per student)
- Total student enrollments = 1200 students × 5 classes per student = 6000 student-class slots

Since each class holds 30 students:
- Number of classes = 6000 ÷ 30 = 200 classes

**Step 4: Find the number of teachers**
Since each class has 1 teacher:
- Minimum teachers needed = 200 teachers

But wait—each teacher teaches 4 classes, so we need to account for this:
- Number of teachers = (total classes) ÷ (classes per teacher)
- Number of teachers = 200 ÷ 4 = 50 teachers

**Step 5: Verify the answer**
- 50 teachers × 4 classes each = 200 classes ✓
- 200 classes × 30 students each = 6000 student enrollments ✓
- 6000 student enrollments ÷ 1200 students = 5 classes per student ✓

**Step 6: Rule out wrong choices**
- A (30): Too small; 30 teachers × 4 classes = 120 classes, which can only serve 120 × 30 = 3600 students
- B (32): Too small; 32 × 4 = 128 classes → 3840 students max
- C (40): Too small; 40 × 4 = 160 classes → 4800 students max
- D (45): Too small; 45 × 4 = 180 classes → 5400 students max
- E (50): Correct; 50 × 4 = 200 classes → 6000 student enrollments ✓

ANSWER: E

1. We need to find the number of teachers.

2. Total class-enrollments by students: 1200 students × 5 classes = 6000 student-class slots.

3. Each class has 30 students, so number of classes = 6000 / 30 = 200 classes.

4. Each teacher teaches 4 classes, so teachers = 200 / 4 = 50.

5. Ruling out others: 30, 32, 40, 45 all result from miscounting; the correct chain gives 50.

ANSWER: E

I'll work through this step-by-step.

**Step 1: Restate the problem.**
Find the number of teachers at King Middle School given:
- 1200 students total
- Each student takes 5 classes per day
- Each teacher teaches 4 classes per day
- Each class has 30 students and 1 teacher

**Step 2: Find the total number of "student-class slots" per day.**
Each of the 1200 students takes 5 classes, so:
$$1200 \times 5 = 6000 \text{ student-class slots}$$

**Step 3: Find the total number of classes offered per day.**
Each class holds 30 students, so the number of classes is:
$$6000 \div 30 = 200 \text{ classes}$$

**Step 4: Find the number of teachers.**
Each teacher teaches 4 classes per day, so the number of teachers needed is:
$$200 \div 4 = 50 \text{ teachers}$$

**Step 5: Eliminate wrong answers.**
- A (30), B (32), C (40), D (45) are all too small based on the calculation above.
- E (50) matches our answer.

ANSWER: E

1. **Restate the question.**  
   We need to find the total number of teachers at the school.

2. **Count the total number of student-class enrollments.**  
   There are 1200 students, and each student takes 5 classes.  
   So the total number of student spots filled in all classes is
   \[
   1200 \times 5 = 6000.
   \]

3. **Use the class size to find the total number of classes.**  
   Each class has 30 students.  
   So the number of classes is
   \[
   6000 \div 30 = 200.
   \]

4. **Use the teaching load to find the number of teachers.**  
   Each teacher teaches 4 classes.  
   So the number of teachers is
   \[
   200 \div 4 = 50.
   \]

5. **Check against the choices.**  
   The choices are:
   - A. 30  
   - B. 32  
   - C. 40  
   - D. 45  
   - E. 50  

   We found \(50\), which matches **E**.

ANSWER: E

1. We are asked to find the total number of teachers at King Middle School.

2. First, find how many total class enrollments there are:
   - There are 1200 students.
   - Each student takes 5 classes a day.
   - So the total number of student-class seats is  
     \(1200 \times 5 = 6000\).

3. Each class has 30 students, so the number of classes needed is:
   \[
   6000 \div 30 = 200
   \]

4. Each teacher teaches 4 classes, so the number of teachers is:
   \[
   200 \div 4 = 50
   \]

5. Check the choices:
   - A. 30 is too small
   - B. 32 is too small
   - C. 40 is too small
   - D. 45 is too small
   - E. 50 matches our result

ANSWER: E

To find the number of teachers at King Middle School, we follow these steps:

1. **Determine the total number of "student-class spots" filled daily:**
   Each of the 1200 students takes 5 classes per day. Therefore, the total number of student-class enrollments is:
   $1200 \text{ students} \times 5 \text{ classes/student} = 6000 \text{ student-class spots}$.

2. **Determine the total number of classes held per day:**
   We are told that each class has 30 students. To find the total number of classes held throughout the day, we divide the total number of student-class spots by the number of students per class:
   $6000 \text{ student-class spots} \div 30 \text{ students/class} = 200 \text{ classes per day}$.

3. **Determine the number of teachers required:**
   We are told that each teacher teaches 4 classes per day. Since there are 200 classes that need to be taught, and each teacher covers 4 of those classes, we divide the total number of classes by the number of classes per teacher:
   $200 \text{ classes} \div 4 \text{ classes/teacher} = 50 \text{ teachers}$.

4. **Evaluate the choices:**
   - A (30): Incorrect; this would only account for 120 classes.
   - B (32): Incorrect; this would only account for 128 classes.
   - C (40): Incorrect; this would only account for 160 classes.
   - D (45): Incorrect; this would only account for 180 classes.
   - E (50): Correct; this accounts for exactly 200 classes.

By calculating the total demand for class instruction (200 classes) and dividing by the capacity of each teacher (4 classes), we find that 50 teachers are required.

ANSWER: E

To find the number of teachers at King Middle School, we can follow these step-by-step calculations:

**Step 1: Restate what is being asked.**
We need to determine the total number of teachers at the school, given the student population, the number of classes each student takes, the number of classes each teacher teaches, and the number of students in each class. 

**Step 2: Set up the math.**
First, we need to find the total number of "seats" filled in all the classes combined (also known as student-classes). We do this by multiplying the total number of students by the number of classes each student takes. 
Next, we divide this total number of seats by the number of students in each class to find the total number of individual classes offered at the school.
Finally, since each class requires exactly 1 teacher, we divide the total number of classes by the number of classes each teacher teaches to find the total number of teachers.

**Step 3: Do the computation.**
* Calculate total student-classes (seats): 
  1200 students × 5 classes/student = 6000 student-classes

* Calculate the total number of classes: 
  6000 student-classes ÷ 30 students/class = 200 classes

* Calculate the total number of teachers:
  200 classes ÷ 4 classes/teacher = 50 teachers

**Step 4: Rule out the wrong choices.**
We can verify our answer by working backwards for the other choices to see how many students they would accommodate:
* **A. 30:** 30 teachers × 4 classes = 120 classes. 120 classes × 30 students = 3600 seats. 3600 seats ÷ 5 classes/student = 720 students. (Incorrect)
* **B. 32:** 32 teachers × 4 classes = 128 classes. 128 classes × 30 students = 3840 seats. 3840 seats ÷ 5 classes/student = 768 students. (Incorrect)
* **C. 40:** 40 teachers × 4 classes = 160 classes. 160 classes × 30 students = 4800 seats. 4800 seats ÷ 5 classes/student = 960 students. (Incorrect)
* **D. 45:** 45 teachers × 4 classes = 180 classes. 180 classes × 30 students = 5400 seats. 5400 seats ÷ 5 classes/student = 1080 students. (Incorrect)
* **E. 50:** 50 teachers × 4 classes = 200 classes. 200 classes × 30 students = 6000 seats. 6000 seats ÷ 5 classes/student = 1200 students. (Correct)

ANSWER: E

To find the number of teachers at King Middle School, we can break the problem down into the following steps:

**Step 1: Restate what is being asked**
We need to determine the total number of teachers at the school, given the number of students, the number of classes each student takes, the number of classes each teacher teaches, and the size of each class.

**Step 2: Calculate the total number of student-class seats filled per day**
There are $1200$ students, and each student takes $5$ classes a day. 
$$\text{Total student-class seats} = 1200 \text{ students} \times 5 \text{ classes/student} = 6000 \text{ student-class seats per day}$$

**Step 3: Calculate the total number of classes offered per day**
Each class has exactly $30$ students. Since there are $6000$ student-class seats to be filled, we divide the total seats by the number of students per class:
$$\text{Total classes per day} = \frac{6000 \text{ seats}}{30 \text{ students/class}} = 200 \text{ classes per day}$$

**Step 4: Calculate the number of teachers needed**
Each of these $200$ classes has exactly $1$ teacher. Since each teacher teaches $4$ classes a day, we can find the number of teachers ($T$) by dividing the total number of classes by the number of classes taught per teacher:
$$T = \frac{200 \text{ classes}}{4 \text{ classes/teacher}} = 50 \text{ teachers}$$

**Step 5: Rule out the wrong choices**
* **A. 30:** If there were 30 teachers, they could only teach $30 \times 4 = 120$ classes, which is too few.
* **B. 32:** If there were 32 teachers, they could only teach $32 \times 4 = 128$ classes, which is too few.
* **C. 40:** If there were 40 teachers, they could only teach $40 \times 4 = 160$ classes, which is too few.
* **D. 45:** If there were 45 teachers, they could only teach $45 \times 4 = 180$ classes, which is too few.
* **E. 50:** This matches our calculation of 50 teachers.

ANSWER: E

The problem requires finding the total number of teachers at the school, given the following information:
- There are 1200 students.
- Each student takes 5 classes per day.
- Each class consists of exactly 30 students and exactly 1 teacher.
- Each teacher teaches exactly 4 classes (per day).

To solve, first compute the total number of student-class enrollments per day across the entire school:
\[
1200 \times 5 = 6000.
\]
This is the total number of "student-slots" that must be filled each day.

Next, determine how many classes are needed to accommodate these student-slots. Since each class holds 30 students,
\[
\frac{6000}{30} = 200
\]
classes are required each day.

Finally, determine how many teachers are needed to staff these classes. Since each teacher is responsible for 4 classes,
\[
\frac{200}{4} = 50
\]
teachers are required.

This calculation is consistent with all given constraints (student load, class size, and teacher load) and produces an integer result that matches one of the answer choices.

The other choices can be ruled out as follows:
- 40 arises from the incomplete calculation \(1200/30\) (which ignores that students take multiple classes and that teachers teach multiple classes).
- 30, 32, and 45 do not arise from any consistent combination of the given numbers and are therefore extraneous.

ANSWER: E

1. **Restate the problem**: We need to determine the number of teachers at King Middle School given that there are 1200 students, each student takes 5 classes, each teacher teaches 4 classes, and each class has 30 students and 1 teacher.

2. **Set up the math for the total number of classes taken by students**: Since each student takes 5 classes a day and there are 1200 students, the total number of classes taken by all students is $1200 \times 5 = 6000$.

3. **Account for the fact that each class is counted multiple times**: Since each class has 30 students, the total count of 6000 classes includes repetitions. To find the actual number of classes, we need to divide this total by the number of students per class, which is 30. So, the number of classes is $6000 / 30 = 200$.

4. **Verify the number of classes using the information about teachers**: We know each teacher teaches 4 classes. If we let $T$ be the number of teachers, then the total number of classes taught by teachers is $T \times 4$. Since each class has 1 teacher, the number of classes taught equals the number of classes. So, $T \times 4 = 200$ because there are 200 classes.

5. **Solve for the number of teachers**: To find $T$, we divide both sides of the equation $T \times 4 = 200$ by 4. Thus, $T = 200 / 4 = 50$.

6. **Compare the solution to the given choices and rule out incorrect options**:
   - A. 30: Too low because $30 \times 4 = 120$, which is less than the 200 classes.
   - B. 32: Still too low since $32 \times 4 = 128$, less than 200.
   - C. 40: This gives $40 \times 4 = 160$, which is also less than 200.
   - D. 45: This results in $45 \times 4 = 180$, still short of 200.
   - E. 50: This yields $50 \times 4 = 200$, matching our calculation.

7. **Conclusion**: The number of teachers at King Middle School is 50.

ANSWER: E

We need to find the number of teachers.

**Step 1: Understand what each “class” means.**  
Each class consists of 30 students and 1 teacher. Every student takes 5 classes per day.

**Step 2: Find the total number of “student-classes” per day.**  
There are 1200 students.  
Each student takes 5 classes.  
Total student-classes per day = \( 1200 \times 5 = 6000 \).

**Step 3: Relate student-classes to number of classes offered.**  
Each “class” in the schedule is one meeting with 1 teacher and 30 students. So 1 class accounts for 30 student-classes.  
Number of class meetings per day = \( \frac{6000 \text{ student-classes}}{30 \text{ students per class}} = 200 \) classes.

**Step 4: Find the number of teachers.**  
Each teacher teaches 4 classes per day.  
If there are \( T \) teachers, total classes taught = \( 4T \).  
We need this to equal 200 classes:  
\[
4T = 200 \quad\Rightarrow\quad T = 50.
\]

**Step 5: Check the choices.**  
A. 30 → \( 4\times30=120 \) classes, not enough to cover 200.  
B. 32 → 128 classes, still too few.  
C. 40 → 160 classes, still short.  
D. 45 → 180 classes, still short.  
E. 50 → 200 classes, matches.

ANSWER: E

**Step 1: Understand the problem.**  
We need to find the number of teachers at King Middle School given:
- 1200 students
- Each student takes 5 classes per day
- Each teacher teaches 4 classes per day
- Each class contains 30 students and 1 teacher

**Step 2: Find total student-class enrollments (student-periods).**  
Each student takes 5 classes, so total “student-classes” attended per day = \(1200 \times 5 = 6000\).

**Step 3: Find the number of class sections.**  
Each class holds 30 students, so the number of class sections offered daily = total student-classes ÷ class size = \(6000 \div 30 = 200\).

**Step 4: Find the number of teachers.**  
Each teacher teaches 4 classes, so the number of teachers needed = total class sections ÷ classes per teacher = \(200 \div 4 = 50\).

**Step 5: Verify with alternative method.**  
Each teacher handles \(4 \times 30 = 120\) student-classes daily. Total student-classes = 6000, so teachers = \(6000 \div 120 = 50\). The answer is 50.

**Step 6: Check choices and rule out wrong ones.**  
The given choices are A. 30, B. 32, C. 40, D. 45. These are all too small; 50 is not listed among A–D, so the intended complete list (as in the original AMC 8 problem) must have E. 50. Therefore, the correct choice is E.

ANSWER: E

1. **Understand what is being asked**:  
   We are given information about the total number of students, how many classes each student takes, class size, and how many classes each teacher teaches. We are to find the total number of **teachers** at the school.

2. **Find total number of class slots from the students' perspective**:  
   Each of the 1200 students takes 5 classes per day.  
   So total "student-class enrollments" =  
   $$
   1200 \times 5 = 6000
   $$

3. **Use class size to find total number of class sections**:  
   Each class has 30 students.  
   So the total number of distinct class **sections** is:  
   $$
   \frac{6000}{30} = 200
   $$  
   So there are 200 class sections being held (each with 30 students and 1 teacher).

4. **Use teacher workload to find number of teachers**:  
   Each teacher teaches 4 classes (i.e., 4 class sections).  
   So if there are 200 class sections, and each teacher handles 4 of them, the number of teachers is:  
   $$
   \frac{200}{4} = 50
   $$

5. **Check consistency of assumptions**:  
   - Each class has exactly 1 teacher → OK.  
   - Each class has 30 students → matches our division.  
   - Each student in 5 classes → accounted for in total enrollments.  
   - Each teacher teaches 4 classes → used to get number of teachers.  
   All conditions are satisfied.

6. **Review answer choices**:  
   A. 30  
   B. 32  
   C. 40  
   D. 45  
   E. 50 ← matches our result.

ANSWER: E

I need to find when the pattern AJHSME 1989 repeats for the first time.

**Step 1: Understand the cycling pattern**

The letters AJHSME cycle: each step shifts left by one position (with wraparound).
The digits 1989 cycle: each step shifts left by one position (with wraparound).

**Step 2: Find the period of the letter cycle**

Starting: AJHSME
- After 1 cycle: JHSMEA
- After 2 cycles: HSMEAJ
- After 3 cycles: SMEAJH
- After 4 cycles: MEAJHS
- After 5 cycles: EAJHSM
- After 6 cycles: AJHSME (back to start!)

The letters have a period of **6**.

**Step 3: Find the period of the digits cycle**

Starting: 1989
- After 1 cycle: 9891
- After 2 cycles: 8919
- After 3 cycles: 9198
- After 4 cycles: 1989 (back to start!)

The digits have a period of **4**.

**Step 4: Find when both return to AJHSME 1989 simultaneously**

We need the least common multiple (LCM) of 6 and 4.

Finding LCM(6, 4):
- 6 = 2 × 3
- 4 = 2²
- LCM = 2² × 3 = 12

**Step 5: Verify**

After 12 cycles:
- Letters: 12 ÷ 6 = 2 complete cycles → back to AJHSME ✓
- Digits: 12 ÷ 4 = 3 complete cycles → back to 1989 ✓

**Step 6: Determine the line number**

The problem states line 0 is "AJHSME 1989" (the initial state).
Then:
- Line 1: JHSMEA 9891
- Line 2: HSMEAJ 8919
- ...
- Line 12: This would be the 12th cycling operation

However, looking at the numbering convention, if we start counting from line 1 as the first entry:
- Line 1 is after 1 cycle
- Line 12 is after 12 cycles

So AJHSME 1989 appears on line **12** (after 12 cycles from the original).

**Ruling out other choices:**
- A (6): After 6 cycles, letters return to AJHSME but digits are at 1989 ÷ 1.5 = incomplete cycle (would be at position 2 in digit cycle = 8919)
- B (10): Neither 10 ÷ 6 nor 10 ÷ 4 gives whole numbers
- C (12): LCM(6,4) = 12 ✓
- D (18): Too large; 12 is the first repeat
- E (24): This is 2 × LCM, but we want the first repeat

ANSWER: C

1. The letters cycle with period 6 (6 letters), digits cycle with period 4.

2. Both return to original simultaneously at LCM(6,4) = 12.

3. So at line 12 both are back to AJHSME 1989.

4. Check: 12 is divisible by 6 and 4. Yes.

5. Rule out: 6 (digits not back), 10 (neither), 18 (digits not back since 18 not div by 4), 24 (works but not first).

ANSWER: C

I need to find when both the letter cycle and digit cycle return to their original positions simultaneously.

## Step 1: Understand the cycling process

**Letters:** AJHSME cycles by moving the first letter to the end each step.
- The letter cycle has period **6** (6 letters, so after 6 steps we're back to AJHSME)

**Digits:** 1989 cycles by moving the first digit to the end each step.
- The digit cycle has period **4** (4 digits, so after 4 steps we're back to 1989)

## Step 2: Track the letter cycle

Starting position: AJHSME (line 1)
- Line 1: AJHSME
- Line 2: JHSMEA
- Line 3: HSMEAJ
- Line 4: SMEAJH
- Line 5: MEAJHS
- Line 6: EAJHSM
- Line 7: AJHSME ← returns to original!

So letters return to AJHSME at line **1, 7, 13, 19, 25...** (every 6 lines, starting at line 1)

## Step 3: Track the digit cycle

Starting position: 1989 (line 1)
- Line 1: 1989
- Line 2: 9891
- Line 3: 8919
- Line 4: 9198
- Line 5: 1989 ← returns to original!

So digits return to 1989 at line **1, 5, 9, 13, 17, 21, 25...** (every 4 lines, starting at line 1)

## Step 4: Find when both return simultaneously

I need the smallest line number > 1 where both cycles are at their original position.

Letters are at original position at lines: 1, 7, 13, 19, 25, **31**...
Digits are at original position at lines: 1, 5, 9, 13, **17**...

I need a line number of the form:
- 1 + 6k (for some integer k ≥ 1): 7, 13, 19, 25, 31...
- 1 + 4m (for some integer m ≥ 1): 5, 9, 13, 17, 21...

Both equal **13**: 1 + 6(2) = 13 ✓ and 1 + 4(3) = 13 ✓

## Step 5: Verify line 13

- Letters: After 12 cycles from start → 12 ÷ 6 = 2 complete cycles → back to AJHSME ✓
- Digits: After 12 cycles from start → 12 ÷ 4 = 3 complete cycles → back to 1989 ✓

The LCM of 6 and 4 is 12, so after 12 steps (line 13) both return to original.

## Step 6: Rule out wrong answers

- **A (6):** Line 6 has letters EAJHSM (not back yet) — wrong
- **B (10):** Line 10: letters at position 9 steps → 9 mod 6 = 3 → SMEAJH; digits at 9 mod 4 = 1 → 9891 — wrong
- **C (12):** Line 12: letters at 11 steps → 11 mod 6 = 5 → EAJHSM; digits at 11 mod 4 = 3 → 9198 — wrong
- **D (18):** This would work too but 13 comes first — wrong
- **E (24):** Too large — wrong

The answer is line **13**, which corresponds to choice **C**.

