🔬 Model Lab

I need to find which of these four-digit numbers is a multiple of another four-digit number that also uses 2, 4, 5, and 7 exactly once.

**Step 1: Understand the problem**
- There are 24 permutations of digits 2, 4, 5, 7
- Exactly one of the given answer choices is a multiple of another permutation
- I need to find which one

**Step 2: Strategy**
If one number is a multiple of another, both using the same four digits, then the smaller number divides the larger. Since both are four-digit numbers using digits {2,4,5,7}, the ratio can't be too large. Let me check if any answer choice is divisible by another permutation of {2,4,5,7}.

**Step 3: Test each answer choice**

For a four-digit number to be a multiple of another four-digit number using the same digits, the divisor must be relatively small. The smallest possible number from {2,4,5,7} is 2457, and the largest is 7542.

Let me check if any answer is a multiple of numbers in this range:

**Choice A: 5724**
- Check divisibility by small permutations: 2457, 2475, 2547, 2574, 2745, 2754...
- 5724 ÷ 2457 ≈ 2.33 (not integer)
- 5724 ÷ 2475 ≈ 2.31 (not integer)

**Choice B: 7245**
- 7245 ÷ 2457 ≈ 2.95 (not integer)
- 7245 ÷ 2475 ≈ 2.93 (not integer)

**Choice C: 7254**
- 7254 ÷ 2457 ≈ 2.95 (not integer)
- 7254 ÷ 2475 ≈ 2.93 (not integer)

**Choice D: 7425**
- 7425 ÷ 2475 = 3 ✓
- Check: 2475 × 3 = 7425
- Verify 2475 uses digits {2,4,5,7}: Yes! ✓
- Verify 7425 uses digits {2,4,5,7}: Yes! ✓

**Choice E: 7542**
- 7542 ÷ 2457 ≈ 3.07 (not integer)
- 7542 ÷ 2475 ≈ 3.05 (not integer)

**Step 4: Verify the answer**
- 2475 × 3 = 7425
- 2475 uses digits 2, 4, 7, 5 ✓
- 7425 uses digits 7, 4, 2, 5 ✓
- 7425 is indeed a multiple of 2475

ANSWER: D

1. We need a number among the choices that is a multiple of another permutation of digits 2,4,5,7.

2. Check divisibility. Try small factors. The smallest such number from these digits is 2457. Let's test if any choice is 2 times some permutation.

3. 5724 / 2 = 2862, not a permutation. Test 5724 = 2 × 2862, no. Try other multipliers.

4. Actually, let's check 7254 / 2 = 3627, no. 7245 / 3 = 2415, no. 7425 / 3 = 2475, which uses digits 2,4,7,5 — yes! 2475 is a permutation. So 7425 = 3 × 2475.

5. Verify: 2475 × 3 = 7425. Digits of 2475: 2,4,7,5 ✓ — valid permutation.

6. Check others don't work easily: 5724 = 4×1431 no; 7245, 7254, 7542 — none divides cleanly by another permutation as nicely. 7425 works.

ANSWER: D

## Step-by-Step Solution

### Step 1: Understand what we're looking for
We need to find which of the given four-digit numbers (using digits 2, 4, 5, 7 exactly once) is a multiple of another four-digit number that also uses digits 2, 4, 5, 7 exactly once.

### Step 2: Identify the range of numbers
All 24 four-digit numbers using digits 2, 4, 5, 7 once range from **2457** (smallest) to **7542** (largest).

For one number to be a multiple of another, we need:
$$\frac{\text{larger}}{\text{smaller}} = \text{integer}$$

Since both numbers are 4-digit numbers using the same digits, the ratio must be between 1 and about 3 (since 7542/2457 ≈ 3.07).

So the ratio must be exactly **2** or **3**.

### Step 3: Check ratio = 2
If larger = 2 × smaller, then larger/2 = smaller.

Check each answer choice divided by 2:
- 5724 ÷ 2 = **2862** → digits are 2,8,6,2 ✗ (not our digit set)
- 7245 ÷ 2 = **3622.5** → not integer ✗
- 7254 ÷ 2 = **3627** → digits 3,6,2,7 ✗
- 7425 ÷ 2 = **3712.5** → not integer ✗
- 7542 ÷ 2 = **3771** → digits 3,7,7,1 ✗

None work for ratio = 2.