ANSWER: C

1. **Restate the problem.**  
We cycle the letters in **AJHSME** and the digits in **1989** separately, moving the first character to the end each time.

   We want to know: **On what numbered line will the original arrangement**
   \[
   \text{AJHSME 1989}
   \]
   **appear again for the first time?**

2. **Find the cycle length of the letters.**  
The word **AJHSME** has 6 letters, all in a fixed order. Rotating it repeatedly returns to the start after 6 shifts.

   So the letters repeat every **6** lines.

3. **Find the cycle length of the digits.**  
The number string is **1989**, which has 4 digits:
   \[
   1989 \to 9891 \to 8919 \to 9198 \to 1989
   \]
   So the digits repeat every **4** lines.

4. **Both must match at the same time.**  
We need the first line where both the letters and digits are back to their starting positions.

   So we need the least common multiple of 6 and 4:
   \[
   6=2\cdot 3,\qquad 4=2^2
   \]
   \[
   \operatorname{lcm}(6,4)=2^2\cdot 3=12
   \]

5. **Check the numbering carefully.**  
The original arrangement is shown before line 1. Then line 1 is after one cycle, line 2 after two cycles, etc.

   Therefore the original arrangement appears again after **12** cycles, which is on **line 12**.

6. **Rule out the choices.**
   - **A. 6**: letters repeat, but digits do not.
   - **B. 10**: neither both repeat together.
   - **C. 12**: yes, both repeat together for the first time.
   - **D. 18**: not the first time, since 12 happens earlier.
   - **E. 24**: also repeats, but not first.

ANSWER: C

1. We are asked to find when the sequence returns to **AJHSME 1989** for the **first time**.

2. The letters are cycled separately:
   - AJHSME
   - JHSMEA
   - HSMEAJ
   - SMEAJH
   - MEAJHS
   - EAJHSM
   - then back to AJHSME

   So the letters have a cycle length of **6**.

3. The digits 1, 9, 8, 9 are cycled separately:
   - 1989
   - 9891
   - 8919
   - 9198
   - then back to 1989

   So the digits have a cycle length of **4**.

4. The whole line repeats only when **both** parts return to their starting positions at the same time.

5. So we need the **least common multiple** of 6 and 4:
   - LCM(6, 4) = 12

6. That means the original line will appear again for the first time on **line 12**.

7. Check the choices:
   - A. 6 — too soon; digits are not back yet
   - B. 10 — not a common multiple of 6 and 4
   - C. 12 — yes
   - D. 18 — later than needed
   - E. 24 — later than needed

ANSWER: C

To find the number of the line on which the sequence "AJHSME 1989" appears for the first time, we must determine the cycle lengths of the two components and find their least common multiple.

**Step 1: Determine the cycle length of the letters.**
The sequence of letters is:
1. AJHSME
2. JHSMEA
3. HSMEAJ
4. SMEAJH
5. MEAJHS
6. EAJHSM
7. AJHSME
The letters are shifted one position to the left each time. Since there are 6 letters, the pattern repeats every 6 lines.

**Step 2: Determine the cycle length of the digits.**
The sequence of digits is:
1. 1989
2. 9891
3. 8919
4. 9198
5. 1989
The digits are shifted one position to the left each time. Since there are 4 digits, the pattern repeats every 4 lines.

**Step 3: Find the period of the combined sequence.**
The combined sequence (Letters + Digits) will repeat when both the letter cycle and the digit cycle return to their starting positions simultaneously. This occurs at the least common multiple (LCM) of the two cycle lengths.
- Cycle length of letters = 6
- Cycle length of digits = 4
- LCM(6, 4) = 12

**Step 4: Verify the cycle.**
- Line 1: AJHSME 1989
- Line 2: JHSMEA 9891
- Line 3: HSMEAJ 8919
- Line 4: SMEAJH 9198
- Line 5: MEAJHS 1989
- Line 6: EAJHSM 9891
- Line 7: AJHSME 8919
- Line 8: JHSMEA 9198
- Line 9: HSMEAJ 1989
- Line 10: SMEAJH 9891
- Line 11: MEAJHS 8919
- Line 12: EAJHSM 9198
- Line 13: AJHSME 1989

The sequence returns to the original state "AJHSME 1989" at line 13. Wait, let's re-examine the question. The question asks for the number of the line on which it appears *for the first time*. It appears at line 1. However, the choices are 6, 10, 12, 18, 24. This implies the question is asking for the *next* time it appears after the start, or there is a misunderstanding of the cycle.

Looking at the provided lines:
1. AJHSME 1989
...
13. AJHSME 1989

If the sequence repeats every 12 lines, then line 13 is the same as line 1. The question asks for the number of the line on which it appears *for the first time* (which is 1), but since 1 is not an option, we look for the next occurrence in the cycle, which is 13. Since 13 is not an option, let's re-check the digit cycle:
1: 1989, 2: 9891, 3: 8919, 4: 9198, 5: 1989. The digit cycle is 4.
The letter cycle is 6.
LCM(6, 4) = 12.
The sequence repeats every 12 lines. The sequence at line 13 is the same as line 1. The sequence at line 12 is the end of the first full cycle. The question likely implies the first time *after* the initial state, or there is a typo in the provided options. Given the options, 12 is the period of the cycle.

ANSWER: C

To determine the line number on which "AJHSME 1989" will appear for the first time, we need to analyze the cycling patterns of the letters and the digits separately. 

**Step 1: Understand the cycling process**
In each new line, the first character of the letter group is moved to the end of the letter group, and the first digit of the number group is moved to the end of the number group. We are looking for the smallest line number where both groups return to their original, unnumbered starting state ("AJHSME" and "1989").

**Step 2: Determine the cycle length for the letters**
The letter group "AJHSME" consists of 6 distinct letters. 
If we shift the first letter to the end, it will take exactly 6 shifts for the letters to return to their original order. Therefore, the letters will appear as "AJHSME" on lines that are multiples of 6 (Line 6, 12, 18, 24, etc.).

**Step 3: Determine the cycle length for the digits**
The digit group "1989" consists of 4 digits. Let's look at its shifts to ensure it doesn't repeat earlier due to the repeating '9's:
- Shift 1: 9891
- Shift 2: 8919
- Shift 3: 9198
- Shift 4: 1989
None of the intermediate shifts match the original number, so the digits return to their original order exactly every 4 shifts. Therefore, the digits will appear as "1989" on lines that are multiples of 4 (Line 4, 8, 12, 16, 20, 24, etc.).

**Step 4: Find the Least Common Multiple (LCM)**
For both the letters and the digits to match their original state at the same time, the line number must be a multiple of both 6 and 4. To find the *first* time this happens, we find the Least Common Multiple (LCM) of 6 and 4.
- Multiples of 6: 6, **12**, 18, 24, 30...
- Multiples of 4: 4, 8, **12**, 16, 20, 24...
The LCM of 6 and 4 is 12. 

**Step 5: Rule out the wrong choices**
- **A. 6:** At line 6, the letters have completed one full cycle and match ("AJHSME"), but the digits have completed 1.5 cycles and display "8919".
- **B. 10:** At line 10, neither matches. The letters are "MEAJHS" and the digits are "8919".
- **D. 18:** At line 18, the letters match (3 full cycles), but the digits are at "8919" again (4.5 cycles).
- **E. 24:** At line 24, both the letters and digits match! However, the question asks for the *first* time they appear together, which already happened at line 12. 

ANSWER: C

An elegant way to solve this problem is to analyze the cycling behavior of the letters and digits separately.

### Step 1: Restate what is being asked
We are asked to find the line number in a numbered list where the combination "AJHSME 1989" appears for the first time. The list starts cycling from the initial state "AJHSME 1989", with the first line of the list (Line 1) being "JHSMEA 9891".

### Step 2: Analyze the letter cycle
The word **AJHSME** consists of $6$ letters. 
- In each step, the first letter is moved to the end of the word.
- The letters will return to their original order "AJHSME" after any number of steps that is a multiple of $6$.
- Thus, the letters are "AJHSME" on lines $n$ where $n \equiv 0 \pmod 6$.

### Step 3: Analyze the digit cycle
The number **1989** consists of $4$ digits.
- In each step, the first digit is moved to the end of the number.
- The digits will return to their original order "1989" after any number of steps that is a multiple of $4$.
- Thus, the digits are "1989" on lines $n$ where $n \equiv 0 \pmod 4$.

### Step 4: Find the first common line
For both the letters and digits to return to their original state at the same time, the line number $n$ must be a positive multiple of both $6$ and $4$. 

The first time this happens is at the least common multiple (LCM) of $6$ and $4$:
$$\text{LCM}(6, 4) = 12$$

### Step 5: Verify the line and rule out wrong choices
Let's check the state at line $12$:
- Since $12$ is a multiple of $6$ ($12 = 6 \times 2$), the letters are in their original state: **AJHSME**.
- Since $12$ is a multiple of $4$ ($12 = 4 \times 3$), the digits are in their original state: **1989**.
- Therefore, line 12 is indeed **AJHSME 1989**.

Let's rule out the other choices:
- **A. 6**: At line 6, the letters are "AJHSME", but the digits are "8919" (since $6 \equiv 2 \pmod 4$).
- **B. 10**: At line 10, the letters are "MEAJHS" ($10 \equiv 4 \pmod 6$) and the digits are "8919" ($10 \equiv 2 \pmod 4$).
- **D. 18**: At line 18, the letters are "AJHSME" ($18 \equiv 0 \pmod 6$), but the digits are "8919" ($18 \equiv 2 \pmod 4$).
- **E. 24**: This is a multiple of both 6 and 4, so "AJHSME 1989" will appear here, but it is the *second* time it appears, not the first.

The correct choice is **C**.

ANSWER: C

The strings of letters and digits are rotated independently (left rotation by one position at each step). The goal is to find the smallest positive integer \(k\) such that both strings simultaneously return to their initial configurations after exactly \(k\) steps.

The initial letter string is AJHSME (length 6, all characters distinct). Its successive rotations are:
- AJHSME
- JHSMEA
- HSMEAJ
- SMEAJH
- MEAJHS
- EAJHSM
- AJHSME (and then the cycle repeats).

Thus, the letter string returns to its initial form if and only if the number of rotations is a multiple of 6. In other words, the period of the letter string is exactly 6.

The initial digit string is 1989 (length 4). Its successive rotations are:
- 1989
- 9891
- 8919
- 9198
- 1989 (and then the cycle repeats).

Thus, the digit string returns to its initial form if and only if the number of rotations is a multiple of 4. In other words, the period of the digit string is exactly 4.

The desired \(k\) must therefore be a common multiple of 6 and 4. The smallest such positive integer is the least common multiple:
\[
\operatorname{lcm}(4,6)=12.
\]
Verification at multiples of 12 confirms both strings return simultaneously, while no smaller positive common multiple exists (e.g., at step 6 the digits are in configuration 8919, not 1989).

Hence the first reappearance occurs on line 12.

ANSWER: C

1. The problem requires finding the line number in a list where the sequence "AJHSME 1989" first appears. The list is generated by cycling the letters "A, J, H, S, M, E" and the digits "1, 9, 8, 9" separately and then combining them.

2. To solve this, we first need to understand the cycling pattern for both the letters and the digits. The letters cycle every 6 steps because there are 6 different letters, and the digits cycle every 4 steps because there are 4 digits.

3. The sequence "AJHSME 1989" will reappear when both the letter cycle and the digit cycle return to their original positions. This means we need to find the least common multiple (LCM) of 6 (the number of letters) and 4 (the number of digits), as this will give us the number of steps required for both cycles to realign.

4. To find the LCM of 6 and 4, we first list the multiples of each: Multiples of 6 are 6, 12, 18, 24, ... and multiples of 4 are 4, 8, 12, 16, 20, 24, ...

5. The smallest number that appears in both lists is 12, so the LCM of 6 and 4 is 12. This means that "AJHSME 1989" will reappear every 12 lines.

6. Since "AJHSME 1989" is on line 1, it will next appear on line 1 + 12 = 13. However, the question asks for the number of the line on which "AJHSME 1989" will appear for the first time. It is already on line 1, so we are actually looking for the next occurrence, but the choices and the context imply we are looking for the first occurrence after the initial setup, or simply verifying if it matches the given options directly.

7. Let's directly check the given choices to see if any match our calculation or reasoning:
- A. 6: This is not the LCM of 6 and 4.
- B. 10: This is not the LCM of 6 and 4.
- C. 12: This is the LCM of 6 and 4, so it's a potential answer because it represents the cycle length.
- D. 18: This is not the LCM of 6 and 4.
- E. 24: This is a common multiple of 6 and 4 but not the least.

8. Since "AJHSME 1989" first appears on line 1 and the LCM of 6 and 4 is 12, the next time it appears is on line 13. However, our task is to identify the correct answer from the given choices, and it seems we need to directly relate our findings to the options.

9. Upon closer inspection, we realize that the actual task is to understand that the first time it appears is indeed on line 1, and the next appearance is after 12 steps, making it line 13. However, our choices don't directly include 13, so we need to think about what's being asked: the number of the line for the first appearance. The first appearance is line 1, and none of the choices directly offer "1". The closest interpretation related to our LCM calculation is to verify if any of the given choices directly relate to our findings or if we misunderstood the question's requirement for "first time."

10. Re-evaluating the question and our steps, we realize the question might be testing understanding of the cycle. The correct interpretation should be about understanding the cycle and its implications.

Given that "AJHSME 1989" appears on line 1 and reappears every 12 lines, the correct line number for its first reappearance after line 1 is 13, but that's not an option. The actual task seems to be verifying the understanding of the cycle.

ANSWER: C

We are told that the string "AJHSME" is cycled (rotated left by one letter each step) and the digits "1989" are also cycled (rotated left by one digit each step). They start together on line 0 (or line 1 as given).  
We need the smallest positive line number where both sequences return to their original order simultaneously.

**Step 1: Determine cycle length for letters**  
The word "AJHSME" has 6 distinct letters. Rotating left by one each time, it will return to "AJHSME" after 6 steps.

**Step 2: Determine cycle length for digits**  
The digits are "1989". That’s 4 digits, but note the digit 9 appears twice.  
Let’s check the cycle:  
1989 → 9891 → 8919 → 9198 → 1989.  
So it takes 4 steps to return to "1989".

**Step 3: Find when both return to start together**  
We need the smallest positive integer \( n \) such that  
\( n \) is a multiple of 6 (for letters) and also a multiple of 4 (for digits).  
That is the least common multiple:  
LCM(6, 4) = 12.

**Step 4: Interpret line numbering**  
The given list starts:  
1. JHSMEA 9891  
2. HSMEAJ 8919  
3. SMEAJH 9198  
...  
So line 1 is after 1 rotation of each.  
Line 0 would be the original "AJHSME 1989".  
The first time it reappears is after 12 rotations, which would be line 12.

**Step 5: Check choices**  
A. 6 → letters would be back, but digits not yet (only 4,8,12…).  
B. 10 → not multiple of 4 or 6.  
C. 12 → correct.  
D. 18 → multiple of 6 but not 4.  
E. 24 → also works but not the first time.

ANSWER: C

1. The problem describes two independent cycles: one for the letters A, J, H, S, M, E (6 distinct letters) and one for the digits 1, 9, 8, 9 (4 digits, with repetition). Each step shifts the letters one position to the left (first letter moves to the end) and shifts the digits one position to the left (first digit moves to the end).

2. The initial string is "AJHSME 1989", which is shown without a line number. The list then numbers the subsequent shifts as line 1, line 2, line 3, etc. The question asks for the number of the line on which "AJHSME 1989" will appear for the *first time* – this means the first time it reappears after the initial unnumbered step, i.e., the smallest positive line number where the original combination returns.

3. The letters cycle with period 6 because there are 6 distinct letters:
   0: AJHSME
   1: JHSMEA
   2: HSMEAJ
   3: SMEAJH
   4: MEAJHS
   5: EAJHSM
   6: AJHSME  (repeats)

4. The digits cycle with period 4:
   0: 1989
   1: 9891
   2: 8919
   3: 9198
   4: 1989  (repeats)

5. For the whole string "AJHSME 1989" to appear, both the letters must be back to the starting arrangement and the digits must be back to 1989 simultaneously. This happens exactly when the step number is a multiple of both 6 and 4.

6. The least common multiple of 6 and 4 is 12. Therefore, after 12 shifts, the string reappears. Since the numbered lines start at shift 1, line 12 is the first numbered line showing "AJHSME 1989".

7. Checking the choices: 6 is a multiple of 6 but not 4 (digits would be 8919); 10 is not a multiple of 6 or 4; 18 is a multiple of 6 but not 4; 24 is a multiple but not the first; only 12 is the least common multiple.

ANSWER: C

 end**.  
   For example:  
   - Step 0: AJHSME 1989  
   - Step 1: JHSMEA 9891  
   - Step 2: HSMEAJ 8919  
   - etc.

   We are to find the **first step number** (starting count at 1 for the first rearrangement, but note the original is at position "0" or line "0"? Wait — actually, looking at the problem:

   > AJHSME 1989 → line **?**  
   > JHSMEA 9891 → **1**  
   > HSMEAJ 8919 → **2**  
   > ...

   So **the original AJHSME 1989 is NOT labeled as line 1** — it appears **before** line 1. But the question asks:  
   > "What is the number of the line on which AJHSME 1989 will appear **for the first time**?"

   Wait — that seems contradictory. It appears at the **start**, so why isn't the answer 0 or not listed?

   Let's read carefully:  
   The problem says:  
   > "The letters A, J, H, S, M, E and the digits 1, 9, 8, 9 are 'cycled' separately as follows and put together in a numbered list:  
   > AJHSME 1989  
   > 1. JHSMEA 9891  
   > 2. HSMEAJ 8919  
   > 3. SMEAJH 9198  
   > ........."

   Wait! Look again: The **original** "AJHSME 1989" is shown **above** the numbered list. Then the **first line of the numbered list is labeled "1."**

   But the question says:  
   > "What is the number of the line on which AJHSME 1989 will appear **for the first time**?"

   Since it's **not** in the numbered list at the start, we are to find **when it reappears** in the **numbered list**. That is, after how many cyclic shifts do both parts **simultaneously** return to their original form, and what is that line number?

   So:  
   - Line 1 = after 1 cycle  
   - Line 2 = after 2 cycles  
   - ...  
   - Line n = after n cycles  

   We are to find the **smallest positive integer n** such that after **n cycles**, the letter part is back to **AJHSME** and the digit part is back to **1989**. Then the answer is **n**, which is the line number.

2. **Model the cycling**:  
   - The 6-letter string cycles every **6** shifts (since all letters are distinct? Wait — are they? A, J, H, S, M, E — yes, all distinct). So the **letter part** has period **6**.
   - The 4-digit string is **1989**. Digits: 1, 9, 8, 9 — note that **9 appears twice**, so we must be careful: does that affect the period?

   Let's examine the digit cycling:

   Original: 1 9 8 9  
   Cycle 1: 9 8 9 1  
   Cycle 2: 8 9 1 9  
   Cycle 3: 9 1 9 8  
   Cycle 4: 1 9 8 9 ← back to original

   So even though there are repeated digits, after **4** cycles, it returns to original. Could it return earlier?

   Check:
   - After 1 cycle: 9891 ≠ 1989  
   - After 2 cycles: 8919 ≠ 1989  
   - After 3 cycles: 9198 ≠ 1989  
   - After 4 cycles: 1989 = original

   So **period is 4**.

   In general, for a string of length $ k $, cycling (rotating left by 1 each time), the period is **the smallest positive $ p $** such that rotating $ p $ times gives original. This always divides $ k $, and if all rotations are distinct, period = $ k $. Even with repeated symbols, the period **divides** $ k $, and in this case, it is **4**.

   So:
   - Letter cycle period = 6  
   - Digit cycle period = 4  

3. **Find when both align**:  
   We need the smallest $ n > 0 $ such that:
   - $ n \equiv 0 \pmod{6} $ (letters back to start)  
   - $ n \equiv 0 \pmod{4} $ (digits back to start)

   So $ n = \text{LCM}(6, 4) $

   Compute LCM:
   - Prime factors:  
     - 6 = 2 × 3  
     - 4 = 2²  
   - LCM = 2² × 3 = **12**

4. **Conclusion**:  
   After **12** cycles, both parts return to original. Since line **n** corresponds to **n cycles**, the original string **AJHSME 1989** will appear again on **line 12**.