### Step 4: Check ratio = 3
If larger = 3 × smaller, then larger/3 = smaller.

Check each answer choice divided by 3:
- 5724 ÷ 3 = **1908** → digits 1,9,0,8 ✗
- 7245 ÷ 3 = **2415** → digits 2,4,1,5 ✗
- 7254 ÷ 3 = **2418** → digits 2,4,1,8 ✗
- 7425 ÷ 3 = **2475** → digits **2,4,7,5** ✓ This uses exactly 2,4,5,7!
- 7542 ÷ 3 = **2514** → digits 2,5,1,4 ✗

### Step 5: Verify the answer
**7425 ÷ 3 = 2475**

- 7425 uses digits: 7, 4, 2, 5 ✓
- 2475 uses digits: 2, 4, 7, 5 ✓
- 3 × 2475 = 7425 ✓

Both numbers use each of 2, 4, 5, 7 exactly once, and 7425 = 3 × 2475.

### Step 6: Rule out wrong choices
All other answer choices, when divided by 2 or 3, do not produce a four-digit number using exactly the digits 2, 4, 5, 7. Only **7425** satisfies the condition.

ANSWER: D

1. We need to find the one four-digit number, made from digits 2, 4, 5, and 7 exactly once, that is a multiple of another such number.

2. The numbers in the choices are:
   - A: 5724
   - B: 7245
   - C: 7254
   - D: 7425
   - E: 7542

3. Since one number must be a multiple of another, we test divisibility among the choices.

4. Notice that:
   - 5724 ÷ 7542 is not an integer
   - 7245 ÷ 5 is not relevant since 5 is not one of the listed numbers
   - So we should look for a pair where one is exactly a whole-number multiple of another among the choices.

5. Check the most promising pair:
   - \( 7245 = 3 \times 2415 \), but 2415 is not in the list.
   - \( 7542 \) is even, so it is divisible by 2, but no listed number is 2.
   - \( 7425 \) is divisible by 3 since \(7+4+2+5=18\), but again we need another listed number.

6. Let’s compare directly:
   - \( 7254 \times 2 = 14508 \), too large.
   - \( 5724 \times 1 = 5724 \), not another choice.
   - \( 7542 \div 2 = 3771 \), not a choice.
   - \( 7425 \div 3 = 2475 \), not a choice.
   - \( 7245 \div 5 = 1449 \), not a choice.

7. The key observation is that \( 7254 = 2 \times 3627 \), and \( 7245 \) is close, but not a multiple relationship. The only clean exact multiple relationship among the options is:
   - \( 7542 = 3 \times 2514 \), not a choice
   So we need a different strategy.

8. Since the problem says exactly one of the 24 numbers is a multiple of another one, we identify the number that can be written as a multiple of another permutation of the same digits. The only one that works is:
   - \( 7425 = 3 \times 2475 \), and 2475 is one of the numbers formed by these digits? No, 2475 is not using 7,2,4,5 once each? It does use 2,4,5,7 exactly once.

9. Thus 7425 is a multiple of 2475, so it is the number asked for.

10. Therefore the correct choice is D.

ANSWER: D

An elegant way to solve this problem is by setting up the mathematical relationship between the two four-digit numbers and testing the possible multipliers.

### Step 1: Restate what is being asked
We are looking for a four-digit number $A$ from the given choices such that $A$ is a multiple of another four-digit number $B$, where both $A$ and $B$ are formed by using the digits $2, 4, 5,$ and $7$ exactly once.

### Step 2: Set up the math
Let $A = k \cdot B$ for some integer $k > 1$. 
The smallest possible four-digit number using the digits $\{2, 4, 5, 7\}$ is $2457$, and the largest is $7542$.
Therefore, we must have:
$$2457 \le B < A \le 7542$$

This limits the possible values of the multiplier $k$:
$$k = \frac{A}{B} \le \frac{7542}{2457} \approx 3.07$$

Since $k$ must be an integer greater than $1$, the only possible values for $k$ are $2$ and $3$.

### Step 3: Test $k = 2$
If $A = 2B$, then $A$ must be an even number. The only even digits available are $2$ and $4$, so $A$ must end in $2$ or $4$.
- If $B$ ends in $2$, then $2B$ ends in $4$. The possible values for $B$ ending in $2$ are $4572$ and $4752$ (since any larger number starting with $5$ or $7$ would result in a five-digit number when doubled). 
  - $2 \times 4572 = 9144$ (not a permutation of $\{2, 4, 5, 7\}$)
  - $2 \times 4752 = 9504$ (not a permutation of $\{2, 4, 5, 7\}$)
- If $B$ ends in $7$, then $2B$ ends in $4$. The possible values for $B$ ending in $7$ are $2457, 2547, 4257,$ and $4527$.
  - $2 \times 2457 = 4914$ (no)
  - $2 \times 2547 = 5094$ (no)
  - $2 \times 4257 = 8514$ (no)
  - $2 \times 4527 = 9054$ (no)

Thus, there are no solutions for $k = 2$.

### Step 4: Test $k = 3$
If $A = 3B$, we can analyze the units digit of $B$ (denoted as $d_B$) and the units digit of $A$ (denoted as $d_A$):
- If $d_B = 4$, then $d_A = 2$ (since $3 \times 4 = 12$).
- If $d_B = 5$, then $d_A = 5$ (since $3 \times 5 = 15$).

Let's test the case where $d_B = 5$ and $d_A = 5$.
Since $A \le 7542$, we must have:
$$B \le \frac{7542}{3} = 2514$$

The only permutation of $\{2, 4, 5, 7\}$ that ends in $5$ and is less than $2514$ is $2475$.
Let's calculate $3 \times 2475$:
$$3 \times 2475 = 7425$$

The number $7425$ uses the digits $7, 4, 2,$ and $5$ exactly once. Thus, $7425$ is a multiple of $2475$, and both are valid permutations of the given digits.

### Step 5: Rule out the wrong choices
- **A. 5724**: $5724 / 3 = 1908$ (not a permutation of $\{2,4,5,7\}$)
- **B. 7245**: $7245 / 3 = 2415$ (not a permutation)
- **C. 7254**: $7254 / 3 = 2418$ (not a permutation)
- **D. 7425**: $7425 / 3 = 2475$ (this is the correct pair)
- **E. 7542**: $7542 / 3 = 2514$ (not a permutation)

ANSWER: D

The problem requires identifying the single number among all 24 permutations of the digits 2, 4, 5, and 7 that is a multiple of another such permutation.

All 24 permutations lie in the range from 2457 to 7542. If one such number \(A\) is an integer multiple \(k \cdot B\) of another (\(k \geq 2\)), then necessarily \(B \leq A/2 \leq 3771\). The only permutations that can possibly serve as the smaller number \(B\) are therefore those beginning with 2:
\[
2457,\ 2475,\ 2547,\ 2574,\ 2745,\ 2754.
\]
(The next-smallest permutation is 4257, which already exceeds 3771.)

The only feasible multipliers are \(k=2\) and \(k=3\), because \(4 \times 2457 = 9828\) already lies outside the four-digit range of the permutations. Direct computation yields:
\[
\begin{align*}
2\times2457&=4914,\\
2\times2475&=4950,\\
2\times2547&=5094,\\
2\times2574&=5148,\\
2\times2745&=5490,\\
2\times2754&=5508.
\end{align*}
\]
None of these products is a permutation of 2, 4, 5, 7.

For \(k=3\):
\[
\begin{align*}
3\times2457&=7371,\\
3\times2475&=7425,\\
3\times2547&=7641,\\
3\times2574&=7722,\\
3\times2745&=8235,\\
3\times2754&=8262.
\end{align*}
\]
The single product that is a permutation of 2, 4, 5, 7 is \(7425=3\times2475\). Thus the only pair is \(7425\) and \(2475\).

Among the five given choices, the only number that participates in this pair is 7425.

(The problem statement asserts that exactly one such multiple exists; the exhaustive enumeration above confirms that claim.)

ANSWER: D

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We are told:  
- We have all 24 four-digit numbers formed by using the digits 2, 4, 5, 7 each exactly once.  
- Only one of them is a multiple of another one from the list.  
- We are to identify which number among the choices is that multiple.  

---

**Step 1 – Understanding the problem**  
We have the set S of 24 distinct four-digit numbers formed by permuting (2, 4, 5, 7).  
We need to find the *unique* pair (A, B) in S such that A is a multiple of B (A > B, since it's a multiple and not equal).  
The question gives us five choices — we must find which of those is the larger one in that unique pair.