5. **Check answer choices**:  
   - A. 6 → too small (digits not back: 6 mod 4 = 2 → digit string would be at cycle 2 → 8919 ≠ 1989)  
   - B. 10 → 10 mod 6 = 4 → letters not back; 10 mod 4 = 2 → digits not back  
   - C. 12 → LCM → both back → ✅  
   - D. 18 → 18 mod 4 = 2 → digits not back  
   - E. 24 → works, but **not first time** (12 is smaller)

   So **12 is the first time**.

ANSWER: C

I need to find the probability that the sum of two randomly selected balls is even.

**Step 1: Understand when a sum is even**

A sum of two numbers is even when:
- Both numbers are even, OR
- Both numbers are odd

**Step 2: Identify the even and odd numbered balls**

Balls numbered 1 to 10:
- Odd: 1, 3, 5, 7, 9 (5 balls)
- Even: 2, 4, 6, 8, 10 (5 balls)

**Step 3: Calculate total possible outcomes**

Jack picks one ball from 10, then Jill picks one ball from the remaining 9.
- Total outcomes = 10 × 9 = 90

**Step 4: Count favorable outcomes (sum is even)**

**Case 1: Both balls are odd**
- Jack picks an odd ball: 5 choices
- Jill picks an odd ball from remaining: 4 choices
- Outcomes: 5 × 4 = 20

**Case 2: Both balls are even**
- Jack picks an even ball: 5 choices
- Jill picks an even ball from remaining: 4 choices
- Outcomes: 5 × 4 = 20

**Total favorable outcomes = 20 + 20 = 40**

**Step 5: Calculate probability**

$$P(\text{sum is even}) = \frac{40}{90} = \frac{4}{9}$$

**Step 6: Check against the answer choices**

- A. 4⁄9 ✓ This matches our answer
- B. 9⁄19 ≠ 4⁄9
- C. 1⁄2 ≠ 4⁄9
- D. 10⁄19 ≠ 4⁄9
- E. 5⁄9 ≠ 4⁄9

ANSWER: A

1. We need probability that sum of two distinct balls (from 1-10) is even.

2. Sum is even when both even or both odd. There are 5 even, 5 odd.

3. Total ways to pick 2 distinct: C(10,2) = 45.

4. Both even: C(5,2) = 10. Both odd: C(5,2) = 10. Total favorable = 20.

5. Probability = 20/45 = 4/9.

6. Check choices: A is 4/9. Others don't match.

ANSWER: A

## Step-by-Step Solution

**Step 1: Understand what's being asked.**
I need to find the probability that when two balls are drawn (without replacement) from balls numbered 1–10, the sum of the two numbers is even.

**Step 2: Determine when a sum is even.**
The sum of two integers is even when:
- Both numbers are even, OR
- Both numbers are odd

**Step 3: Identify even and odd balls.**
- Even balls: 2, 4, 6, 8, 10 → **5 even balls**
- Odd balls: 1, 3, 5, 7, 9 → **5 odd balls**

**Step 4: Count favorable outcomes.**
I'll count ordered pairs (Jack's pick, Jill's pick) since the draws are sequential.

**Total ordered pairs:** 10 × 9 = 90

**Both even:** Jack picks even (5 ways) × Jill picks even from remaining 4 (4 ways) = 5 × 4 = **20**

**Both odd:** Jack picks odd (5 ways) × Jill picks odd from remaining 4 (4 ways) = 5 × 4 = **20**

**Favorable outcomes:** 20 + 20 = **40**

**Step 5: Calculate the probability.**
$$P(\text{sum is even}) = \frac{40}{90} = \frac{4}{9}$$

**Step 6: Rule out wrong answers.**
- **(A) 4/9** ✓ — matches our calculation
- **(B) 9/19** — This would arise if we mistakenly computed C(10,2) = 45 as denominator... let's check: C(5,2) + C(5,2) = 10 + 10 = 20 favorable combinations out of C(10,2) = 45 total → 20/45 = **4/9**, same answer. So 9/19 is wrong.
- **(C) 1/2** — Would require equal likelihood, but the discrete counting gives 4/9 ≠ 1/2.
- **(D) 10/19** — Incorrect count.
- **(E) 5/9** — Would correspond to an odd sum (mixed parity), not even.

ANSWER: A

1. **Restate the question.**  
We have balls numbered \(1\) through \(10\). Jack removes one ball, then Jill removes a different ball. We want the probability that the **sum** of the two numbers is **even**.

2. **Figure out when a sum is even.**  
A sum is even in exactly these cases:
- even + even = even
- odd + odd = even

So the two chosen balls must be **both even** or **both odd**.

3. **Count how many odd and even balls there are.**  
From \(1\) to \(10\):
- Odd numbers: \(1,3,5,7,9\) → \(5\) odd balls
- Even numbers: \(2,4,6,8,10\) → \(5\) even balls

4. **Count total ways to choose 2 different balls.**  
There are \(10\) balls total, and we choose \(2\):
\[
\binom{10}{2}=45
\]

5. **Count favorable ways.**  
- Choose 2 odd balls from 5:
\[
\binom{5}{2}=10
\]
- Choose 2 even balls from 5:
\[
\binom{5}{2}=10
\]

So total favorable ways:
\[
10+10=20
\]

6. **Compute the probability.**  
\[
\frac{20}{45}=\frac{4}{9}
\]

7. **Check against the choices.**  
\[
\frac{4}{9}
\]
matches **Choice A**.

8. **Rule out the other choices.**  
- **B. \(9/19\)**: wrong denominator; probabilities here naturally come from 45 total pairs or equivalent ordered counting.
- **C. \(1/2\)**: close, but not correct; actually there are slightly fewer mixed-parity same-sum cases than half? More precisely, favorable is \(20/45=4/9\), not \(1/2\).
- **D. \(10/19\)**: again wrong denominator structure.
- **E. \(5/9\)**: too large; favorable count would have to be \(25/45\), but actual favorable count is only \(20/45\).

ANSWER: A

1. We are choosing **2 different balls** from the 10 balls numbered 1 through 10.
   - We want the probability that the **sum** of the two numbers is **even**.

2. A sum is even when:
   - both numbers are **even**, or
   - both numbers are **odd**.

3. Count the balls:
   - Odd numbers: 1, 3, 5, 7, 9 → **5 odd balls**
   - Even numbers: 2, 4, 6, 8, 10 → **5 even balls**

4. Find favorable outcomes.

   **Case 1: both odd**
   \[
   \binom{5}{2} = 10
   \]

   **Case 2: both even**
   \[
   \binom{5}{2} = 10
   \]

   So total favorable outcomes:
   \[
   10+10=20
   \]

5. Find total possible ways to choose 2 different balls from 10:
   \[
   \binom{10}{2}=45
   \]

6. Therefore, the probability is
   \[
   \frac{20}{45}=\frac{4}{9}
   \]

7. Check the choices:
   - A. \( \frac{4}{9} \) ✅
   - B. \( \frac{9}{19} \)
   - C. \( \frac{1}{2} \)
   - D. \( \frac{10}{19} \)
   - E. \( \frac{5}{9} \)

The correct choice is A.

ANSWER: A

To find the probability that the sum of two balls numbered 1 to 10 is even, we follow these steps:

1. **Identify the set of numbers:** The jar contains balls numbered $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. There are 5 odd numbers $\{1, 3, 5, 7, 9\}$ and 5 even numbers $\{2, 4, 6, 8, 10\}$.

2. **Determine the total number of ways to pick two balls:** Jack picks one ball (10 choices), and then Jill picks a different ball (9 choices). The total number of ordered outcomes is $10 \times 9 = 90$.

3. **Determine the condition for an even sum:** A sum of two numbers is even if both numbers are odd or both numbers are even.
   - **Case 1: Both are odd.** Jack picks one of the 5 odd balls, and Jill picks one of the remaining 4 odd balls. The number of ways is $5 \times 4 = 20$.
   - **Case 2: Both are even.** Jack picks one of the 5 even balls, and Jill picks one of the remaining 4 even balls. The number of ways is $5 \times 4 = 20$.

4. **Calculate the total favorable outcomes:** The total number of favorable outcomes is $20 + 20 = 40$.

5. **Calculate the probability:** The probability is the ratio of favorable outcomes to total outcomes:
   $$P = \frac{40}{90} = \frac{4}{9}$$

6. **Evaluate the choices:**
   - A: $4/9$ matches our calculation.
   - B: $9/19$ is incorrect (this would arise from a different sample space).
   - C: $1/2$ is incorrect (this would be the probability if replacement were allowed, or if we were picking from a set where odd/even counts were equal and we were looking at parity).
   - D: $10/19$ is incorrect.
   - E: $5/9$ is incorrect (this would be the probability of an odd sum).

Since the calculation yields $4/9$, we conclude that choice A is the correct answer.

ANSWER: A

To find the probability that the sum of the two numbers on the balls removed is even, we can follow these step-by-step calculations:

**Step 1: Restate what is being asked**
We have a jar with 10 balls numbered 1 through 10. Two balls are drawn without replacement (one by Jack, then a different one by Jill). We need to find the probability that the sum of the numbers on these two balls is an even number. 

**Step 2: Set up the math for an even sum**
The sum of two integers is even if and only if they share the same parity. This means we have two successful scenarios:
- Both balls drawn are odd.
- Both balls drawn are even.

**Step 3: Count the odds and evens in the jar**
In the set of numbers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, there are:
- 5 odd numbers (1, 3, 5, 7, 9)
- 5 even numbers (2, 4, 6, 8, 10)

**Step 4: Calculate the probabilities**
We can solve this using conditional probability based on Jack's first draw:
- **Scenario 1 (Both Odd):** The probability that Jack draws an odd number is 5/10 (or 1/2). Because he keeps the ball, there are now 9 balls left in the jar, 4 of which are odd. The probability that Jill then draws an odd number is 4/9. 
  Probability of Both Odd = (1/2) × (4/9) = 4/18 = 2/9.
  
- **Scenario 2 (Both Even):** The probability that Jack draws an even number is 5/10 (or 1/2). There are now 9 balls left, 4 of which are even. The probability that Jill then draws an even number is 4/9.
  Probability of Both Even = (1/2) × (4/9) = 4/18 = 2/9.

**Step 5: Add the probabilities of the successful scenarios**
Total Probability = (Probability of Both Odd) + (Probability of Both Even)
Total Probability = 2/9 + 2/9 = 4/9.

*(Alternatively, using combinations: There are "10 Choose 2" = 45 total ways to draw two balls. There are "5 Choose 2" = 10 ways to draw two odds, and "5 Choose 2" = 10 ways to draw two evens. 20 successful ways / 45 total ways = 4/9).*

**Step 6: Rule out the wrong choices**
- **B (9/19)** and **D (10/19):** These fractions have a denominator of 19, which does not mathematically align with our total possible outcomes of 90 permutations or 45 combinations.
- **C (1/2):** This would be the correct answer if the balls were drawn *with replacement* (where the second draw would still have a 5/10 chance of matching the parity of the first). Because the first ball is removed, the odds change. 
- **E (5/9):** This is the probability that the sum is *odd*. An odd sum requires one even and one odd ball (5 odds × 5 evens = 25 combinations out of 45 total combinations; 25/45 = 5/9). 

ANSWER: A

To find the probability that the sum of the two numbers on the balls removed is even, we can follow these steps:

**Step 1: Restate what is being asked**
We need to find the probability that the sum of two different numbers randomly selected from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ is even.

**Step 2: Identify the odd and even numbers in the set**
In the set of numbers from 1 to 10, there are:
*   **5 odd numbers:** $\{1, 3, 5, 7, 9\}$
*   **5 even numbers:** $\{2, 4, 6, 8, 10\}$

**Step 3: Determine the condition for the sum to be even**
The sum of two integers is even if and only if:
1.  Both numbers are odd (Odd + Odd = Even), OR
2.  Both numbers are even (Even + Even = Even).

If one number is odd and the other is even, their sum is odd (Odd + Even = Odd).

**Step 4: Calculate the total number of ways to choose 2 balls**
The total number of ways to choose 2 balls from a pool of 10 without replacement is given by the combination formula $\binom{n}{k}$:
$$\text{Total Outcomes} = \binom{10}{2} = \frac{10 \times 9}{2} = 45$$

**Step 5: Calculate the number of successful outcomes**
*   **Case 1: Both balls are odd**
    We choose 2 balls from the 5 odd balls:
    $$\text{Ways to choose 2 odds} = \binom{5}{2} = \frac{5 \times 4}{2} = 10$$

*   **Case 2: Both balls are even**
    We choose 2 balls from the 5 even balls:
    $$\text{Ways to choose 2 evens} = \binom{5}{2} = \frac{5 \times 4}{2} = 10$$

*   **Total Successful Outcomes:**
    $$\text{Successful Outcomes} = 10 + 10 = 20$$

**Step 6: Compute the probability**
The probability is the ratio of successful outcomes to total outcomes:
$$\text{Probability} = \frac{\text{Successful Outcomes}}{\text{Total Outcomes}} = \frac{20}{45} = \frac{4}{9}$$

**Step 7: Compare with the choices**
*   **A. 4⁄9** — This matches our calculated probability of $\frac{4}{9}$.
*   **B. 9⁄19** — Incorrect.
*   **C. 1⁄2** — Incorrect (this would be the probability if the first ball was replaced before drawing the second).
*   **D. 10⁄19** — Incorrect.
*   **E. 5⁄9** — Incorrect (this is the probability that the sum is odd).

ANSWER: A

The sum of two numbers is even if and only if both numbers are even or both numbers are odd. Among the integers from 1 to 10 there are exactly five odd numbers (1, 3, 5, 7, 9) and five even numbers (2, 4, 6, 8, 10).

Because the selections are made sequentially without replacement, the sample space consists of all ordered pairs \((J, K)\) where \(J\) is Jack’s choice and \(K\) is Jill’s choice. There are 10 possibilities for \(J\) and then 9 remaining possibilities for \(K\), giving a total of \(10 \times 9 = 90\) equally likely ordered pairs.

The favorable outcomes are the ordered pairs in which both numbers are odd or both numbers are even:
- Both odd: 5 choices for \(J\) and then 4 remaining choices for \(K\), giving \(5 \times 4 = 20\) pairs.
- Both even: likewise \(5 \times 4 = 20\) pairs.

Thus there are \(20 + 20 = 40\) favorable ordered pairs. The desired probability is therefore
\[
\frac{40}{90} = \frac{4}{9}.
\]

(The same fraction is obtained if unordered pairs are used instead: there are \(\binom{10}{2} = 45\) possible pairs and \(\binom{5}{2} + \binom{5}{2} = 20\) favorable pairs, so \(20/45 = 4/9\).)

The result can also be derived conditionally. The probability that the first ball is odd is \(1/2\); given that it is odd, the probability that the second is also odd is \(4/9\). The contribution from the “both odd” case is therefore \((1/2) \times (4/9) = 2/9\). The “both even” case contributes the same amount, and the total is again \(4/9\).

Hence the probability is exactly \(4/9\), which corresponds to choice A. All other listed fractions arise only from counting errors (for example, treating selections as dependent in an inconsistent way or miscounting the number of odd/even balls remaining).

ANSWER: A

1. **Restate the problem**: We are asked to find the probability that the sum of the numbers on two balls randomly drawn by Jack and Jill from a jar containing balls numbered 1 to 10 is even.

2. **Understand the condition for an even sum**: The sum of two numbers is even if both numbers are either both odd or both even.

3. **Count the odd and even balls**: There are 5 odd numbered balls (1, 3, 5, 7, 9) and 5 even numbered balls (2, 4, 6, 8, 10) in the jar.

4. **Calculate the probability of both balls being odd**: 
   - The probability that Jack picks an odd ball is 5/10.
   - After Jack picks an odd ball, there are 4 odd balls left out of 9 balls. So, the probability that Jill picks an odd ball is 4/9.
   - The probability that both pick an odd ball is (5/10) * (4/9) = 20/90.

5. **Calculate the probability of both balls being even**: 
   - The probability that Jack picks an even ball is 5/10.
   - After Jack picks an even ball, there are 4 even balls left out of 9 balls. So, the probability that Jill picks an even ball is 4/9.
   - The probability that both pick an even ball is (5/10) * (4/9) = 20/90.

6. **Calculate the total probability of having an even sum**: 
   - This is the sum of the probabilities of both being odd and both being even: 20/90 + 20/90 = 40/90 = 4/9.

7. **Compare with the given choices**:
   - Our calculation gives us 4/9.
   - Checking the choices: A. 4⁄9, B. 9⁄19, C. 1⁄2, D. 10⁄19, E. 5⁄9, we see that our result matches A.

8. **Ruling out other choices**:
   - B (9/19) is approximately 0.4737, which is not equal to 4/9.
   - C (1/2) is 0.5, which is not equal to 4/9.
   - D (10/19) is approximately 0.5263, which is not equal to 4/9.
   - E (5/9) is approximately 0.5556, which is not equal to 4/9.

9. **Conclusion**: The probability that the sum of the two numbers on the balls removed is even is 4/9.

ANSWER: A

The probability that the sum of the two numbers is even equals the probability that both numbers have the same parity (both even or both odd). There are 5 even numbers (2,4,6,8,10) and 5 odd numbers (1,3,5,7,9).

- Ordered selection: total outcomes = 10 × 9 = 90. Favorable: first even and second even = 5 × 4 = 20; first odd and second odd = 5 × 4 = 20; total 40. Probability = 40/90 = 4/9.
- Unordered selection: total pairs = C(10,2)=45. Favorable: both even = C(5,2)=10; both odd = C(5,2)=10; total 20. Probability = 20/45 = 4/9.

Thus the answer is 4/9, which corresponds to choice A.

ANSWER: A

1. **Understand the problem:** There are 10 balls numbered 1 through 10. Jack draws one ball, then Jill draws a different ball. We want the probability that the sum of the two numbers is even.

2. **Identify the parity of the balls:** The even-numbered balls are 2, 4, 6, 8, 10 (5 even). The odd-numbered balls are 1, 3, 5, 7, 9 (5 odd).

3. **Condition for an even sum:** A sum of two integers is even if both are even or both are odd.

4. **Calculate the probability using ordered draws (Jack then Jill):**
   - Total possible ordered outcomes: Jack has 10 choices, Jill has 9 choices → \(10 \times 9 = 90\) equally likely outcomes.
   - **Both even:** Jack picks even (5/10), then Jill picks a different even (4/9) → probability = \(\frac{5}{10} \times \frac{4}{9} = \frac{20}{90}\).
   - **Both odd:** Jack picks odd (5/10), then Jill picks a different odd (4/9) → probability = \(\frac{5}{10} \times \frac{4}{9} = \frac{20}{90}\).
   - Add them: \(\frac{20}{90} + \frac{20}{90} = \frac{40}{90} = \frac{4}{9}\).

5. **Alternative check using combinations:**
   - Total number of 2-ball combinations: \(\binom{10}{2} = 45\).
   - Number of both-even combinations: \(\binom{5}{2} = 10\).
   - Number of both-odd combinations: \(\binom{5}{2} = 10\).
   - Favorable combinations: \(10 + 10 = 20\).
   - Probability = \(\frac{20}{45} = \frac{4}{9}\).

6. **Match with answer choices:**
   - A: \(\frac{4}{9}\) ✓
   - B: \(\frac{9}{19}\) ✗
   - C: \(\frac{1}{2}\) ✗
   - D: \(\frac{10}{19}\) ✗
   - E: \(\frac{5}{9}\) ✗

The correct answer is clearly \(\frac{4}{9}\).

ANSWER: A

1. **Understand the problem**:  
   There are 10 balls numbered 1 through 10. Jack picks one at random, then Jill picks a different one at random (so no replacement). We are to find the probability that the **sum of the two numbers is even**.

2. **Key fact about even sums**:  
   A sum of two integers is even **if and only if** both numbers are **even** or both are **odd**.

3. **Count even and odd numbers from 1 to 10**:  
   - Odd numbers: 1, 3, 5, 7, 9 → 5 odds  
   - Even numbers: 2, 4, 6, 8, 10 → 5 evens

4. **Total number of possible outcomes**:  
   Jack picks any of the 10 balls, then Jill picks any of the remaining 9.  
   So total number of **ordered** pairs = $10 \times 9 = 90$

   (We use ordered pairs because Jack and Jill are distinct — who picks which matters in counting total outcomes, but since we're computing probability as favorable/total with consistent counting, this is fine.)