---

**Step 2 – Divisibility check approach**  
If one number is a multiple of another, the ratio must be an integer. Possible small integer ratios: 2, 3, 4, etc.  
Since the digits are fixed and all digits are used once in each number, the numbers are all between 2457 and 7542.  

The only possible ratio values are small integers: 2, 3, maybe 4 — because if ratio ≥ 5, the smaller number would be less than ~1500, but smallest is 2457 (impossible).

---

**Step 3 – Check possible ratio = 2**  
We check choices given:

- **A. 5724**  
  Half: 5724 ÷ 2 = 2862 — does 2862 consist of digits 2,4,5,7? No (has 8 and 6). Not possible.

- **B. 7245**  
  Half: 7245 ÷ 2 = 3622.5, not integer — ratio 2 impossible.

- **C. 7254**  
  Half: 7254 ÷ 2 = 3627 — digits include 3,6; not from {2,4,5,7}.

- **D. 7425**  
  Half: 7425 ÷ 2 = 3712.5, not integer — impossible.

- **E. 7542**  
  Half: 7542 ÷ 2 = 3771 — digits 3,7,7,1 — not valid.

So ratio 2 seems not possible for these choices.

---

**Step 4 – Check ratio = 3**  
Divide each by 3:

- **A. 5724** ÷ 3 = 1908 — digits 1,9,0,8 → no.  
- **B. 7245** ÷ 3 = 2415 — digits 2,4,1,5 → 1 is not allowed (we need 2,4,5,7).  
- **C. 7254** ÷ 3 = 2418 — digits 2,4,1,8 → no.  
- **D. 7425** ÷ 3 = 2475 — digits 2,4,7,5 — yes! This is a permutation of the required digits.  
- **E. 7542** ÷ 3 = 2514 — digits 2,5,1,4 → no.

Only **D** works with ratio 3: 7425 = 3 × 2475.

---

**Step 5 – Check if any other ratio works for remaining choices**  
Check D for ratio 4: 7425 ÷ 4 = 1856.25 → no. Ratio > 4 impossible because the smaller number would be below the minimum possible.

---

**Step 6 – Conclusion**  
The only pair is (7425, 2475). The multiple asked is 7425.

---

ANSWER: D

# Finding the Four-Digit Multiple

## Step 1: Restate the Problem
We have 24 permutations of the digits 2, 4, 5, 7 forming four-digit numbers. Exactly one of these numbers is a multiple of another. I need to find which one from the given choices.

## Step 2: Determine Possible Multipliers
The smallest number is 2457 and the largest is 7542.
The maximum possible ratio is:
$$\frac{7542}{2457} \approx 3.07$$

So the multiplier can only be **2 or 3** (it can't be 1 since the numbers must be different, and 4 or more would exceed the range).

## Step 3: Check Multiplier = 2
If $N \times 2$ also uses digits {2, 4, 5, 7}, then $N \leq 7542/2 = 3771$, so $N$ must start with 2.

Testing all permutations starting with 2:
- $2457 \times 2 = 4914$ → contains 9, 1 ✗
- $2475 \times 2 = 4950$ → contains 9, 0 ✗
- $2547 \times 2 = 5094$ → contains 0, 9 ✗
- $2574 \times 2 = 5148$ → contains 1, 8 ✗
- $2745 \times 2 = 5490$ → contains 9, 0 ✗
- $2754 \times 2 = 5508$ → contains 0, 8 ✗

**No solution with multiplier 2.**

## Step 4: Check Multiplier = 3
If $N \times 3$ uses digits {2, 4, 5, 7}, then $N \leq 7542/3 = 2514$, so $N$ must start with 2 and have second digit ≤ 5.

Testing valid permutations:
- $2457 \times 3 = 7371$ → contains 3, 1 ✗
- $2475 \times 3 = 7425$ → digits are **7, 4, 2, 5** ✓ **MATCH!**

Let me verify: $2475 \times 3 = 6000 + 1200 + 210 + 15 = 7425$ ✓

Both 2475 and 7425 use exactly the digits {2, 4, 5, 7}.

## Step 5: Confirm Uniqueness
Checking remaining candidates starting with 2:
- $2547 \times 3 = 7641$ → contains 6, 1 ✗
- $2574 \times 3 = 7722$ → repeated digits ✗
- $2745 \times 3 = 8235$ → contains 8, 3 ✗
- $2754 \times 3 = 8262$ → contains 8, 6 ✗

The **only** pair is $2475 \times 3 = 7425$.

## Step 6: Match to Choices
The number that is a multiple of another is **7425**, which corresponds to choice **D**.

ANSWER: D

 **Step 1: Restate the problem.**  
We have 24 four-digit numbers, each using the digits 2, 4, 5, and 7 exactly once. We must determine which of the five answer choices equals an integer multiple (greater than 1) of some *other* number in that same set.

**Step 2: Bound the possible multiplier.**  
The smallest permutation is 2457 and the largest is 7542.  
The ratio of any two distinct numbers is less than  

\[
\frac{7542}{2457}\approx 3.07.
\]

Therefore the only possible integer multipliers greater than 1 are **2** and **3**.

**Step 3: Examine multiplier 2.**  
If a number is twice another, it must be even.  
Among the choices, the even numbers are:

- A. \(5724 \div 2 = 2862\) (uses digits 2,8,6,2 — not allowed)  
- C. \(7254 \div 2 = 3627\) (uses digits 3,6,2,7 — not allowed)  
- E. \(7542 \div 2 = 3771\) (uses digits 3,7,7,1 — not allowed)

None of the even choices work, and no odd choice can be 2 times an integer. So **multiplier 2 is impossible** for every answer choice.

**Step 4: Examine multiplier 3.**  
If a number is 3 times another permutation, the other permutation must be at least the smallest one, 2457. Hence the multiple must be at least  

\[
3 \times 2457 = 7371.
\]

This immediately eliminates choices **A (5724)**, **B (7245)**, and **C (7254)** because they are all smaller than 7371 and therefore cannot be 3 times any valid permutation.

**Step 5: Test the remaining candidates for multiplier 3.**  
Only **D (7425)** and **E (7542)** are large enough.

- **D.** \(7425 \div 3 = 2475\).  
  The digits of 2475 are 2, 4, 7, 5 — exactly the set {2,4,5,7}.  
  Thus \(7425 = 3 \times 2475\), and 2475 is indeed one of the 24 numbers.

- **E.** \(7542 \div 3 = 2514\).  
  This contains the digit 1, so it is **not** a permutation of {2,4,5,7}.

**Step 6: Rule out the remaining wrong choices.**  
- **A, C:** Even, but fail the divisor-2 test and are too small for divisor-3.  
- **B:** Odd, but \(7245 \div 3 = 2415\) (contains 1) and it is too small to be 3 times a valid permutation anyway.  
- **E:** Large enough for divisor-3, but its quotient 2514 is invalid.

**Step 7: Conclusion.**  
The unique number among the choices that is an integer multiple of another valid permutation is **7425**, because \(7425 = 3 \times 2475\).

ANSWER: D

**Step 1: Understand the Problem**
We are given the set of digits $\{2, 4, 5, 7\}$. There are 24 permutations of these digits forming four-digit numbers. We need to find which of the five given choices (A-E) is a multiple of another number from this set of 24 permutations. In other words, we are looking for a number $X$ from the choices such that $X = k \cdot Y$, where $Y$ is another number formed by the digits $\{2, 4, 5, 7\}$ and $k$ is an integer greater than 1.

**Step 2: Analyze the Constraints**
*   **Digits:** The available digits are 2, 4, 5, 7.
*   **Range:** The smallest number is 2457. The largest number is 7542.
*   **Possible Multipliers ($k$):**
    *   If $k=2$, then $Y = X/2$. For $X$ to be a valid 4-digit number in our range, $Y$ must be between roughly 1228 and 3771. Since $Y$ must use our digits, $Y$ would likely start with 2.
    *   If $k=3$, then $Y = X/3$. For $X$ to be valid, $Y$ must be between roughly 819 and 2514. $Y$ would have to be small, starting with 2.
    *   If $k=4$, then $4Y = X$. Even the smallest $Y$ (2457) gives $4 \times 2457 = 9828$, which is larger than the maximum possible number (7542). Thus, $k$ cannot be 4 or higher.
    *   **Conclusion:** We only need to check if any choice is divisible by 2 or 3 resulting in a permutation of $\{2, 4, 5, 7\}$. Note that since the sum of digits is $2+4+5+7=18$, every permutation is divisible by 9 (and thus 3).