5. **Count favorable outcomes (sum is even)**:  
   This happens in two cases:

   - **Both numbers are odd**:  
     Jack picks one of 5 odd balls, then Jill picks one of the remaining 4 odd balls.  
     Number of ordered pairs = $5 \times 4 = 20$

   - **Both numbers are even**:  
     Jack picks one of 5 even balls, then Jill picks one of the remaining 4 even balls.  
     Number of ordered pairs = $5 \times 4 = 20$

   So total favorable outcomes = $20 + 20 = 40$

6. **Compute probability**:  
   $$
   \text{Probability} = \frac{\text{favorable}}{\text{total}} = \frac{40}{90} = \frac{4}{9}
   $$

7. **Check answer choices**:  
   Choice A is $ \frac{4}{9} $, which matches our result.

8. **Double-check reasoning**:  
   Alternative approach: Use combinations (unordered pairs).  
   Total number of ways to choose 2 distinct balls: $ \binom{10}{2} = 45 $  
   Favorable: choose 2 odds → $ \binom{5}{2} = 10 $, or 2 evens → $ \binom{5}{2} = 10 $, total = 20  
   Then probability = $ \frac{20}{45} = \frac{4}{9} $ — same result.

   So answer is consistent.

9. **Eliminate other choices**:  
   - B: $9/19 \approx 0.474$, close but not equal to $4/9 \approx 0.444$  
   - C: $1/2 = 0.5$ — too high  
   - D: $10/19 \approx 0.526$ — too high  
   - E: $5/9 \approx 0.556$ — too high  
   So only A matches.

ANSWER: A

I need to find how many Dallas-bound buses a Houston-bound bus passes on the highway during its 5-hour journey.

## Step 1: Understand the setup
- Dallas→Houston buses depart every hour on the hour (0:00, 1:00, 2:00, etc.)
- Houston→Dallas buses depart every hour on the half-hour (0:30, 1:30, 2:30, etc.)
- Trip duration: 5 hours
- We're counting passes on the highway only (not in stations)

## Step 2: Set up a specific scenario
Let me consider a Houston-bound bus that departs Houston at time t = 0:30.
- It travels from t = 0:30 to t = 5:30

## Step 3: Identify which Dallas-bound buses are on the highway during this interval
A Dallas-bound bus is on the highway during the Houston-bound bus's journey if:
- The Dallas bus departs before or at time 5:30 (otherwise it hasn't started)
- The Dallas bus arrives after time 0:30 (otherwise it's already done)

Since Dallas buses take 5 hours:
- A Dallas bus departing at time D arrives at time D + 5

For the Dallas bus to be on the highway while our Houston bus is traveling (0:30 to 5:30):
- The Dallas bus must arrive after 0:30: D + 5 > 0:30, so D > -4:30
- The Dallas bus must depart before 5:30: D < 5:30

Dallas buses depart at: ..., -3:00, -2:00, -1:00, 0:00, 1:00, 2:00, 3:00, 4:00, 5:00, 6:00, ...

The buses that satisfy D > -4:30 and D < 5:30 are those departing at:
- -4:00 (arrives at 1:00) ✓
- -3:00 (arrives at 2:00) ✓
- -2:00 (arrives at 3:00) ✓
- -1:00 (arrives at 4:00) ✓
- 0:00 (arrives at 5:00) ✓
- 1:00 (arrives at 6:00) ✓
- 2:00 (arrives at 7:00) ✓
- 3:00 (arrives at 8:00) ✓
- 4:00 (arrives at 9:00) ✓
- 5:00 (arrives at 10:00) ✓

## Step 4: Verify they actually pass (both on highway simultaneously)
Our Houston bus: 0:30 to 5:30

- Bus at -4:00: on road from -4:00 to 1:00 → overlaps [0:30, 1:00] ✓
- Bus at -3:00: on road from -3:00 to 2:00 → overlaps [0:30, 2:00] ✓
- Bus at -2:00: on road from -2:00 to 3:00 → overlaps [0:30, 3:00] ✓
- Bus at -1:00: on road from -1:00 to 4:00 → overlaps [0:30, 4:00] ✓
- Bus at 0:00: on road from 0:00 to 5:00 → overlaps [0:30, 5:00] ✓
- Bus at 1:00: on road from 1:00 to 6:00 → overlaps [1:00, 5:30] ✓
- Bus at 2:00: on road from 2:00 to 7:00 → overlaps [2:00, 5:30] ✓
- Bus at 3:00: on road from 3:00 to 8:00 → overlaps [3:00, 5:30] ✓
- Bus at 4:00: on road from 4:00 to 9:00 → overlaps [4:00, 5:30] ✓
- Bus at 5:00: on road from 5:00 to 10:00 → overlaps [5:00, 5:30] ✓

All 10 buses have overlapping highway time with our Houston-bound bus.

## Step 5: Check boundary cases
- Bus at -5:00 arrives at 0:00, before our bus departs at 0:30 ✗
- Bus at 6:00 departs at 6:00, after our bus arrives at 5:30 ✗

So we count exactly 10 buses.

ANSWER: D

1. **Restate the problem:** A bus leaves Dallas heading to Houston. Buses from Houston to Dallas leave on the half hour. The trip takes 5 hours. Count how many oncoming (Houston→Dallas) buses our Houston-bound (Dallas→Houston) bus passes on the highway.

2. **Set up:** Our bus departs Dallas at time 0:00 (on the hour), arrives Houston at 5:00. It's on the road during the interval (0:00, 5:00).

3. **Houston-bound buses on the road:** Houston→Dallas buses leave Houston at the half-hour marks (..., -4:30, -3:30, ..., 4:30, ...). Each takes 5 hours, so a bus departing Houston at time T is on the road during (T, T+5).

4. **Condition to meet on highway:** Our bus is on road (0,5). An oncoming bus departing Houston at time T occupies the highway during (T, T+5). They meet if these intervals overlap AND they actually cross on the road (not at a station).

   A bus departing Houston at T is on the road interval (T, T+5). Our bus interval is (0,5). They travel toward each other, so they cross if both are simultaneously on the highway, i.e. intervals overlap: T < 5 and T+5 > 0, so -5 < T < 5.

5. **Half-hour departure times in (-5, 5):**
   -4:30, -3:30, -2:30, -1:30, -0:30, 0:30, 1:30, 2:30, 3:30, 4:30.

   That's 10 buses.

6. **Check the endpoints (station issue):** At T = -5:00 a bus would arrive Houston exactly at 0:00 — that's a half-hour bus? No, departures are on half hours, so T=-5:00 isn't a departure anyway. At T=5:00, that bus departs as ours arrives — also not a half-hour time. So no station-only encounters to worry about; all 10 are genuine highway passes.

7. **Verify count:** The departures from -4:30 to 4:30 in half-hour... actually they're spaced 1 hour apart (each hour on the half hour). From -4:30 to 4:30 inclusive: -4:30, -3:30, -2:30, -1:30, -0:30, +0:30, +1:30, +2:30, +3:30, +4:30 = 10 buses.

8. **Rule out wrong answers:**
   - A (5), B (6): too few — these forget buses already en route or about to leave.
   - C (9): off by one, perhaps double-counting an endpoint as a station meeting.
   - E (11): would require an extra half-hour slot like -5:30 or 5:30, which is outside the overlap window.
   - D (10): matches our careful count.

ANSWER: D

I need to find how many Dallas-bound buses a Houston-bound bus passes while traveling from Houston to Dallas.

## Step 1: Set Up the Problem

A Houston-bound bus leaves Houston at some time and takes 5 hours to reach Dallas. Let me instead track a Dallas-bound bus (going from Dallas to Houston) and count how many Houston-bound buses it passes.

Actually, let me track one specific Houston-bound bus and count Dallas-bound buses it encounters ON THE HIGHWAY (not in stations).

## Step 2: Choose a Specific Bus

Let a Houston-bound bus leave Houston at 12:30 (noon). It arrives in Dallas at 5:30 PM.

## Step 3: Identify Dallas-bound Buses on the Highway

Dallas-bound buses leave Dallas every hour ON THE HOUR and take 5 hours to reach Houston.

A Dallas-bound bus is on the highway between Dallas and Houston during the 5-hour trip. I need to find which Dallas-bound buses are somewhere on the highway (not yet arrived, not yet departed) during the time our Houston-bound bus is traveling (12:30 to 5:30).

## Step 4: Find Dallas-bound Buses That Will Be Encountered

The Houston-bound bus travels from 12:30 to 5:30. It will meet a Dallas-bound bus if their paths overlap on the highway.

A Dallas-bound bus leaving Dallas at time T is on the highway from time T to time T+5 hours.

For the Houston-bound bus (traveling 12:30 to 5:30) to **pass** a Dallas-bound bus on the highway, they must be simultaneously on the highway AND moving toward each other (which they always are, since they travel opposite directions).

The Houston-bound bus meets a Dallas-bound bus if:
- The Dallas-bound bus has **departed Dallas** before our bus **arrives in Dallas**: T < 5:30
- The Dallas-bound bus has **not yet arrived in Houston** when our bus **departs Houston**: T + 5 > 12:30, meaning T > 7:30 AM

So Dallas-bound buses leaving between 7:30 AM and 5:30 PM (exclusive of endpoints since "not in station" means we exclude buses exactly at the terminal when we pass through).

## Step 5: Count the Buses

Dallas-bound buses leave on the hour. I need T strictly between 7:30 AM and 5:30 PM.

Hours that qualify: 8:00 AM, 9:00 AM, 10:00 AM, 11:00 AM, 12:00 PM, 1:00 PM, 2:00 PM, 3:00 PM, 4:00 PM, 5:00 PM

That's **10 buses**.

## Step 6: Check the Boundary Cases

- Bus leaving at 7:00 AM arrives Houston at 12:00 PM — it arrives in Houston **before** our bus departs at 12:30, so they don't meet on highway. ✓ (excluded)
- Bus leaving at 8:00 AM arrives Houston at 1:00 PM — our bus departs 12:30, so at 12:30 the Dallas bus is still on highway (30 min from Houston). They meet. ✓
- Bus leaving at 5:00 PM departs Dallas at 5:00 PM — our bus arrives Dallas at 5:30 PM, so at 5:00 PM our bus is still 30 min away. They meet on the highway. ✓
- Bus leaving at 6:00 PM — our bus arrives Dallas at 5:30, so the 6:00 PM bus hasn't left yet when we arrive. Don't meet. ✓

## Step 7: Verify the Count

Buses at: 8, 9, 10, 11, 12, 1, 2, 3, 4, 5 = **10 buses**

## Step 8: Rule Out Other Choices

- The intuitive "5 hours = 5 buses" ignores buses already en route when you depart
- The correct answer accounts for buses already traveling (those that left up to 5 hours before your departure) plus buses that leave during your 5-hour trip
- That's roughly 5 + 5 = 10, confirming our count

ANSWER: D

1. **Restate the problem.**  
We want to know: during its 5-hour trip, how many buses going **to Dallas** will one bus going **to Houston** meet on the highway?

2. **Set up a specific Houston-bound bus.**  
Pick one bus that leaves **Dallas for Houston** at exactly **12:00 noon**.

   Since the trip takes 5 hours, this bus is on the highway from **12:00 to 5:00**.

3. **Figure out which Dallas-bound buses are on the road during that time.**  
Buses from **Houston to Dallas** leave every hour on the half hour:
- 7:30
- 8:30
- 9:30
- 10:30
- 11:30
- 12:30
- 1:30
- 2:30
- 3:30
- 4:30
- 5:30
- etc.

   Each of these buses is also on the road for 5 hours.

4. **Determine which of these Dallas-bound buses meet the noon Houston-bound bus.**  
A Dallas-bound bus will meet the noon Houston-bound bus if the two buses are both on the highway at the same time.

   - The bus leaving Houston at **7:30** is on the road from **7:30 to 12:30**, so it is still on the road after 12:00. They meet.
   - The bus leaving Houston at **8:30** is on the road from **8:30 to 1:30**. They meet.
   - Similarly, buses leaving at **9:30, 10:30, 11:30, 12:30, 1:30, 2:30, 3:30, 4:30** all overlap in time with the noon bus, so they meet.
   - The bus leaving at **5:30** starts exactly when the noon bus arrives in Houston, so they meet only in the station, **not on the highway**, so do **not** count it.
   - The bus leaving at **6:30** is too late.
   - The bus leaving at **6:30 AM** would have arrived by 11:30 AM, before the noon bus left, so no meeting there.

5. **Count them.**  
The meeting buses are the ones leaving Houston at:
- 7:30
- 8:30
- 9:30
- 10:30
- 11:30
- 12:30
- 1:30
- 2:30
- 3:30
- 4:30

That is **10 buses**.

6. **Why the other choices are wrong.**
- **A. 5**: too few; this ignores buses already on the road before noon.
- **B. 6**: still too few.
- **C. 9**: misses one valid overlapping bus.
- **D. 10**: correct.
- **E. 11**: counts one extra bus, probably the 5:30 bus met at the station, which should not be counted.

ANSWER: D

1. **Restate the question.**  
   We want to know how many **Dallas-bound buses** a **Houston-bound bus** meets/passes **on the highway** during the 5-hour trip. We are told:
   - Dallas to Houston buses leave **every hour on the hour**
   - Houston to Dallas buses leave **every hour on the half hour**
   - The trip takes **5 hours**

2. **Set up the timing.**  
   Let the Houston-bound bus leave Dallas at **12:00** on the hour.  
   Since the trip takes 5 hours, it will arrive in Houston at **5:00**.

3. **Figure out which Dallas-bound buses it can pass.**  
   Dallas-bound buses leave Houston at:
   - 12:30
   - 1:30
   - 2:30
   - 3:30
   - 4:30
   - 5:30

   But the Houston-bound bus is only on the highway from **12:00 to 5:00**, so it can only pass the Dallas-bound buses that are also on the highway during that time.

4. **Check each Dallas-bound bus.**
   - The bus leaving Houston at **12:30** is on the road from 12:30 to 5:30, so it meets the Houston-bound bus.
   - The bus leaving at **1:30** also meets it.
   - The bus leaving at **2:30** also meets it.
   - The bus leaving at **3:30** also meets it.
   - The bus leaving at **4:30** also meets it.
   - The bus leaving at **5:30** leaves after the Houston-bound bus has already arrived, so it does **not** meet it.

   That gives **5** buses.

5. **Rule out the other choices.**  
   - **6** would include the 5:30 bus, but that bus is too late.
   - **9, 10, 11** are far too many because only one bus leaves each half hour, and the trip is only 5 hours long.
   - So the only possible answer is **5**.

ANSWER: A

To solve this problem, we need to determine how many Dallas-bound buses a Houston-bound bus encounters while on the road.

1. **Define the travel parameters:**
   - A Houston-bound bus leaves Dallas at time $T$. Let's assume it leaves at $T = 0:00$. It arrives in Houston at $T = 5:00$.
   - Dallas-bound buses leave Houston at $H:30$ for every hour $H$.
   - The trip duration is 5 hours.

2. **Identify the Dallas-bound buses on the road:**
   - A bus leaving Houston at $H:30$ will be on the road between $H:30$ and $(H+5):30$.
   - We are looking for Dallas-bound buses that are on the road at the same time as our Houston-bound bus (which is on the road from $0:00$ to $5:00$).
   - A Dallas-bound bus is on the road during the interval $[H:30, H+5:30]$.
   - For a bus to be on the road at any point during our bus's trip $[0:00, 5:00]$, the intervals must overlap.

3. **List the Dallas-bound buses:**
   - Bus 1: Leaves Houston at 11:30 (previous day). On road: 11:30 to 4:30.
   - Bus 2: Leaves Houston at 12:30. On road: 12:30 to 5:30.
   - Bus 3: Leaves Houston at 1:30. On road: 1:30 to 6:30.
   - Bus 4: Leaves Houston at 2:30. On road: 2:30 to 7:30.
   - Bus 5: Leaves Houston at 3:30. On road: 3:30 to 8:30.
   - Bus 6: Leaves Houston at 4:30. On road: 4:30 to 9:30.
   - Bus 7: Leaves Houston at 5:30. On road: 5:30 to 10:30. (This bus is not on the road when our bus is on the road, as it starts after our bus arrives).

4. **Verify the "passing" condition:**
   - Our bus is on the road from $0:00$ to $5:00$.
   - Bus 1 (11:30–4:30): On road during our trip.
   - Bus 2 (12:30–5:30): On road during our trip.
   - Bus 3 (1:30–6:30): On road during our trip.
   - Bus 4 (2:30–7:30): On road during our trip.
   - Bus 5 (3:30–8:30): On road during our trip.
   - Bus 6 (4:30–9:30): On road during our trip.
   - Bus 7 (5:30–10:30): Starts after our bus arrives in Houston.

   Wait, let's re-evaluate. A bus leaves Houston every hour on the half-hour. In a 5-hour window, there are 5 or 6 buses.
   If our bus leaves at 0:00 and arrives at 5:00, the Dallas-bound buses that are on the road are those that left Houston between 11:30 (the one that arrives at 4:30) and 4:30 (the one that arrives at 9:30).
   The buses that left at 11:30, 12:30, 1:30, 2:30, 3:30, and 4:30 are all on the road at some point between 0:00 and 5:00. This gives 6 buses.

5. **Ruling out other choices:**
   - If we check the math: The bus leaving at 11:30 is on the road until 4:30. The bus leaving at 4:30 is on the road starting at 4:30. Since they pass on the highway (not in the station), we count all buses that are on the road during the interval $(0, 5)$. There are 6 such buses.

ANSWER: B

To determine how many Dallas-bound buses a Houston-bound bus passes on the highway, we can track the journey of a single Houston-bound bus and find how many Dallas-bound buses overlap with its time on the road.

**Step 1: Restate what is being asked**
We need to find the number of Dallas-bound buses that a Houston-bound bus passes strictly on the highway (not at the stations). The Houston-bound buses leave on the hour, the Dallas-bound buses leave on the half-hour, and the trip takes exactly 5 hours.

**Step 2: Set up the timeline for the Houston-bound bus**
Let's assume our Houston-bound bus leaves Dallas at **12:00**. 
Since the trip takes exactly 5 hours, it will arrive in Houston at **17:00** (5:00 PM). 
Therefore, our bus is on the highway strictly between 12:00 and 17:00.

**Step 3: Determine the condition for passing a Dallas-bound bus**
For our bus to pass a Dallas-bound bus on the highway, the Dallas-bound bus must be on the road at the same time as our bus. This means:
- The Dallas-bound bus must arrive in Dallas *after* 12:00 (otherwise, it finishes its trip before we even leave).
- The Dallas-bound bus must leave Houston *before* 17:00 (otherwise, it starts its trip after we have already arrived).

**Step 4: List the schedules of the Dallas-bound buses**
Dallas-bound buses leave Houston every hour on the half-hour. Let's look at their departure and arrival times (Departure $\rightarrow$ Arrival):
- 6:30 $\rightarrow$ 11:30 (Arrives before we leave; we do not pass it)
- **7:30 $\rightarrow$ 12:30** (On the road until 12:30; we pass it)
- **8:30 $\rightarrow$ 13:30** (We pass it)
- **9:30 $\rightarrow$ 14:30** (We pass it)
- **10:30 $\rightarrow$ 15:30** (We pass it)
- **11:30 $\rightarrow$ 16:30** (We pass it)
- **12:30 $\rightarrow$ 17:30** (We pass it)
- **13:30 $\rightarrow$ 18:30** (We pass it)
- **14:30 $\rightarrow$ 19:30** (We pass it)
- **15:30 $\rightarrow$ 20:30** (We pass it)
- **16:30 $\rightarrow$ 21:30** (Leaves before we arrive; we pass it)
- 17:30 $\rightarrow$ 22:30 (Leaves after we arrive; we do not pass it)

**Step 5: Count the buses and verify**
Counting the valid departure times (7:30, 8:30, 9:30, 10:30, 11:30, 12:30, 13:30, 14:30, 15:30, and 16:30), we find exactly **10 buses**. 

*Alternative Relative Speed Method:* 
Both buses travel at a rate of 1 trip per 5 hours. Because they travel in opposite directions, their relative speed is $1/5 + 1/5 = 2/5$ of the total distance per hour. Over our 5-hour trip, the relative distance covered is $(2/5) \times 5 = 2$ full trip distances. Since the Dallas-bound buses are spaced exactly 1 hour apart (which is $1/5$ of a trip distance), our bus will pass $2 / (1/5) = 10$ buses. Because they leave on the half-hour and we leave on the hour, no buses are met exactly at the stations, keeping the count cleanly at 10.

**Step 6: Rule out the wrong choices**
- **A (5):** This is just the number of hours the trip takes, which only counts the buses that *depart* Houston while we are driving, ignoring the 5 buses already on the road when we leave.
- **B (6):** This counts the buses departing Houston during our trip plus one boundary error.
- **C (9) & E (11):** These would be the result of an off-by-one error, such as incorrectly excluding the 7:30 or 16:30 bus, or mistakenly including the 6:30 or 17:30 bus.

ANSWER: D

To find the number of Dallas-bound buses a Houston-bound bus passes on the highway, we can set up a coordinate system and analyze the schedules of the buses.

### **Step 1: Restate what is being asked**
We want to find how many Dallas-bound buses (traveling from Houston to Dallas) are passed by a single Houston-bound bus (traveling from Dallas to Houston) on the highway. The trip takes 5 hours. Dallas-bound buses leave Houston every hour on the half hour, and Houston-bound buses leave Dallas every hour on the hour. "On the highway (not in the station)" means the passing must occur strictly between the departure and arrival times of our bus.

### **Step 2: Set up the math**
Let our Houston-bound bus leave Dallas (position $x = 0$) at time $t = 0$. It will arrive in Houston (position $x = D$) at $t = 5$. 
The position of our bus at any time $t$ (for $0 \le t \le 5$) is given by:
$$x(t) = \frac{D}{5} t$$

A Dallas-bound bus leaves Houston ($x = D$) at some time $t_d$ and arrives in Dallas ($x = 0$) at $t_d + 5$. 
Its position at time $t$ (for $t_d \le t \le t_d + 5$) is given by:
$$y(t) = D - \frac{D}{5}(t - t_d)$$

Since Dallas-bound buses leave every hour on the half hour, their departure times $t_d$ must be of the form:
$$t_d = n + 0.5 \quad \text{for integers } n$$

### **Step 3: Solve for the passing time**
The two buses pass each other when their positions are equal ($x(t) = y(t)$):
$$\frac{D}{5} t = D - \frac{D}{5}(t - t_d)$$

We can divide both sides by $D$ and multiply by $5$:
$$t = 5 - (t - t_d)$$
$$2t = 5 + t_d$$
$$t = \frac{5 + t_d}{2}$$

### **Step 4: Apply the "on the highway" constraint**
For the passing to occur on the highway (not at either station), the passing time $t$ must be strictly between the departure and arrival times of our bus:
$$0 < t < 5$$

Substituting $t = \frac{5 + t_d}{2}$:
$$0 < \frac{5 + t_d}{2} < 5$$
$$0 < 5 + t_d < 10$$
$$-5 < t_d < 5$$

### **Step 5: Count the number of valid departure times**
Since $t_d = n + 0.5$, we substitute this into our inequality:
$$-5 < n + 0.5 < 5$$
$$-5.5 < n < 4.5$$

Since $n$ must be an integer, the possible values for $n$ are:
$$n \in \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4\}$$

Counting these integers:
$$4 - (-5) + 1 = 10 \text{ buses}$$

### **Step 6: Rule out the wrong choices**
* **A (5):** Incorrect. This only counts the buses that depart after our bus has already left Dallas.
* **B (6):** Incorrect.
* **C (9):** Incorrect. This misses one of the boundary buses (either the one that left Houston at $t = -4.5$ or at $t = 4.5$).
* **D (10):** Correct. All 10 buses are passed strictly on the highway.
* **E (11):** Incorrect. This would include a bus passed at one of the stations, but the schedules do not align for any station passings.

ANSWER: D

The problem requires determining the number of oncoming (Dallas-bound) buses met by a single Houston-bound bus during its journey. Buses depart Dallas on the hour (at integer times \(L\)) and Houston on the half-hour (at half-integer times \(S\)). Each trip takes exactly 5 hours.

Two buses, one departing Dallas at time \(L\) and one departing Houston at time \(S\), meet at time
\[
M = \frac{5 + L + S}{2},
\]
provided the meeting occurs strictly between the departure and arrival times of both buses (i.e., they meet on the highway, not at a terminal). This condition simplifies to the open interval
\[
S-5 < L < S+5.
\]

Consider without loss of generality the Houston-bound bus that departs at \(S = 0.5\). The admissible departure times \(L\) are the integers satisfying
\[
-4.5 < L < 5.5,
\]
which are exactly the ten values
\[
L = -4,-3,-2,-1,0,1,2,3,4,5.
\]
For each of these ten values the corresponding meeting time \(M\) lies strictly inside the open interval \((S,S+5)\), so every meeting occurs on the highway.

No admissible \(L\) produces a meeting at either terminal, because Houston departures occur only on the half-hour while Dallas departures occur only on the hour. Consequently the Houston-bound bus meets exactly ten Dallas-bound buses.

(The same count of ten is obtained for any other half-integer departure time \(S\).)

Thus the correct choice is D.

ANSWER: D

1. **Understanding the Problem**: We need to determine how many buses traveling from Houston to Dallas a bus traveling from Dallas to Houston will pass on the highway. Buses from Dallas to Houston leave every hour on the hour, and buses from Houston to Dallas leave every hour on the half hour. The travel time between the two cities is 5 hours.

2. **Setting Up the Math**: Let's consider a bus leaving Dallas at a certain hour (say, $t = 0$ hours). This bus will travel for 5 hours, so it will be on the highway from $t = 0$ to $t = 5$. We need to figure out how many buses from Houston it will encounter during this time.

3. **Identifying Encountering Buses**: Buses from Houston leave on the half hour. So, the first bus a Dallas-bound bus could encounter would be the one that left Houston at $t = -0.5$ (half an hour before our reference Dallas-bound bus leaves). The last bus it could encounter would be the one leaving Houston at $t = 4.5$ (since the bus leaving at $t = 5.5$ would not be on the highway at the same time as our Dallas-bound bus, which arrives at Houston at $t = 5$).

4. **Calculating the Number of Buses Encountered**: 
   - The bus that left Houston at $t = -0.5$ is on the highway.
   - Buses leave Houston at $t = 0.5, 1.5, 2.5, 3.5, 4.5$.
   - So, we count the buses leaving at $t = -0.5, 0.5, 1.5, 2.5, 3.5, 4.5$. That makes a total of 6 buses (including the one that left at $t = -0.5$ and the one that leaves at $t = 4.5$).

5. **Ruling Out Wrong Choices**:
   - **A. 5**: This is incorrect because we counted 6 buses.
   - **C. 9**, **D. 10**, and **E. 11**: These are all higher than our count, so they are incorrect as well.
   - Our count matches **B. 6**, making it the correct choice.

6. **Conclusion**: Based on our step-by-step analysis, the Houston-bound bus will pass 6 Dallas-bound buses on the highway.

ANSWER: B

We need to determine how many Dallas-bound buses are passed by a single Houston-bound bus during its 5-hour trip, not counting buses at the stations.  

**Step 1: Understand the schedule**  
- Dallas → Houston: leaves every hour on the hour (e.g., 1:00, 2:00, …).  
- Houston → Dallas: leaves every hour on the half hour (e.g., 1:30, 2:30, …).  
- Travel time: 5 hours in either direction.

**Step 2: Pick a specific departure for the Houston-bound bus**  
Let’s say our bus leaves Houston at 12:30 PM (noon). It will arrive in Dallas at 5:30 PM.

**Step 3: Determine which Dallas-bound buses it will encounter**  
Dallas-bound buses leave Dallas every hour on the hour. A bus leaving Dallas at time \( t \) will be on the road from \( t \) to \( t+5 \).  
Our Houston-bound bus is on the road from 12:30 PM to 5:30 PM.  
They meet if their time intervals overlap on the highway.

**Step 4: List Dallas departures and their travel windows**  

| Departure (Dallas) | On road from | On road until |
|--------------------|--------------|---------------|
| 7:00 AM            | 7:00         | 12:00         |
| 8:00 AM            | 8:00         | 1:00 PM       |
| 9:00 AM            | 9:00         | 2:00 PM       |
| 10:00 AM           | 10:00        | 3:00 PM       |
| 11:00 AM           | 11:00        | 4:00 PM       |
| 12:00 PM           | 12:00        | 5:00 PM       |
| 1:00 PM            | 1:00         | 6:00 PM       |
| 2:00 PM            | 2:00         | 7:00 PM       |
| 3:00 PM            | 3:00         | 8:00 PM       |
| 4:00 PM            | 4:00         | 9:00 PM       |
| 5:00 PM            | 5:00         | 10:00 PM      |

Our bus is on road 12:30–5:30. Overlap occurs with any bus whose window intersects this interval.

- 8:00 AM bus: on road until 1:00 PM → meets from 12:30 to 1:00 (overlap).  
- 9:00 AM bus: until 2:00 PM → meets.  
- 10:00 AM bus: until 3:00 PM → meets.  
- 11:00 AM bus: until 4:00 PM → meets.  
- 12:00 PM bus: until 5:00 PM → meets.  
- 1:00 PM bus: from 1:00 to 6:00 → meets from 1:00 to 5:30.  
- 2:00 PM bus: from 2:00 to 7:00 → meets from 2:00 to 5:30.  
- 3:00 PM bus: from 3:00 to 8:00 → meets from 3:00 to 5:30.  
- 4:00 PM bus: from 4:00 to 9:00 → meets from 4:00 to 5:30.  
- 5:00 PM bus: from 5:00 to 10:00 → meets from 5:00 to 5:30.  

That’s 10 buses: 8 AM, 9 AM, 10 AM, 11 AM, 12 PM, 1 PM, 2 PM, 3 PM, 4 PM, 5 PM.

**Step 5: Check endpoints**  
The 7:00 AM bus finishes at 12:00 PM, before our bus starts at 12:30 — no meet.  
The 6:00 PM bus starts at 6:00 PM, after our bus arrives at 5:30 — no meet.  
So exactly 10.

**Step 6: Rule out wrong choices**  
- 5 (A): Too few — only counts buses going same direction or something.  
- 6 (B): Might forget the ones at the start/end.  
- 9 (C): Might exclude the one at the departure or arrival boundary.  
- 11 (E): Might count one extra at a station incorrectly.  

Thus the correct answer is 10.

ANSWER: D

**Step-by-step reasoning:**

1. **Restate the problem:** We have a Houston-bound bus (Dallas → Houston) that leaves Dallas on the hour. Dallas-bound buses (Houston → Dallas) leave Houston on the half hour. The trip takes 5 hours. We must count how many Dallas-bound buses are passed by our Houston-bound bus strictly on the highway, not inside a station.

2. **Set a timeline:** Let our Houston-bound bus leave Dallas at time \(t = 0\) (on the hour). It travels for 5 hours and arrives in Houston at \(t = 5\).  
   Dallas-bound buses leave Houston at times \( \dots, -1.5, -0.5, 0.5, 1.5, 2.5, 3.5, 4.5, \dots \) (every hour on the half hour).  
   Each Dallas-bound bus is on the highway during the open interval \((T_H, T_H+5)\), where \(T_H\) is its departure time from Houston.

3. **Find overlapping journeys:** A Dallas-bound bus will meet our bus on the highway if their time intervals overlap with \((0,5)\). That requires:
   \[
   T_H < 5 \quad \text{and} \quad T_H + 5 > 0 \quad\Rightarrow\quad -5 < T_H < 5.
   \]
   Since \(T_H\) must be a half-hour time, the valid departure times are:
   \[
   -4.5,\; -3.5,\; -2.5,\; -1.5,\; -0.5,\; 0.5,\; 1.5,\; 2.5,\; 3.5,\; 4.5.
   \]
   That is exactly 10 buses.

4. **Check for station meetings:**  
   - A bus leaving Houston at \(-5\) would arrive at Dallas exactly at \(t = 0\), meeting at the Dallas station. But \(-5\) is not a half-hour time, so this does not happen.  
   - A bus leaving Houston at \(5\) would depart exactly when our bus arrives at the Houston station (\(t = 5\)). But \(5\) is not a half-hour time, so this does not happen.  
   - The first actual bus leaves at \(-4.5\); it arrives at Dallas at \(0.5\). Our bus leaves at \(0\) and meets it at \(t = 0.25\) on the open road.  
   - The last actual bus leaves at \(4.5\); our bus arrives at \(5\) and meets it at \(t = 4.75\) on the open road.  
   All 10 meetings occur strictly on the highway.

5. **Cross-check with relative motion:**  
   Buses travel at speed \(v\) and distance \(D = 5v\). Dallas-bound buses are spaced 1 hour apart in departure, so their distance on the road is \(v \times 1\) hour. Relative speed is \(2v\), so our bus encounters an oncoming bus every \( \frac{v}{2v} = 0.5 \) hours. In a 5-hour trip, \(5 / 0.5 = 10\) buses are encountered. The half-hour offset guarantees no encounters exactly at the stations.

6. **Conclusion:** The Houston-bound bus passes **10** Dallas-bound buses on the highway.

**ANSWER: D**

at Dallas.

Meeting at Houston: our bus arrives at t=5. A Dallas-bound bus is at Houston at its departure time T. So if T = 5, then it’s at Houston at t=5. But T=5 is **on the hour**, not half-hour → no Dallas-bound bus departs at t=5. The closest are t=4.5 and t=5.5. So no Dallas-bound bus is at Houston at t=5.

Therefore, **none of the 10 buses are met at a station** — all meetings happen **on the highway**.

6. **Double-check with a timeline or relative motion idea**:

Alternative approach:  
From the perspective of our Houston-bound bus (leaving Dallas at 12:00, arriving 5:00), any Dallas-bound bus that is **en route** during that time will be encountered.

Dallas-bound buses take 5 hours, and leave every hour at half-past.

So, which Dallas-bound buses are en route between 12:00 and 5:00?

- The bus that left Houston at **7:30 AM** arrives in Dallas at **12:30 PM** → so between 12:00 and 12:30, it’s still on the road → our bus (which starts at 12:00) will meet it shortly after departure.

- The bus that left Houston at **8:30 AM** arrives 1:30 PM → definitely overlaps.

- ...
- The bus that left Houston at **11:30 AM** arrives 4:30 PM → overlaps.

- The bus that leaves Houston at **12:30 PM** arrives 5:30 PM → our bus is on road until 5:00, so they overlap from 12:30 to 5:00 → meet.

- ...
- The bus that leaves Houston at **4:30 PM** arrives 9:30 PM → our bus is still on road until 5:00, so they overlap between 4:30 and 5:00 → meet.

- The next bus leaves at 5:30 PM — after our bus arrives — so no.

Now list those departure times from Houston:
7:30, 8:30, 9:30, 10:30, 11:30, 12:30, 1:30, 2:30, 3:30, 4:30

That’s **10 buses**.

All these are met on the highway — not at stations.

7. **Check answer choices**:  
A. 5  
B. 6  
C. 9  
D. 10  
E. 11  

We got **10**.

But wait — let’s make sure we didn’t misinterpret direction.

Re-read problem:

> Buses from Dallas to Houston leave every hour on the hour.  
> Buses from Houston to Dallas leave every hour on the half hour.  
> ... how many Dallas-bound buses does a Houston-bound bus pass...

Yes:  
- Houston-bound = Dallas → Houston (leaves on hour)  
- Dallas-bound = Houston → Dallas (leaves on half-hour)

We analyzed a Dallas→Houston bus, counted opposite-direction buses it meets: 10.

But some might think: what about the bus that left Houston at **7:30 AM** — at 12:00 noon, it's only 30 minutes from Dallas. Our bus just leaves Dallas. Do they meet?

Yes: in the next 30 minutes, our bus heads toward Houston, the other heads toward Dallas — they are 30 minutes apart (in time), but since both take 5 hours for full trip, their speeds are same.

Let’s do a quick position check.

Let distance between cities = 5 units → speed = 1 unit/hour.

At t=0 (12:00):
- Our bus is at position 0 (Dallas), heading toward Houston (position +5).
- The Dallas-bound bus that left Houston at 7:30 (t = –4.5) has been traveling for 4.5 hours → it's 4.5 units from Houston → so at position 5 – 4.5 = 0.5 (i.e., 0.5 units from Dallas).

So at t=0, that bus is at x=0.5, heading toward Dallas (decreasing x).  
Our bus is at x=0, heading toward Houston (increasing x).

They are moving toward each other: relative speed = 2 units/hour, distance between them = 0.5 → meet in 0.25 hours (15 minutes) at x=0.25 — **on the highway**, not at station.

Similarly, the bus that leaves Houston at 4:30 PM (t=4.5):  
At t=4.5, our bus has been traveling 4.5 hours → at x=4.5  
The Dallas-bound bus just leaves Houston → x=5, heading toward Dallas.

Distance = 0.5, closing at 2 units/hour → meet at t=4.75, x=4.75 — again, on highway.

So both endpoints are fine.

Thus, all 10 buses are passed on the highway.

8. **Could there be an 11th?**  
What about the bus that left Houston at **6:30 AM** (t = –5.5)? Arrives Dallas at 11:30 AM (t = –0.5) → before our bus departs → no overlap.

Bus that leaves Houston at **5:30 PM** (t=5.5): departs after our bus arrives → no.

So only 10.

Therefore, correct answer is **D. 10**

ANSWER: D

I need to find how many values of n can be added to {3, 6, 9, 10} such that the mean equals the median of the resulting set of 5 numbers.

**Step 1: Set up the constraint equation.**

For 5 numbers, the median is the 3rd number when ordered.
The mean equals (3 + 6 + 9 + 10 + n)/5 = (28 + n)/5

**Step 2: Consider cases based on where n falls in the ordered list.**

The original ordered set is {3, 6, 9, 10}. When I add n, I need to consider where it lands.

**Case 1: n ≤ 3**
Ordered set: {n, 3, 6, 9, 10}
Median = 6
Mean = Median: (28 + n)/5 = 6
28 + n = 30
n = 2 ✓ (valid since 2 ≤ 3)

**Case 2: 3 < n ≤ 6**
Ordered set: {3, n, 6, 9, 10}
Median = 6
Mean = Median: (28 + n)/5 = 6
n = 2 (not in this range, so no solution here)

**Case 3: 6 < n ≤ 9**
Ordered set: {3, 6, n, 9, 10}
Median = n
Mean = Median: (28 + n)/5 = n
28 + n = 5n
28 = 4n
n = 7 ✓ (valid since 6 < 7 ≤ 9)

**Case 4: 9 < n ≤ 10**
Ordered set: {3, 6, 9, n, 10}
Median = 9
Mean = Median: (28 + n)/5 = 9
28 + n = 45
n = 17 (not in range 9 < n ≤ 10, so no solution here)

**Case 5: n > 10**
Ordered set: {3, 6, 9, 10, n}
Median = 9
Mean = Median: (28 + n)/5 = 9
28 + n = 45
n = 17 ✓ (valid since 17 > 10)

**Step 3: Verify the solutions.**

- n = 2: Set {2, 3, 6, 9, 10}, median = 6, mean = 30/5 = 6 ✓
- n = 7: Set {3, 6, 7, 9, 10}, median = 7, mean = 35/5 = 7 ✓
- n = 17: Set {3, 6, 9, 10, 17}, median = 9, mean = 45/5 = 9 ✓

**Step 4: Count the possible values.**

I found exactly 3 possible values: n ∈ {2, 7, 17}

**Step 5: Check why other answers don't work.**

- Choice A (1 value): Too small; we found 3 solutions
- Choice B (2 values): Too small; we found 3 solutions
- Choice D (4 values): Too large; we found 3 solutions
- Choice E (more than 4): Too large; we found exactly 3 solutions

ANSWER: C

1. We have the set {3, 6, 9, 10} and add a fifth number n. We need the mean to equal the median.

2. Sum of the four numbers = 3+6+9+10 = 28. With n, sum = 28+n, mean = (28+n)/5.

3. The median of five numbers is the 3rd value when sorted. Consider cases based on where n falls.

**Case A: n ≤ 6.** Sorted order examples: if n ≤ 3: {n,3,6,9,10}, median = 6. If 3≤n≤6: {3,n,6,9,10}, median = 6.
Set mean = 6: (28+n)/5 = 6 → 28+n = 30 → n = 2.
Check: n=2 ≤ 6 ✓. Sorted {2,3,6,9,10}, median 6, mean 30/5=6 ✓.

**Case B: 6 ≤ n ≤ 9.** Sorted: {3,6,n,9,10}, median = n.
Set mean = n: (28+n)/5 = n → 28+n = 5n → 28 = 4n → n = 7.
Check: 7 is between 6 and 9 ✓. Sorted {3,6,7,9,10}, median 7, mean 35/5=7 ✓.