**Step 3: Evaluate the Choices**

*   **Choice A: 5724**
    *   Check division by 2: $5724 / 2 = 2862$. Digits are $\{2, 6, 8\}$. Not a match.
    *   Check division by 3: $5724 / 3 = 1908$. Digits contain 0, 1, 8, 9. Not a match.

*   **Choice B: 7245**
    *   Ends in 5, so not divisible by 2.
    *   Check division by 3: $7245 / 3 = 2415$. Digits are $\{1, 2, 4, 5\}$. Contains '1', missing '7'. Not a match.

*   **Choice C: 7254**
    *   Check division by 2: $7254 / 2 = 3627$. Contains 3, 6. Not a match.
    *   Check division by 3: $7254 / 3 = 2418$. Contains 1, 8. Not a match.

*   **Choice D: 7425**
    *   Ends in 5, so not divisible by 2.
    *   Check division by 3: 
        $$7425 \div 3$$
        $7000$ is not div by 3 easily, let's do long division or simple arithmetic.
        $7425 = 7200 + 225$.
        $7200 / 3 = 2400$.
        $225 / 3 = 75$.
        Total quotient = $2400 + 75 = 2475$.
    *   **Check Digits of Quotient:** The quotient is 2475.
    *   Does 2475 use the digits $\{2, 4, 5, 7\}$ exactly once? Yes: 2, 4, 7, 5.
    *   This is a valid pair: $3 \times 2475 = 7425$.

*   **Choice E: 7542**
    *   Check division by 2: $7542 / 2 = 3721$. Contains 3, 1. Not a match.
    *   Check division by 3: $7542 / 3 = 2514$. Contains 1. Not a match.

**Step 4: Conclusion**
Only Choice D yields a quotient (2475) that is a permutation of the original digits $\{2, 4, 5, 7\}$.

ANSWER: D

**Step 1 – Understand the problem**  
We have the four digits \(\{2,4,5,7\}\). All \(4! = 24\) four‑digit numbers formed by using each digit exactly once are possible. Among these 24 numbers exactly one is an integer multiple of another one. We have to pick that number from the given choices.

**Step 2 – List all 24 permutations**  

\[
\begin{aligned}
&2457,\,2475,\,2547,\,2574,\,2745,\,2754,\\
&4257,\,4275,\,4527,\,4572,\,4725,\,4752,\\
&5247,\,5274,\,5427,\,5472,\,5724,\,5742,\\
&7245,\,7254,\,7425,\,7452,\,7524,\,7542 .
\end{aligned}
\]

**Step 3 – Determine which integer factors could produce a multiple**  

All numbers lie between 2457 and 7542.  
- A factor of 2 would require the larger number ≤ 8000 and the smaller ≥ 2458. Only numbers beginning with 2 (the six 2 xxx numbers) can be the smaller, but none of those doubled stays within the set (e.g. \(2457\times2=4914\) uses a 9). Hence **no pair with factor 2**.  
- A factor of 3: the smaller must be ≤ 2666 so that the product stays ≤ 8000. Only the 2 xxx numbers satisfy this. We test each:

\[
\begin{array}{c|c}
\text{Smaller} & 3\times\text{smaller} \\ \hline
2457 & 7371\;(7,3,7,1) \\
2475 & 7425\;(7,4,2,5) \\
2547 & 7641\;(7,6,4,1) \\
2574 & 7722\;(7,7,2,2) \\
2745 & 8235\;(>8000) \\
2754 & 8262\;(>8000)
\end{array}
\]

Only \(2475\times3 = 7425\) uses the digits \(\{2,4,5,7\}\). Both 2475 and 7425 belong to the 24 numbers.  

- A factor larger than 3 would make the product exceed 8000 for any number in the set, so no further multiples are possible.

**Step 4 – Verify uniqueness**  

No other 2 xxx number gives a valid product, and no 4 xxx or 5 xxx number can be doubled or tripled to stay within the allowed range. Thus the only multiple relation among the 24 numbers is  

\[
2475 \times 3 = 7425 .
\]

The number that **is** a multiple is 7425.

**Step 5 – Select the answer**  

Among the choices, 7425 corresponds to option **D**.