**Case C: n ≥ 9.** Sorted: {3,6,9,n,10} or {3,6,9,10,n}, median = 9.
Set mean = 9: (28+n)/5 = 9 → 28+n = 45 → n = 17.
Check: 17 ≥ 9 ✓. Sorted {3,6,9,10,17}, median 9, mean 45/5=9 ✓.

4. So we found three valid values: n = 2, 7, 17.

5. Ruling out wrong choices: We found exactly 3 distinct values, so it's not 1, 2, 4, or more than 4. Answer is 3.

ANSWER: C

I need to find all values of n such that when added to {3, 6, 9, 10}, the mean equals the median of the five-number set.

## Step 1: Set Up the Mean Condition

The five numbers are {3, 6, 9, 10, n}.
- Sum = 3 + 6 + 9 + 10 + n = 28 + n
- Mean = (28 + n)/5

## Step 2: Identify the Median

The median of 5 numbers is the 3rd number when sorted. I need to consider different cases based on where n falls in the ordering.

## Step 3: Case 1 — n ≤ 3

Sorted order: n, 3, 6, 9, 10 → Median = 6

Set mean = median: (28 + n)/5 = 6
28 + n = 30
n = 2

Check: n = 2 ≤ 3 ✓. Valid.

## Step 4: Case 2 — 3 ≤ n ≤ 6

Sorted order: 3, n, 6, 9, 10 → Median = 6

Same equation: (28 + n)/5 = 6 → n = 2

Check: n = 2 is NOT in [3, 6]. ✗ Invalid.

## Step 5: Case 3 — 6 ≤ n ≤ 9

Sorted order: 3, 6, n, 9, 10 → Median = n

Set mean = median: (28 + n)/5 = n
28 + n = 5n
28 = 4n
n = 7

Check: 6 ≤ 7 ≤ 9 ✓. Valid.

## Step 6: Case 4 — 9 ≤ n ≤ 10

Sorted order: 3, 6, 9, n, 10 → Median = 9

Set mean = median: (28 + n)/5 = 9
28 + n = 45
n = 17

Check: n = 17 is NOT in [9, 10]. ✗ Invalid.

## Step 7: Case 5 — n ≥ 10

Sorted order: 3, 6, 9, 10, n → Median = 9

Same equation: (28 + n)/5 = 9 → n = 17

Check: n = 17 ≥ 10 ✓. Valid.

## Step 8: Count Valid Solutions

The valid values of n are: **2, 7, 17** — that's 3 values.

## Ruling Out Wrong Answers

- A (1) and B (2): Too few — we found 3 solutions.
- D (4) and E (more than 4): Too many — we systematically checked all cases.

ANSWER: C

1. **Restate the question.**  
   We start with the set \(\{3,6,9,10\}\). We add a fifth number \(n\). We want the **mean** of the five numbers to be equal to the **median** of the five numbers. We must find how many possible values of \(n\) work.

2. **Find the mean in terms of \(n\).**  
   The sum of the original four numbers is
   \[
   3+6+9+10=28.
   \]
   So the sum of all five numbers is
   \[
   28+n.
   \]
   Therefore the mean is
   \[
   \frac{28+n}{5}.
   \]

3. **Think about the median.**  
   With 5 numbers, the median is the 3rd number when the numbers are arranged from least to greatest.

   Since the original numbers are
   \[
   3,\ 6,\ 9,\ 10,
   \]
   the position of \(n\) determines the median.

4. **Case 1: \(n \le 6\).**  
   Then the ordered list looks like one of these:
   - \(n,3,6,9,10\) if \(n\le 3\)
   - \(3,n,6,9,10\) if \(3<n\le 6\)

   In either situation, the 3rd number is \(6\).  
   So the median is \(6\).

   Set mean = median:
   \[
   \frac{28+n}{5}=6.
   \]
   Then
   \[
   28+n=30 \quad \Rightarrow \quad n=2.
   \]
   Check the case condition: \(2\le 6\), so this works.

5. **Case 2: \(6<n<9\).**  
   Then the ordered list is
   \[
   3,6,n,9,10.
   \]
   The median is now \(n\).

   Set mean = median:
   \[
   \frac{28+n}{5}=n.
   \]
   Multiply by 5:
   \[
   28+n=5n.
   \]
   So
   \[
   28=4n,\quad n=7.
   \]
   Check the case condition: \(6<7<9\), so this works.

6. **Case 3: \(n \ge 9\).**  
   Then the ordered list looks like one of these:
   - \(3,6,n,9,10\) if \(n=9\) does not happen this way exactly, but the median still becomes \(9\)
   - \(3,6,9,n,10\) if \(9<n\le 10\)
   - \(3,6,9,10,n\) if \(n>10\)

   In all these situations, the 3rd number is \(9\).  
   So the median is \(9\).

   Set mean = median:
   \[
   \frac{28+n}{5}=9.
   \]
   Then
   \[
   28+n=45 \quad \Rightarrow \quad n=17.
   \]
   Check the case condition: \(17\ge 9\), so this works.

7. **List all possible values of \(n\).**  
   The values we found are
   \[
   n=2,\ 7,\ 17.
   \]
   So there are exactly
   \[
   3
   \]
   possible values.

8. **Match with the choices.**  
   - A. 1 — no  
   - B. 2 — no  
   - C. 3 — yes  
   - D. 4 — no  
   - E. more than 4 — no

ANSWER: C

1. We are adding a fifth number \(n\) to the set \(\{3,6,9,10\}\).

2. The new mean must equal the median of the five-number set.

3. Since the original four numbers are already in order,
   \[
   3,\ 6,\ 9,\ 10
   \]
   and after adding \(n\), the median of five numbers is the **middle (3rd) number** when the five numbers are sorted.

4. The mean of the five numbers is
   \[
   \frac{3+6+9+10+n}{5}=\frac{28+n}{5}.
   \]

5. Now test where \(n\) can go in the ordered list:

   - If \(n \le 6\), then the sorted list is
     \[
     3,\ n,\ 6,\ 9,\ 10
     \]
     so the median is \(6\).

     Set mean = median:
     \[
     \frac{28+n}{5}=6
     \]
     \[
     28+n=30
     \]
     \[
     n=2.
     \]
     This works because \(2 \le 6\).

   - If \(6 < n < 9\), then the sorted list is
     \[
     3,\ 6,\ n,\ 9,\ 10
     \]
     so the median is \(n\).

     Set mean = median:
     \[
     \frac{28+n}{5}=n
     \]
     \[
     28+n=5n
     \]
     \[
     28=4n
     \]
     \[
     n=7.
     \]
     This works because \(7\) is between \(6\) and \(9\).

   - If \(n \ge 9\), then the median is \(9\).
     Set mean = median:
     \[
     \frac{28+n}{5}=9
     \]
     \[
     28+n=45
     \]
     \[
     n=17.
     \]
     This works because \(17 \ge 9\).

6. So the possible values of \(n\) are
   \[
   2,\ 7,\ 17,
   \]
   which is **3 values**.

7. Therefore, the correct choice is **C**.

ANSWER: C

To find the number of possible values for $n$ such that the mean of the set $\{3, 6, 9, 10, n\}$ equals its median, we follow these steps:

1. **Set up the mean equation:**
   The mean of the five numbers is $\frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$.
   We are given that the mean equals the median, so $\frac{28 + n}{5} = \text{median}$.
   This implies $28 + n = 5 \times (\text{median})$, or $n = 5 \times (\text{median}) - 28$.

2. **Analyze the median based on the value of $n$:**
   Let the ordered set be $S = \{3, 6, 9, 10, n\}$. The median depends on where $n$ falls relative to the existing numbers $\{3, 6, 9, 10\}$.

   *   **Case 1: $n \le 6$**
       The ordered set is $\{n, 3, 6, 9, 10\}$ or $\{3, n, 6, 9, 10\}$ or $\{3, 6, n, 9, 10\}$. In all these cases, the median is 6.
       Using the formula: $n = 5(6) - 28 = 30 - 28 = 2$.
       Check: If $n=2$, the set is $\{2, 3, 6, 9, 10\}$. Median is 6. Mean is $\frac{30}{5} = 6$. This is a valid solution.

   *   **Case 2: $6 < n < 9$**
       The ordered set is $\{3, 6, n, 9, 10\}$. The median is $n$.
       Using the formula: $n = 5(n) - 28 \implies 4n = 28 \implies n = 7$.
       Check: If $n=7$, the set is $\{3, 6, 7, 9, 10\}$. Median is 7. Mean is $\frac{35}{5} = 7$. This is a valid solution.

   *   **Case 3: $n \ge 9$**
       The ordered set is $\{3, 6, 9, n, 10\}$ (if $9 \le n \le 10$) or $\{3, 6, 9, 10, n\}$ (if $n > 10$). In both cases, the median is 9.
       Using the formula: $n = 5(9) - 28 = 45 - 28 = 17$.
       Check: If $n=17$, the set is $\{3, 6, 9, 10, 17\}$. Median is 9. Mean is $\frac{45}{5} = 9$. This is a valid solution.

3. **Verify the solutions:**
   We found three potential values for $n$: 2, 7, and 17.
   - For $n=2$: Median is 6, Mean is 6. (Valid)
   - For $n=7$: Median is 7, Mean is 7. (Valid)
   - For $n=17$: Median is 9, Mean is 9. (Valid)

4. **Rule out other possibilities:**
   - If we assumed the median was 3, $n = 5(3) - 28 = -13$. If $n = -13$, the set is $\{-13, 3, 6, 9, 10\}$, median is 6, not 3. (Invalid)
   - If we assumed the median was 10, $n = 5(10) - 28 = 22$. If $n = 22$, the set is $\{3, 6, 9, 10, 22\}$, median is 9, not 10. (Invalid)

There are exactly 3 possible values for $n$.

ANSWER: C

To find the number of possible values for $n$, we need to analyze the conditions under which the mean of the new set of five numbers equals its median. 

**Step 1: Restate what is being asked and set up the mean**
We are given the set $\{3, 6, 9, 10\}$ and a fifth number $n$. We want to find how many different values of $n$ make the mean of these five numbers equal to their median.
The sum of the original four numbers is $3 + 6 + 9 + 10 = 28$. 
When $n$ is added, the new sum is $28 + n$. 
Thus, the mean of the five numbers is:
$$\text{Mean} = \frac{28 + n}{5}$$

**Step 2: Determine the possible medians**
For a set of five numbers, the median is the 3rd number when they are arranged in ascending order. Depending on the value of $n$, there are three possible cases for the median:
*   **Case 1:** If $n \le 6$, the ordered set is either $\{n, 3, 6, 9, 10\}$ or $\{3, n, 6, 9, 10\}$. In both scenarios, the 3rd number (the median) is **6**.
*   **Case 2:** If $6 < n < 9$, the ordered set is $\{3, 6, n, 9, 10\}$. Here, the 3rd number (the median) is **$n$**.
*   **Case 3:** If $n \ge 9$, the ordered set is either $\{3, 6, 9, n, 10\}$ or $\{3, 6, 9, 10, n\}$. In both scenarios, the 3rd number (the median) is **9**.

*(Note: The median cannot be 3 or 10 because there are not enough numbers smaller than 3 or larger than 10 in the set to push them to the middle position).*

**Step 3: Solve for $n$ in each case**
Now we set the mean equal to the median for each of our three cases:

*   **Case 1 (Median = 6):**
    $$\frac{28 + n}{5} = 6$$
    $$28 + n = 30$$
    $$n = 2$$
    *Check:* Is $2 \le 6$? Yes. So $n = 2$ is a valid solution. (Set: $\{2, 3, 6, 9, 10\}$, Mean = 6, Median = 6).

*   **Case 2 (Median = $n$):**
    $$\frac{28 + n}{5} = n$$
    $$28 + n = 5n$$
    $$28 = 4n$$
    $$n = 7$$
    *Check:* Is $6 < 7 < 9$? Yes. So $n = 7$ is a valid solution. (Set: $\{3, 6, 7, 9, 10\}$, Mean = 7, Median = 7).

*   **Case 3 (Median = 9):**
    $$\frac{28 + n}{5} = 9$$
    $$28 + n = 45$$
    $$n = 17$$
    *Check:* Is $17 \ge 9$? Yes. So $n = 17$ is a valid solution. (Set: $\{3, 6, 9, 10, 17\}$, Mean = 9, Median = 9).

**Step 4: Rule out the wrong choices**
We have found exactly 3 valid values for $n$ ($2$, $7$, and $17$). 
*   Choices A (1) and B (2) are incorrect because they underestimate the number of solutions by missing one or more of the valid median cases.
*   Choices D (4) and E (more than 4) are incorrect because the median of these five numbers can only ever be 6, $n$, or 9. Each of these three possible medians creates exactly one simple linear equation, yielding exactly one unique solution each. There are no other mathematical possibilities.

ANSWER: C

To find the number of possible values of $n$, we can break the problem down into steps by analyzing how the position of $n$ affects the median of the five-number set.

### **Step 1: Understand the requirements**
We are given the set $\{3, 6, 9, 10\}$ and we add a fifth number $n$. 
* The sum of the first four numbers is $3 + 6 + 9 + 10 = 28$.
* The sum of all five numbers is $28 + n$.
* The mean of the five numbers is $M = \frac{28 + n}{5}$.
* We want the mean to equal the median ($Med$), so:
  $$\frac{28 + n}{5} = Med \implies 28 + n = 5 \cdot Med$$

Since there are 5 numbers, when they are sorted in ascending order, the median will be the 3rd (middle) number. The value of the median depends on where $n$ lies relative to $3, 6, 9,$ and $10$.

---

### **Step 2: Case-by-case analysis**

**Case 1: $n < 3$**
* If $n$ is the smallest number, the sorted set is $\{n, 3, 6, 9, 10\}$.
* The median is $6$.
* Setting the mean equal to the median:
  $$28 + n = 5 \times 6 \implies 28 + n = 30 \implies n = 2$$
* Since $2 < 3$, this is a valid solution. 
* **Verification:** For $\{2, 3, 6, 9, 10\}$, the mean is $\frac{30}{5} = 6$ and the median is $6$. (Valid)

**Case 2: $3 \le n < 6$**
* The sorted set is $\{3, n, 6, 9, 10\}$.
* The median is $6$.
* Setting the mean equal to the median:
  $$28 + n = 30 \implies n = 2$$
* Since $2$ is not in the interval $[3, 6)$, there is no solution in this interval.

**Case 3: $6 \le n < 9$**
* The sorted set is $\{3, 6, n, 9, 10\}$.
* The median is $n$.
* Setting the mean equal to the median:
  $$28 + n = 5n \implies 4n = 28 \implies n = 7$$
* Since $7$ is in the interval $[6, 9)$, this is a valid solution.
* **Verification:** For $\{3, 6, 7, 9, 10\}$, the mean is $\frac{35}{5} = 7$ and the median is $7$. (Valid)

**Case 4: $9 \le n < 10$**
* The sorted set is $\{3, 6, 9, n, 10\}$.
* The median is $9$.
* Setting the mean equal to the median:
  $$28 + n = 5 \times 9 \implies 28 + n = 45 \implies n = 17$$
* Since $17$ is not in the interval $[9, 10)$, there is no solution in this interval.

**Case 5: $n \ge 10$**
* The sorted set is $\{3, 6, 9, 10, n\}$.
* The median is $9$.
* Setting the mean equal to the median:
  $$28 + n = 45 \implies n = 17$$
* Since $17 \ge 10$, this is a valid solution.
* **Verification:** For $\{3, 6, 9, 10, 17\}$, the mean is $\frac{45}{5} = 9$ and the median is $9$. (Valid)

---

### **Step 3: Count the solutions and rule out wrong choices**
The possible values for $n$ are $2$, $7$, and $17$. 
* This gives exactly **3** possible values.
* This rules out choices A (1), B (2), D (4), and E (more than 4).

ANSWER: C

The set consists of the numbers \(3, 6, 9, 10\) together with a fifth number \(n\). The sum of these five numbers is \(28 + n\), so their mean is \((28 + n)/5\).
The median of the five numbers is the third number in the ordered list, which depends on the value of \(n\). The possible cases, determined by the position of \(n\) relative to the fixed numbers, are as follows.
- If \(n \leq 3\), the ordered list is \(n, 3, 6, 9, 10\), so the median is \(6\). Setting the mean equal to the median gives
 \[
 \frac{28 + n}{5} = 6 \implies n = 2.
 \]
 The value \(n = 2\) satisfies \(n \leq 3\), so it is valid.
- If \(3 < n \leq 6\), the ordered list is \(3, n, 6, 9, 10\), so the median is again \(6\). The same equation as above yields \(n = 2\), but this value does not lie in the interval \((3, 6]\). Thus, there is no solution in this case.
- If \(6 < n \leq 9\), the ordered list is \(3, 6, n, 9, 10\), so the median is \(n\). Setting the mean equal to the median gives
 \[
 \frac{28 + n}{5} = n \implies n = 7.
 \]
 The value \(n = 7\) satisfies \(6 < 7 \leq 9\), so it is valid.
- If \(9 < n \leq 10\), the ordered list is \(3, 6, 9, n, 10\), so the median is \(9\). Setting the mean equal to the median gives
 \[
 \frac{28 + n}{5} = 9 \implies n = 17.
 \]
 The value \(n = 17\) does not lie in the interval \((9, 10]\). Thus, there is no solution in this case.
- If \(n > 10\), the ordered list is \(3, 6, 9, 10, n\), so the median is again \(9\). The same equation as above yields \(n = 17\), and this value satisfies \(n > 10\). Thus, it is valid.
No other values of the median are possible. In all cases the third number in the ordered list is at most \(9\), because the original set already contains three numbers strictly less than \(10\).
The three values \(n = 2, 7, 17\) are therefore the only solutions. This rules out all choices other than C.
ANSWER: C

1. **Restate the problem**: We need to find the number of possible values of \( n \) that, when added to the set {3, 6, 9, 10}, makes the mean of the resulting set of five numbers equal to its median.

2. **Understand the concepts involved**:
   - The mean of a set of numbers is the average, found by summing all the numbers and then dividing by the count of numbers.
   - The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since we are adding one number to a set of four, the resulting set will have five numbers, making the median the third number when arranged in ascending order.

3. **Set up the math**:
   - The original set is {3, 6, 9, 10}. Adding \( n \) makes it {3, 6, 9, 10, \( n \)}.
   - The sum of the original four numbers is \( 3 + 6 + 9 + 10 = 28 \). Adding \( n \) makes the sum \( 28 + n \).
   - The mean of the five numbers is \( \frac{28 + n}{5} \).

4. **Consider the median**:
   - To find the median, we need to consider the possible positions of \( n \) in the ordered set.
   - If \( n \leq 6 \), the ordered set is {\( n \), 3, 6, 9, 10} or {3, \( n \), 6, 9, 10}, making the median 6.
   - If \( 6 < n \leq 9 \), the ordered set is {3, 6, \( n \), 9, 10}, making the median \( n \).
   - If \( n > 9 \), the ordered set is {3, 6, 9, 10, \( n \)} or {3, 6, 9, \( n \), 10} if \( 9 < n \leq 10 \), making the median 9.

5. **Analyze the cases for \( n \) and equate mean and median**:
   - **Case 1**: \( n \leq 6 \), median = 6. So, \( \frac{28 + n}{5} = 6 \). Solving this gives \( 28 + n = 30 \), hence \( n = 2 \). This satisfies \( n \leq 6 \), so \( n = 2 \) is a valid solution.
   - **Case 2**: \( 6 < n \leq 9 \), median = \( n \). So, \( \frac{28 + n}{5} = n \). Solving this gives \( 28 + n = 5n \), hence \( 28 = 4n \), and \( n = 7 \). This satisfies \( 6 < n \leq 9 \), so \( n = 7 \) is a valid solution.
   - **Case 3**: \( n > 9 \), median = 9. So, \( \frac{28 + n}{5} = 9 \). Solving this gives \( 28 + n = 45 \), hence \( n = 17 \). This satisfies \( n > 9 \), so \( n = 17 \) is a valid solution.

6. **Count the valid values of \( n \)**:
   - We found three valid values for \( n \): 2, 7, and 17.

7. **Rule out the wrong choices**:
   - A. 1 is incorrect because we found more than one value.
   - B. 2 is incorrect because we found three values.
   - D. 4 is incorrect because we found exactly three values, not more or less.
   - E. more than 4 is incorrect because we found exactly three values.

8. **Conclusion**:
   - Since we have exactly 3 valid values for \( n \), the correct answer is C.

ANSWER: C

We need to find how many values of \(n\) make the mean of the set \(\{3,6,9,10,n\}\) equal to its median. The mean is \(\frac{3+6+9+10+n}{5}=\frac{28+n}{5}\). The median depends on the position of \(n\) when the five numbers are sorted. We consider cases based on where \(n\) falls relative to the existing numbers.

**Case 1:** \(n \le 3\). Sorted order: \(n, 3, 6, 9, 10\). Median is 6. Condition: \(\frac{28+n}{5}=6 \Rightarrow 28+n=30 \Rightarrow n=2\). Since \(2\le3\), valid.

**Case 2:** \(3 \le n \le 6\). Sorted: \(3, n, 6, 9, 10\). Median is 6. Condition: \(\frac{28+n}{5}=6 \Rightarrow n=2\). Not in \([3,6]\), no solution.

**Case 3:** \(6 \le n \le 9\). Sorted: \(3, 6, n, 9, 10\). Median is \(n\). Condition: \(\frac{28+n}{5}=n \Rightarrow 28+n=5n \Rightarrow 28=4n \Rightarrow n=7\). In \([6,9]\), valid.

**Case 4:** \(9 \le n \le 10\). Sorted: \(3, 6, 9, n, 10\). Median is 9. Condition: \(\frac{28+n}{5}=9 \Rightarrow 28+n=45 \Rightarrow n=17\). Not in \([9,10]\), no solution.

**Case 5:** \(n \ge 10\). Sorted: \(3, 6, 9, 10, n\). Median is 9. Condition: \(\frac{28+n}{5}=9 \Rightarrow n=17\). Since \(17\ge10\), valid.

Thus the valid values are \(n=2,7,17\), giving three possible values. The other choices (1,2,4, more than 4) are ruled out because only three values satisfy the condition.

ANSWER: C

We are given the set {3, 6, 9, 10} and a fifth number \(n\). The new set has five numbers, and we need the mean to equal the median. The sum of the original four numbers is \(3+6+9+10 = 28\), so the mean of the five numbers is \(\frac{28+n}{5}\). The median of five numbers is the third number when they are arranged in increasing order. We consider cases based on where \(n\) falls in the sorted list.

1. **If \(n \le 6\)**: The sorted order is \(n, 3, 6, 9, 10\) (if \(n<3\)) or \(3, n, 6, 9, 10\) (if \(3 \le n \le 6\)). In all these, the third number is 6, so the median is 6.  
   Set mean = median: \(\frac{28+n}{5} = 6 \Rightarrow 28+n = 30 \Rightarrow n = 2\).  
   Since \(2 \le 6\), this is valid. So \(n=2\) is a solution.

2. **If \(6 < n < 9\)**: The sorted order is \(3, 6, n, 9, 10\). The median is \(n\).  
   Set mean = median: \(\frac{28+n}{5} = n \Rightarrow 28+n = 5n \Rightarrow 4n = 28 \Rightarrow n = 7\).  
   Since \(6 < 7 < 9\), this is valid. So \(n=7\) is a solution.

3. **If \(n \ge 9\)**: The sorted order is \(3, 6, 9, n, 10\) (if \(9 \le n \le 10\)) or \(3, 6, 9, 10, n\) (if \(n > 10\)). In all these, the third number is 9, so the median is 9.  
   Set mean = median: \(\frac{28+n}{5} = 9 \Rightarrow 28+n = 45 \Rightarrow n = 17\).  
   Since \(17 \ge 9\), this is valid. So \(n=17\) is a solution.

Checking the boundaries: if \(n=6\), median=6 but mean=6.8; if \(n=9\), median=9 but mean=7.4; neither works. No other cases exist because the median can only be 6, \(n\), or 9. Thus there are exactly three possible values for \(n\): 2, 7, and 17.

ANSWER: C

*Case 3: median = n**, which occurs when $ 6 < n < 9 $ (actually $ 6 < n < 9 $, but include endpoints? Let's be precise.)

If $ n = 6 $, the set is {3, 6, 6, 9, 10} → sorted, median = 6 → not equal to $ n $? Well, $ n = 6 $, median = 6, so actually median = n. But in this case, we already considered $ n \leq 6 $ above. Similarly, if $ n = 9 $, set is {3, 6, 9, 9, 10}, median = 9 = n. So endpoints can be considered in adjacent cases.

But in Case 3, the median is the third number, which is $ n $ **only if** $ n $ is between 6 and 9 **inclusive?** Let's check:

- If $ n = 6 $: sorted list is [3, 6, 6, 9, 10] → third = 6 = n → median = n  
- If $ n = 9 $: [3, 6, 9, 9, 10] → third = 9 = n → median = n  

So actually, for $ 6 \leq n \leq 9 $, the median is $ n $? Wait, not quite:

Wait, original numbers: 3, 6, 9, 10

Insert $ n = 6 $: we have two 6s. Sorted: 3, 6, 6, 9, 10 → third = 6 → median = 6 = n ✔️

Insert $ n = 7 $: 3, 6, 7, 9, 10 → median = 7 = n ✔️

Insert $ n = 8 $: 3, 6, 8, 9, 10 → median = 8 = n ✔️

Insert $ n = 9 $: 3, 6, 9, 9, 10 → median = 9 = n ✔️

So actually, **for $ 6 \leq n \leq 9 $, the median is $ n $**.

But earlier, in Cases 1 and 2, we assumed median = 6 when $ n \leq 6 $. That includes $ n = 6 $. So there's overlap at $ n = 6 $. We’ll handle that carefully.

To avoid confusion, better to split into **three distinct median possibilities**:

- Median = 6 → occurs when **at least three numbers are ≥ 6 and the third smallest is 6**. That happens when $ n \leq 6 $
- Median = n → when $ 6 \leq n \leq 9 $
- Median = 9 → when $ n \geq 9 $

But note: at $ n = 6 $, both median = 6 and median = n are same, so it's fine. Similarly at $ n = 9 $. So we can treat:

- Case A: $ n \leq 6 $ → median = 6  
- Case B: $ 6 \leq n \leq 9 $ → median = n  
- Case C: $ n \geq 9 $ → median = 9  

But Cases A and B overlap at $ n = 6 $, B and C at $ n = 9 $. That’s okay as long as we check consistency.

Now proceed with **Case B: median = n**, so set mean = n:

$$
\frac{28 + n}{5} = n \Rightarrow 28 + n = 5n \Rightarrow 28 = 4n \Rightarrow n = 7
$$

Check: Is $ n = 7 $ in [6, 9]? Yes. So valid.  
Sorted set: {3, 6, 7, 9, 10} → median = 7, mean = (28+7)/5 = 35/5 = 7 ✔️

So **second solution**: $ n = 7 $

---

**Case C: median = 9**, which occurs when $ n \geq 9 $

Set mean = 9:

$$
\frac{28 + n}{5} = 9 \Rightarrow 28 + n = 45 \Rightarrow n = 17
$$

Check: Is $ n = 17 \geq 9 $? Yes. Valid.  
Sorted set: {3, 6, 9, 10, 17} → median = 9, mean = (28+17)/5 = 45/5 = 9 ✔️

So **third solution**: $ n = 17 $

---

6. **Now check boundary values to ensure we didn’t miss anything**:

- What about $ n = 6 $?  
Mean = (28 + 6)/5 = 34/5 = 6.8  
Median = 6  
Not equal → so $ n = 6 $ is **not** a solution.  
But wait! Earlier we solved for median = 6 and got $ n = 2 $. So $ n = 6 $ is in the interval for median = 6, but doesn’t satisfy mean = median. So it’s not a solution — fine.

- What about $ n = 9 $?  
Mean = (28 + 9)/5 = 37/5 = 7.4  
Median = 9  
Not equal → so not a solution.

So only the three values we found: **n = 2, 7, 17**

7. **Are there any other cases?**  
We considered all possible positions for $ n $, which affect the median. The median of five numbers must be one of the numbers in the set, and based on where $ n $ falls, the median is either 6, $ n $, or 9. There's no way for median to be 3 or 10:

- Median = 3 would require at least 3 numbers ≤ 3 → we only have one (3), unless $ n \leq 3 $, but even then: if $ n \leq 3 $, the sorted list is [n, 3, 6, 9, 10] → third = 6, not 3. So median can’t be 3.

- Median = 10 would require at least 3 numbers ≥ 10 → we have only one (10), even if $ n \geq 10 $, list is [3,6,9,10,n] → third = 9. So median can’t be 10.

So only possible medians: 6, n, or 9 — covered.

8. **Conclusion**:  
We found **three** values of $ n $: 2, 7, and 17.  
Thus, the number of possible values of $ n $ is **3**.

Check answer choices:  
A. 1  
B. 2  
C. 3  
D. 4  
E. more than 4  

Correct choice is **C**.

ANSWER: C

I need to find the 98th term of the sequence starting with 98, 49, ...

**Step 1: Understand the rules**
- If n < 10: multiply by 9
- If n is even and n > 9: divide by 2
- If n is odd and n > 9: subtract 5

**Step 2: Generate the sequence starting from 98**

Term 1: 98
- 98 is even and > 9 → 98 ÷ 2 = 49

Term 2: 49
- 49 is odd and > 9 → 49 - 5 = 44

Term 3: 44
- 44 is even and > 9 → 44 ÷ 2 = 22

Term 4: 22
- 22 is even and > 9 → 22 ÷ 2 = 11

Term 5: 11
- 11 is odd and > 9 → 11 - 5 = 6

Term 6: 6
- 6 < 10 → 6 × 9 = 54

Term 7: 54
- 54 is even and > 9 → 54 ÷ 2 = 27

Term 8: 27
- 27 is odd and > 9 → 27 - 5 = 22

Term 9: 22
- 22 is even and > 9 → 22 ÷ 2 = 11

Term 10: 11
- 11 is odd and > 9 → 11 - 5 = 6

Term 11: 6
- 6 < 10 → 6 × 9 = 54

Term 12: 54
- 54 is even and > 9 → 54 ÷ 2 = 27

**Step 3: Identify the cycle**

I notice that starting from Term 6, we have: 54, 27, 22, 11, 6, 54, 27, 22, 11, 6, ...

The sequence enters a cycle of length 5:
- Position in cycle 1: 54
- Position in cycle 2: 27
- Position in cycle 3: 22
- Position in cycle 4: 11
- Position in cycle 5: 6

The cycle begins at Term 6.

**Step 4: Find the 98th term**

Terms 1-5 are: 98, 49, 44, 22, 11
Starting at Term 6, the cycle of 5 repeats: 54, 27, 22, 11, 6, 54, 27, 22, 11, 6, ...

To find the 98th term:
- Subtract the first 5 terms: 98 - 5 = 93
- Find position in the cycle: 93 ÷ 5 = 18 remainder 3

This means we go through 18 complete cycles of 5, then need the 3rd element of the 19th cycle.

The 3rd position in the repeating cycle (54, 27, 22, 11, 6) is **22**.

**Step 5: Verify**
- Term 6 = 54 (position 1 in cycle)
- Term 7 = 27 (position 2 in cycle)
- Term 8 = 22 (position 3 in cycle) ✓

For Term 98: 
- Terms 6-10: positions 1-5 of cycle
- Terms 11-15: positions 1-5 of cycle
- ...
- Term 6 + 93 - 1 = Term 98 corresponds to position (93 mod 5) = 3 in the cycle

Position 3 in the cycle is 22.

ANSWER: C

1. Restate: Apply rules starting from 98 to generate sequence, find the 98th term.

2. Generate terms:
- Term 1: 98 (even >9) → 49
- Term 2: 49 (odd >9) → 44
- Term 3: 44 (even >9) → 22
- Term 4: 22 (even >9) → 11
- Term 5: 11 (odd >9) → 6
- Term 6: 6 (<10) → 54
- Term 7: 54 (even >9) → 27
- Term 8: 27 (odd >9) → 22
- Term 9: 22 → 11
- Term 10: 11 → 6
- Term 11: 6 → 54
- Term 12: 54 → 27
- Term 13: 27 → 22

3. Identify cycle: From term 3 onward: 22, 11, 6, 54, 27, then back to 22 at term 8.
Cycle: 22, 11, 6, 54, 27 (length 5) starting at term 3.

4. Find term 98: position in cycle = (98 - 3) mod 5 = 95 mod 5 = 0.
So term 98 corresponds to the last element of cycle (offset 0 = first element 22).

Let me index: term 3 → offset 0 → 22; term 4 → offset 1 → 11; etc.
Offset = (98-3) mod 5 = 95 mod 5 = 0 → 22.

5. So term 98 = 22.

6. Rule out others: 11, 6, 54, 27 occur at other positions in cycle; only 22 matches offset 0.

ANSWER: C

I need to find the 98th term of a sequence starting with 98, 49, ...

## Step 1: Understand the Rules
- If integer < 10: multiply by 9
- If integer is even AND > 9: divide by 2
- If integer is odd AND > 9: subtract 5

## Step 2: Generate the Sequence

Start: 98
- 98: even, >9 → 98/2 = **49**
- 49: odd, >9 → 49-5 = **44**
- 44: even, >9 → 44/2 = **22**
- 22: even, >9 → 22/2 = **11**
- 11: odd, >9 → 11-5 = **6**
- 6: <10 → 6×9 = **54**
- 54: even, >9 → 54/2 = **27**
- 27: odd, >9 → 27-5 = **22**
- 22: even, >9 → 22/2 = **11**
- 11: odd, >9 → 11-5 = **6**
- 6: <10 → 6×9 = **54**
- ... (cycle repeats)

## Step 3: Identify the Cycle

Let me list the terms with their positions:
- Term 1: 98
- Term 2: 49
- Term 3: 44
- Term 4: 22
- Term 5: 11
- Term 6: 6
- Term 7: 54
- Term 8: 27
- Term 9: 22
- Term 10: 11
- Term 11: 6
- Term 12: 54
- Term 13: 27
- Term 14: 22
- ...

## Step 4: Identify the Repeating Cycle

Starting from Term 9, I see: 22, 11, 6, 54, 27, 22, 11, 6, 54, 27, ...

The cycle is: **22, 11, 6, 54, 27** with period 5, starting at Term 9.

## Step 5: Find the 98th Term

For terms from position 9 onward, I need to find where term 98 falls.

Terms 9, 10, 11, 12, 13 correspond to cycle positions 1, 2, 3, 4, 5.

For term n (where n ≥ 9), the cycle position is: (n - 9) mod 5

For n = 98: (98 - 9) mod 5 = 89 mod 5 = **4** (since 89 = 17×5 + 4)

Cycle position 0 → 22, position 1 → 11, position 2 → 6, position 3 → 54, position 4 → 27

Cycle position 4 → **27**

## Step 6: Verify

Term 9: position (0) → 22 ✓
Term 10: position (1) → 11 ✓
Term 13: position (4) → 27 ✓
Term 14: position (0) → 22 ✓ (27-5=22 ✓)

## Step 7: Rule Out Wrong Answers

The cycle contains {22, 11, 6, 54, 27}. The answer is 27, which corresponds to choice D.

ANSWER: D

1. **Restate the problem.**  
We start with the sequence beginning
\[
98,\ 49,\ \dots
\]
and each new term is made from the previous term by these rules:

- If the number is less than 10, multiply by 9.
- If the number is even and greater than 9, divide by 2.
- If the number is odd and greater than 9, subtract 5.

We want the **98th term**.

---

2. **Generate the sequence until a pattern appears.**

Start with the 1st term:

- 1st: \(98\)

Now apply the rules one step at a time:

- \(98\) is even and greater than 9, so divide by 2:  
  \(98 \to 49\)

- \(49\) is odd and greater than 9, so subtract 5:  
  \(49 \to 44\)

- \(44\) is even and greater than 9, so divide by 2:  
  \(44 \to 22\)

- \(22\) is even and greater than 9, so divide by 2:  
  \(22 \to 11\)

- \(11\) is odd and greater than 9, so subtract 5:  
  \(11 \to 6\)

- \(6\) is less than 10, so multiply by 9:  
  \(6 \to 54\)

- \(54\) is even and greater than 9, so divide by 2:  
  \(54 \to 27\)

- \(27\) is odd and greater than 9, so subtract 5:  
  \(27 \to 22\)

Now we have returned to \(22\), which already appeared.

So the sequence is:

\[
98,\ 49,\ 44,\ 22,\ 11,\ 6,\ 54,\ 27,\ 22,\ 11,\ 6,\ 54,\ 27,\dots
\]

---

3. **Identify the repeating cycle.**

Starting at the 4th term, we get:

\[
22,\ 11,\ 6,\ 54,\ 27
\]

and then it repeats back to \(22\).

So the repeating cycle has length \(5\).

---

4. **Figure out where the 98th term falls in the cycle.**

The 4th term starts the cycle, so count from there.

We need the position of the 98th term relative to the 4th term:

\[
98 - 4 = 94
\]

Now divide by the cycle length 5:

\[
94 \div 5 = 18 \text{ remainder } 4
\]

A remainder of 4 means the 98th term is 4 steps after the 4th term in the cycle.

Cycle positions:

- 4th term: \(22\)
- 5th term: \(11\)
- 6th term: \(6\)
- 7th term: \(54\)
- 8th term: \(27\)

So 4 steps after \(22\) is \(27\).

Thus the **98th term is \(27\)**.

---

5. **Check against the answer choices.**

- A. 6  
- B. 11  
- C. 22  
- D. 27  ← correct  
- E. 54  

So the correct choice is **D**.

ANSWER: D

1. **Restate the problem.**  
   We start with the sequence beginning
   \[
   98,\ 49,\ \dots
   \]
   and then repeatedly apply these rules:

   - If the number is **less than 10**, multiply by 9.
   - If the number is **even and greater than 9**, divide by 2.
   - If the number is **odd and greater than 9**, subtract 5.

   We need the **98th term**.

2. **Find the pattern.**  
   Let’s continue the sequence carefully:

   - \(98\) is even and greater than 9, so divide by 2:
     \[
     98 \to 49
     \]
   - \(49\) is odd and greater than 9, so subtract 5:
     \[
     49 \to 44
     \]
   - \(44\) is even and greater than 9, so divide by 2:
     \[
     44 \to 22
     \]
   - \(22\) is even and greater than 9, so divide by 2:
     \[
     22 \to 11
     \]
   - \(11\) is odd and greater than 9, so subtract 5:
     \[
     11 \to 6
     \]
   - \(6\) is less than 10, so multiply by 9:
     \[
     6 \to 54
     \]
   - \(54\) is even and greater than 9, so divide by 2:
     \[
     54 \to 27
     \]
   - \(27\) is odd and greater than 9, so subtract 5:
     \[
     27 \to 22
     \]

3. **Spot the repeating cycle.**  
   From \(22\) onward, we get:
   \[
   22 \to 11 \to 6 \to 54 \to 27 \to 22
   \]
   This is a cycle of length \(5\):
   \[
   22,\ 11,\ 6,\ 54,\ 27
   \]
   and then it repeats.

4. **Locate the 98th term in the cycle.**  
   The terms are:

   - Term 1: \(98\)
   - Term 2: \(49\)
   - Term 3: \(44\)
   - Term 4: \(22\)
   - Term 5: \(11\)
   - Term 6: \(6\)
   - Term 7: \(54\)
   - Term 8: \(27\)
   - Term 9: \(22\)

   So starting at term 4, the cycle repeats every 5 terms:
   \[
   22,\ 11,\ 6,\ 54,\ 27
   \]

   We need term 98. Since
   \[
   98 - 4 = 94
   \]
   and
   \[
   94 \div 5 = 18 \text{ remainder } 4,
   \]
   term 98 is the **4th** number in the cycle.

   The 4th number in the cycle is:
   \[
   54
   \]

5. **Check the answer choices.**  
   The choices are:
   - A. 6
   - B. 11
   - C. 22
   - D. 27
   - E. 54

   So the correct choice is **E**.

ANSWER: E

To find the 98th term of the sequence, we follow the given rules:
1. If $n < 10$, multiply by 9.
2. If $n \ge 10$ and even, divide by 2.
3. If $n \ge 10$ and odd, subtract 5.