ANSWER: D

actly once. We need to find out which pair of these numbers has one number as a multiple of the other.
2. **Set up the math and understand the concept**:
   - Let the two four - digit numbers be $a$ and $b$ such that $a = k\times b$, where $k$ is a positive integer greater than 1. Since $a$ and $b$ are four - digit numbers formed by 2, 4, 5, and 7, we know that $1000\leqslant a,b\leqslant9999$. Also, $k=\frac{a}{b}$. Since $a$ and $b$ are formed by the same set of digits, we can start by looking at the divisibility rules and the relative magnitudes of the numbers.
   - A useful approach is to consider the fact that if $a = k\times b$, then the ratio of the two numbers should be a simple fraction. We can start by dividing each number by the others among the given options to check for integer ratios.
3. **Do the computation**:
   - Let's start by dividing the numbers pairwise.
   - First, consider option A: If we take 5724 and try to divide it by the other numbers.
     - $\frac{5724}{7245}\approx0.79$, $\frac{5724}{7254}\approx0.79$, $\frac{5724}{7425}\approx0.77$, $\frac{5724}{7542}\approx0.76$. So, 5724 is not a multiple of any of the other numbers.
   - Consider option B: If we take 7245 and divide it by the other numbers.
     - $\frac{7245}{5724}\approx1.26$, $\frac{7245}{7254}\approx0.999$, $\frac{7245}{7425}\approx0.976$, $\frac{7245}{7542}\approx0.961$. So, 7245 is not a multiple of any of the other numbers.
   - Consider option C: If we take 7254 and divide it by the other numbers.
     - $\frac{7254}{5724}\approx1.27$, $\frac{7254}{7245}\approx1.001$, $\frac{7254}{7425}\approx0.977$, $\frac{7254}{7542}\approx0.962$. So, 7254 is not a multiple of any of the other numbers.
   - Consider option D: If we take 7425 and divide it by 5724, $\frac{7425}{5724}\approx1.297$. If we divide 7425 by 7245, $\frac{7425}{7245}=\frac{7425\div45}{7245\div45}=\frac{165}{161}\approx1.025$. If we divide 7425 by 7254, $\frac{7425}{7254}=\frac{7425\div9}{7254\div9}=\frac{825}{806}\approx1.024$. But if we divide 7425 by 2475 (we can re - arrange the digits to think about possible factors, and note that $7425 = 3\times3\times3\times5\times5\times11$ and $2475=3\times3\times5\times5\times11$, and also we can check $7425\div297 = 25$ and by brute - force checking the numbers formed by 2,4,5,7 we find that $7425\div297$ is not relevant. However, $7425\div2475 = 3$ and we can form 2475 using the digits 2, 4, 5, 7. In terms of our given options, we can also note that $7425\div2475$ conceptually shows that among the numbers formed by 2, 4, 5, 7, we can check the ratios directly. If we divide 7425 by 2475 (re - arranging digits), and also check $7425\div 297$ (not relevant as 297 is not formed by 2, 4, 5, 7). But if we calculate $\frac{7425}{2475}$ in terms of our set of numbers, we find that $7425 = 3\times2475$ and when we consider the numbers formed by 2, 4, 5, 7, we know that $7425\div2475$ implies that if we consider the numbers in our list, we can check:
       - $\frac{7425}{2475}$ shows that 7425 is a multiple. Among our options, we can also calculate $\frac{7425}{2475}$ in a more brute - force way with the given numbers. We know that $7425\div 2475 = 3$. And if we check the other way around, $2475\times3=7425$. Since we are working with the set of numbers formed by 2, 4, 5, 7, we can confirm that 7425 is a multiple of a number formed by the same digits.
   - Consider option E: If we take 7542 and divide it by the other numbers.
     - $\frac{7542}{5724}\approx1.32$, $\frac{7542}{7245}\approx1.04$, $\frac{7542}{7254}\approx1.04$, $\frac{7542}{7425}\approx1.016$. So, 7542 is not a multiple of any of the other numbers.
4. **Rule out the wrong choices**:
   - By performing the above division operations for each option, we have shown that options A, B, C, and E do not satisfy the condition that one number is a multiple of another among the set of four - digit numbers formed by 2, 4, 5, and 7. Only option D (7425) meets the requirement.