**Step 1: Generate the sequence starting from 98.**
1. $a_1 = 98$ (Even, > 9)
2. $a_2 = 98 / 2 = 49$ (Odd, > 9)
3. $a_3 = 49 - 5 = 44$ (Even, > 9)
4. $a_4 = 44 / 2 = 22$ (Even, > 9)
5. $a_5 = 22 / 2 = 11$ (Odd, > 9)
6. $a_6 = 11 - 5 = 6$ (Less than 10)
7. $a_7 = 6 \times 9 = 54$ (Even, > 9)
8. $a_8 = 54 / 2 = 27$ (Odd, > 9)
9. $a_9 = 27 - 5 = 22$

**Step 2: Identify the cycle.**
We see that $a_4 = 22$ and $a_9 = 22$. This means the sequence enters a repeating cycle starting from the 4th term.
The terms are:
$a_4 = 22$
$a_5 = 11$
$a_6 = 6$
$a_7 = 54$
$a_8 = 27$
The cycle is $(22, 11, 6, 54, 27)$, which has a length of 5.

**Step 3: Determine the 98th term.**
Since the cycle starts at $n=4$, we want to find the position of the 98th term within this cycle.
The number of terms after the first 3 terms is $98 - 3 = 95$.
We need to find the 95th term of the repeating cycle $(22, 11, 6, 54, 27)$.
Since the cycle length is 5, we look at $95 \pmod 5$:
$95 \div 5 = 19$ remainder 0.
A remainder of 0 corresponds to the last element of the cycle.

**Step 4: Verify the terms.**
The cycle is:
Position 1 (mod 5): 22
Position 2 (mod 5): 11
Position 3 (mod 5): 6
Position 4 (mod 5): 54
Position 0/5 (mod 5): 27

Since 95 is a multiple of 5, the 95th term in the sequence starting from $a_4$ is the 5th term of the cycle, which is 27.

**Step 5: Ruling out choices.**
- A (6) is $a_6$, $a_{11}$, etc. (Position 3 in cycle)
- B (11) is $a_5$, $a_{10}$, etc. (Position 2 in cycle)
- C (22) is $a_4$, $a_9$, etc. (Position 1 in cycle)
- D (27) is $a_8$, $a_{13}$, etc. (Position 5 in cycle)
- E (54) is $a_7$, $a_{12}$, etc. (Position 4 in cycle)

The 98th term corresponds to the 5th position in the cycle, which is 27.

ANSWER: D

To find the 98th term of the sequence, we need to generate the first few terms using the given rules until we find a repeating pattern (a cycle). 

**Step 1: Restate the rules for the sequence**
For any term $x$ in the sequence:
- If $x < 10$, the next term is $x \times 9$.
- If $x > 9$ and is even, the next term is $x \div 2$.
- If $x > 9$ and is odd, the next term is $x - 5$.

**Step 2: Calculate the first few terms**
- **Term 1:** 98
- **Term 2:** 98 is even and $> 9$, so $98 \div 2 =$ **49**
- **Term 3:** 49 is odd and $> 9$, so $49 - 5 =$ **44**
- **Term 4:** 44 is even and $> 9$, so $44 \div 2 =$ **22**
- **Term 5:** 22 is even and $> 9$, so $22 \div 2 =$ **11**
- **Term 6:** 11 is odd and $> 9$, so $11 - 5 =$ **6**
- **Term 7:** 6 is $< 10$, so $6 \times 9 =$ **54**
- **Term 8:** 54 is even and $> 9$, so $54 \div 2 =$ **27**
- **Term 9:** 27 is odd and $> 9$, so $27 - 5 =$ **22**

**Step 3: Identify the cycle**
Notice that Term 9 is 22, which is exactly the same as Term 4. Because the rules are deterministic (each term depends only on the previous one), the sequence will now repeat the same values endlessly. 
The repeating cycle starts at Term 4 and consists of 5 numbers: **[22, 11, 6, 54, 27]**.

**Step 4: Find the 98th term**
To find the 98th term, we must determine where it falls within this 5-number cycle. 
First, we subtract the 3 initial terms (98, 49, 44) that occur before the cycle begins:
$98 - 3 = 95$

This means the 98th term of the sequence corresponds to the 95th term of the repeating cycle. Because the cycle is 5 numbers long, we divide 95 by 5:
$95 \div 5 = 19$ with a remainder of $0$.

A remainder of $0$ means that the 98th term lands exactly on the last (5th) number of the cycle. 
Looking at our cycle **[22, 11, 6, 54, 27]**, the 5th number is 27.

**Step 5: Rule out the wrong choices**
We can verify why the other options are incorrect by looking at where they appear in the sequence:
- **A (6):** Appears at Term 6, 11, 16, etc. (Term $n$ where $n$ ends in 1 or 6). 98 does not end in 1 or 6.
- **B (11):** Appears at Term 5, 10, 15, etc. (Term $n$ where $n$ is a multiple of 5). 98 is not a multiple of 5.
- **C (22):** Appears at Term 4, 9, 14, etc. (Term $n$ where $n$ ends in 4 or 9). 98 does not end in 4 or 9.
- **E (54):** Appears at Term 7, 12, 17, etc. (Term $n$ where $n$ ends in 2 or 7). 98 does not end in 2 or 7.
- **D (27):** Appears at Term 8, 13, 18, etc. (Term $n$ where $n$ ends in 3 or 8). 98 ends in 8, confirming our math is correct!

ANSWER: D

An elegant way to solve this problem is to generate the first few terms of the sequence to see if a repeating pattern (cycle) emerges.

### Step 1: Understand the Rules
The rules to find the next term $a_{n+1}$ from the current term $a_n$ are:
* If $a_n < 10$, then $a_{n+1} = 9 \times a_n$.
* If $a_n \ge 10$ and $a_n$ is even, then $a_{n+1} = a_n / 2$.
* If $a_n \ge 10$ and $a_n$ is odd, then $a_{n+1} = a_n - 5$.

### Step 2: Generate the Sequence
Let's compute the terms starting with $a_1 = 98$:
1. $a_1 = 98$ (even, $\ge 10$) $\rightarrow a_2 = 98 / 2 = 49$
2. $a_2 = 49$ (odd, $\ge 10$) $\rightarrow a_3 = 49 - 5 = 44$
3. $a_3 = 44$ (even, $\ge 10$) $\rightarrow a_4 = 44 / 2 = 22$
4. $a_4 = 22$ (even, $\ge 10$) $\rightarrow a_5 = 22 / 2 = 11$
5. $a_5 = 11$ (odd, $\ge 10$) $\rightarrow a_6 = 11 - 5 = 6$
6. $a_6 = 6$ ($< 10$) $\rightarrow a_7 = 6 \times 9 = 54$
7. $a_7 = 54$ (even, $\ge 10$) $\rightarrow a_8 = 54 / 2 = 27$
8. $a_8 = 27$ (odd, $\ge 10$) $\rightarrow a_9 = 27 - 5 = 22$

### Step 3: Identify the Cycle
Since $a_9 = 22$ and $a_4 = 22$, the sequence begins to repeat itself. The repeating cycle is:
$$\{22, 11, 6, 54, 27\}$$
This cycle has a length of $5$.

### Step 4: Determine the 98th Term
Since the cycle of length $5$ starts at the 4th term ($a_4$), we can express any term index $n \ge 4$ in terms of its position in the cycle. 

Specifically, the values of $a_n$ for $n \ge 4$ depend on the remainder of $n$ when divided by $5$:
* If $n \equiv 4 \pmod 5$, then $a_n = 22$ (e.g., $a_4, a_9, a_{14}, \dots$)
* If $n \equiv 0 \pmod 5$, then $a_n = 11$ (e.g., $a_5, a_{10}, a_{15}, \dots$)
* If $n \equiv 1 \pmod 5$, then $a_n = 6$ (e.g., $a_6, a_{11}, a_{16}, \dots$)
* If $n \equiv 2 \pmod 5$, then $a_n = 54$ (e.g., $a_7, a_{12}, a_{17}, \dots$)
* If $n \equiv 3 \pmod 5$, then $a_n = 27$ (e.g., $a_8, a_{13}, a_{18}, \dots$)

We want to find $a_{98}$. We divide $98$ by $5$:
$$98 = 5 \times 19 + 3$$

Since $98 \equiv 3 \pmod 5$, the 98th term will have the same value as $a_8$:
$$a_{98} = a_8 = 27$$

### Step 5: Rule out wrong choices
* **A. 6** corresponds to terms where $n \equiv 1 \pmod 5$ (for $n \ge 6$).
* **B. 11** corresponds to terms where $n \equiv 0 \pmod 5$ (for $n \ge 5$).
* **C. 22** corresponds to terms where $n \equiv 4 \pmod 5$ (for $n \ge 4$).
* **D. 27** corresponds to terms where $n \equiv 3 \pmod 5$ (for $n \ge 8$), which is correct.
* **E. 54** corresponds to terms where $n \equiv 2 \pmod 5$ (for $n \ge 7$).

ANSWER: D

The sequence begins with the terms \(a_1 = 98\) and \(a_2 = 49\). Apply the given rules term-by-term to generate the initial segment of the sequence and identify its long-term behavior:

- \(a_2 = 49\) (odd and greater than 9) gives \(49 - 5 = 44\), so \(a_3 = 44\).
- \(a_3 = 44\) (even and greater than 9) gives \(44 / 2 = 22\), so \(a_4 = 22\).
- \(a_4 = 22\) (even and greater than 9) gives \(22 / 2 = 11\), so \(a_5 = 11\).
- \(a_5 = 11\) (odd and greater than 9) gives \(11 - 5 = 6\), so \(a_6 = 6\).
- \(a_6 = 6\) (less than 10) gives \(6 \times 9 = 54\), so \(a_7 = 54\).
- \(a_7 = 54\) (even and greater than 9) gives \(54 / 2 = 27\), so \(a_8 = 27\).
- \(a_8 = 27\) (odd and greater than 9) gives \(27 - 5 = 22\), so \(a_9 = 22\).
- \(a_9 = 22\) (even and greater than 9) gives \(22 / 2 = 11\), so \(a_{10} = 11\).

From \(a_5 = 11\) onward the terms are \(11, 6, 54, 27, 22, 11, 6, \dots\). This is a repeating cycle of length 5 whose ordered elements are
\[
11,\ 6,\ 54,\ 27,\ 22.
\]
(The next term after 22 is again 11, confirming closure of the cycle.)

The 98th term lies inside this cycle. There are exactly \(98 - 5 + 1 = 94\) terms from position 5 through position 98. Dividing the index 94 by the cycle length 5 yields
\[
94 = 18 \times 5 + 4
\]
(quotient 18, remainder 4). The remainder 4 corresponds to the fourth entry of the cycle, which is 27. Consequently \(a_{98} = 27\).

To confirm, the five possible residues modulo 5 map to the cycle entries as follows:
- remainder 1 \(\mapsto\) 11,
- remainder 2 \(\mapsto\) 6,
- remainder 3 \(\mapsto\) 54,
- remainder 4 \(\mapsto\) 27,
- remainder 0 \(\mapsto\) 22.

The observed remainder 4 therefore selects 27 and excludes every other listed choice.

ANSWER: D

1. The problem requires us to determine the 98th term of a sequence that starts with 98 and follows specific rules to generate the next term based on the current term.

2. The given sequence starts with 98. We need to apply the rules to generate the next terms and identify any pattern or repetition.

3. The rules are as follows:
   - If the integer is less than 10, multiply it by 9.
   - If it is even and greater than 9, divide it by 2.
   - If it is odd and greater than 9, subtract 5.

4. Let's apply these rules to the starting number, 98.
   - 98 is even and greater than 9, so we divide it by 2 to get 49.
   - 49 is odd and greater than 9, so we subtract 5 to get 44.
   - 44 is even and greater than 9, so we divide it by 2 to get 22.
   - 22 is even and greater than 9, so we divide it by 2 to get 11.
   - 11 is odd and greater than 9, so we subtract 5 to get 6.
   - 6 is less than 10, so we multiply it by 9 to get 54.
   - 54 is even and greater than 9, so we divide it by 2 to get 27.
   - 27 is odd and greater than 9, so we subtract 5 to get 22.
   - We notice that 22 has appeared before, so the sequence will now loop: 22, 11, 6, 54, 27, 22, ...

5. The sequence starting from 98 is: 98, 49, 44, 22, 11, 6, 54, 27, 22, ...
   - The sequence enters a loop: 22, 11, 6, 54, 27, 22, ...

6. To find the 98th term, we first identify the length of the sequence before it enters the loop and the length of the loop.
   - The initial sequence before the loop is: 98, 49, 44 (3 terms).
   - The loop is: 22, 11, 6, 54, 27 (5 terms).

7. The total length of the sequence before it starts looping is 3 (initial terms) + 5 (loop length) = 8 terms to reach the start of the loop.
   - The sequence loops every 5 terms after the initial 3 terms.

8. Since the sequence starts looping after the 3rd term, we need to find where the 98th term falls within this loop.
   - The 4th term is the start of the loop, so we consider the position of the 98th term relative to the start of the loop.
   - 98 - 3 = 95. This means we are looking for the 95th term within the loop.

9. To find the term within the loop, we take 95 mod 5 (since the loop length is 5).
   - 95 mod 5 = 0, which means it is exactly at the end of a complete cycle of the loop.

10. Since 95 is a multiple of 5, it indicates the last term of the loop.
    - The loop is: 22, 11, 6, 54, 27.
    - The 5th term in the loop is 27, so the term corresponding to 95 mod 5 = 0 is the last term of the loop, which is 27.

11. Thus, the 98th term of the sequence is 27.

Let's rule out the wrong choices:
- A. 6 is a term in the loop but not the 98th term.
- B. 11 is a term in the loop but not the 98th term.
- C. 22 is a term in the loop but not the 98th term.
- D. 27 is the correct term as calculated.
- E. 54 is a term in the loop but not the 98th term.

ANSWER: D

We are told the sequence begins: 98, 49, …  
We need the 98th term. The rules are:

- If the integer is less than 10, multiply it by 9.
- If it is even and greater than 9, divide it by 2.
- If it is odd and greater than 9, subtract 5.

**Step 1: Generate terms until we detect a cycle.**  
Start:  
Term 1: 98 (even, >9 → divide by 2)  
Term 2: 49 (odd, >9 → subtract 5)  
Term 3: 44 (even, >9 → divide by 2)  
Term 4: 22 (even, >9 → divide by 2)  
Term 5: 11 (odd, >9 → subtract 5)  
Term 6: 6 (less than 10 → multiply by 9)  
Term 7: 54 (even, >9 → divide by 2)  
Term 8: 27 (odd, >9 → subtract 5)  
Term 9: 22 (even, >9 → divide by 2)  

**Step 2: Look for repetition.**  
We see:  
Term 4 = 22, Term 9 = 22.  
From Term 4 onward: 22, 11, 6, 54, 27, 22, …  
Check:  
22 → 11 → 6 → 54 → 27 → 22.  
That’s a cycle of length 5: (22, 11, 6, 54, 27).

**Step 3: Find position of the 98th term.**  
The cycle starts at term 4.  
So terms from 4 onward repeat every 5.  
We want term 98.  
Subtract the first 3 terms: 98 − 3 = 95 terms into the cycle.  
Now find position in cycle: 95 mod 5 = 0.  
A remainder of 0 means it’s the 5th element of the cycle.

**Step 4: Identify the 5th element.**  
Cycle:  
1st: 22  
2nd: 11  
3rd: 6  
4th: 54  
5th: 27  

So term 98 = 27.

**Step 5: Check choices.**  
27 corresponds to choice D.

ANSWER: D

**Step 1: Understand the rules and the sequence start**  
The sequence begins with 98, 49, … and follows these rules:  
- If the integer is less than 10, multiply it by 9.  
- If it is even and greater than 9, divide it by 2.  
- If it is odd and greater than 9, subtract 5.  

We need the 98th term.

**Step 2: Generate terms until a pattern emerges**  
- \(a_1 = 98\) (even, >9) → \(a_2 = 98 \div 2 = 49\)  
- \(a_2 = 49\) (odd, >9) → \(a_3 = 49 - 5 = 44\)  
- \(a_3 = 44\) (even, >9) → \(a_4 = 44 \div 2 = 22\)  
- \(a_4 = 22\) (even, >9) → \(a_5 = 22 \div 2 = 11\)  
- \(a_5 = 11\) (odd, >9) → \(a_6 = 11 - 5 = 6\)  
- \(a_6 = 6\) (<10) → \(a_7 = 6 \times 9 = 54\)  
- \(a_7 = 54\) (even, >9) → \(a_8 = 54 \div 2 = 27\)  
- \(a_8 = 27\) (odd, >9) → \(a_9 = 27 - 5 = 22\)  
- \(a_9 = 22\) → \(a_{10} = 11\) → \(a_{11} = 6\) → \(a_{12} = 54\) → \(a_{13} = 27\) → …

**Step 3: Identify the cycle**  
From \(a_4 = 22\) onward, the sequence repeats: 22, 11, 6, 54, 27, 22, …  
This is a cycle of length 5:  
Position in cycle: 0 → 22, 1 → 11, 2 → 6, 3 → 54, 4 → 27.

**Step 4: Locate the 98th term in the cycle**  
The cycle starts at \(a_4\). For \(n \ge 4\), the term is given by the remainder of \((n-4)\) divided by 5.  
\(n = 98\): \(98 - 4 = 94\)  
\(94 \div 5 = 18\) remainder \(4\) (since \(5 \times 18 = 90\), \(94 - 90 = 4\)).  
Remainder 4 corresponds to the last element of the cycle: 27.

**Step 5: Eliminate wrong choices**  
- A: 6 appears at positions 6, 11, 16, … (remainder 2).  
- B: 11 appears at positions 5, 10, 15, … (remainder 1).  
- C: 22 appears at positions 4, 9, 14, … (remainder 0).  
- E: 54 appears at positions 7, 12, 17, … (remainder 3).  
Only 27 matches remainder 4.

ANSWER: D

1. **Understand the problem and rules**:  
   We are given a sequence that starts with 98, and each next term is generated by the following rules:
   - If the integer is **less than 10**, multiply it by 9.
   - If the integer is **even and ≥10**, divide it by 2.
   - If the integer is **odd and ≥10**, subtract 5.

   We are to find the **98th term** of this sequence.

2. **Start generating terms to detect a cycle**:  
   Begin with the first term:  
   - Term 1: **98** → even and ≥10 → divide by 2 → **49**  
   - Term 2: **49** → odd and ≥10 → subtract 5 → **44**  
   - Term 3: **44** → even and ≥10 → divide by 2 → **22**  
   - Term 4: **22** → even and ≥10 → divide by 2 → **11**  
   - Term 5: **11** → odd and ≥10 → subtract 5 → **6**  
   - Term 6: **6** → less than 10 → multiply by 9 → **54**  
   - Term 7: **54** → even and ≥10 → divide by 2 → **27**  
   - Term 8: **27** → odd and ≥10 → subtract 5 → **22**  
   - Term 9: **22** → we’ve seen this before! (It was Term 4)

3. **Identify the repeating cycle**:  
   From Term 4 onward, we have:
   - Term 4: 22  
   - Term 5: 11  
   - Term 6: 6  
   - Term 7: 54  
   - Term 8: 27  
   - Term 9: 22 ← repeats Term 4

   So the cycle is: **22, 11, 6, 54, 27**, and then back to 22.  
   This is a **cycle of length 5**, starting at **Term 4**.

4. **Determine position within cycle for Term 98**:  
   - Terms 1–3 are **before** the cycle:  
     Term 1: 98  
     Term 2: 49  
     Term 3: 44  
   - Starting at Term 4, the cycle begins.

   So, the cycle includes terms numbered:  
   Term 4 → position 1 in cycle  
   Term 5 → position 2  
   ...  
   Term (3 + k) → position k in cycle

   We want Term 98.  
   So, position in cycle = 98 − 3 = **95th term in the cycle**.

   Since the cycle length is 5, compute:  
   $ 95 \mod 5 = 0 $

   A remainder of 0 means it's the **last element** of the cycle.

   The cycle is:  
   Position 1: 22  
   Position 2: 11  
   Position 3: 6  
   Position 4: 54  
   Position 5: 27  

   So, position 5 → **27**

5. **Match with answer choices**:  
   The 98th term is **27**, which corresponds to choice **D**.

6. **Double-check for errors**:  
   - Rules applied correctly at each step?  
     - 98 → 49 ✔️  
     - 49 → 44 ✔️  
     - 44 → 22 ✔️  
     - 22 → 11 ✔️  
     - 11 → 6 ✔️  
     - 6 → 54 ✔️  
     - 54 → 27 ✔️  
     - 27 → 22 ✔️  
   - Cycle correctly identified? Yes, from 22 back to 22 in 5 steps.  
   - Indexing correct? Term 4 is first in cycle; Term 98 is 95th in cycle; 95 mod 5 = 0 → 5th element → 27 ✔️

ANSWER: D