ANSWER: D

### Step 1: Restate the Problem
We have all 24 permutations of the four digits 2, 4, 5, 7, which form unique four-digit numbers. One of these numbers is an integer multiple of another permutation from the set. We need to identify which of the given options is this larger multiple.
---
### Step 2: Narrow possible multipliers
The smallest permutation is 2457, the largest is 7542. For two distinct permutations $N = k \cdot M$ (where $N$ is the larger multiple, $M$ the smaller number, $k \geq 2$ integer):
- $4 \cdot 2457 = 9828$, which is a 5-digit number exceeding the maximum permutation 7542.  $k$ can only be 2 or 3.
---
### Step 3: Test $k=3$ first (only feasible small range for $M$)
If $k=3$, then $M < 7542/3 \approx 2514$. The only permutations starting with 2 (the only possible leading digit for $M$ to stay under 2514) are 2457 and 2475:
- $3 \cdot 2457 = 7371$, which uses invalid digits 3 and 1, not in the set {2,4,5,7}.
- $3 \cdot 2475 = 7425$, which uses exactly the digits 7,4,2,5 — all four required digits, no repeats.
---
### Step 4: Rule out $k=2$
For $k=2$, $M < 7542/2 \approx 3771$, so $M$ must be a 2xxx permutation. Testing all 2xxx permutations:
- $2 \cdot 2457 = 4914$, $2 \cdot 2475 = 4950$, $2 \cdot 2547 = 5094$, etc. All products include invalid digits (0,1,8,9) not in the original set,  valid pairs exist for $k=2$.
---
### Step 5: Eliminate wrong choices
- A. 5724: $5724/2 = 2862$ (invalid digits, duplicates), $5724/3=1908$ (invalid digits)
- B. 7245: $7245/3=2415$ (has invalid digit 1)
- C. 7254: $7254/3=2418$ (invalid digits 1,8)
- E. 7542: $7542/3=2514$ (invalid digit 1)
Only 7425 is a valid multiple of another permutation in the set.
ANSWER: D

### Step 1: Restate the problem and key constraints
We need to find which 4-digit permutation of 2,4,5,7 is an integer multiple of a *different* permutation of these digits. There are 4! = 24 total permutations, all between the minimum value 2457 and maximum value 7542. For a number \(N\) to be a multiple of another \(M\), \(N = kM\) where \(k\) is an integer ≥2 (since \(N \neq M\)).
---
### Step 2: Bound the possible values of \(k\)
The smallest possible \(M\) is 2457. Calculate the maximum possible \(k\):
- \(4 \times 2457 = 9828\), which is larger than the maximum permutation 7542,  \(k\) can only be 2 or 3.
---
### Step 3: Use divisibility and bounds to narrow candidates
All permutations have digit sum \(2+4+5+7=18\), which is divisible by 9,  all numbers are divisible by 9. For \(k=3\), \(N=3M\) requires \(M = N/3\). For \(k=2\), \(N=2M\) requires \(N\) even (last digit 2 or 4) and \(M=N/2\).
Additionally, \(3 \times 2457 = 7371\),  any \(N < 7371\) cannot be \(3M\) (since \(M\) would need to be smaller than 2457, the minimum permutation). This means:
- Numbers <7371 can only have \(k=2\),  they must be even.
- Numbers ≥7371 can have \(k=2\) or \(k=3\).
---
### Step 4: Evaluate each option
#### Option A: 5724 (even, <7371)
Check \(k=2\): \(5724 / 2 = 2862\). Digits are 2,8,6,2, which are not a permutation of 2,4,5,7. Eliminate A.
#### Option B:7245 (odd, <7371)
Odd numbers cannot be \(2M\) (since \(2M\) is even), and \(N<7371\) rules out \(k=3\). Eliminate B.
#### Option C:7254 (even, <7371)
Check \(k=2\): \(7254 / 2 = 3627\). Digits are 3,6,2,7, which are not a permutation of 2,4,5,7. Eliminate C.
#### Option D:7425 (odd, ≥7371)
Odd,  \(k=2\) is impossible. Check \(k=3\): \(7425 / 3 = 2475\). Digits are 2,4,7,5, which is exactly a permutation of 2,4,5,7. This works.
#### Option E:7542 (even, ≥7371)
Check \(k=2\): \(7542 / 2 = 3771\). Digits are 3,7,7,1, invalid. Check \(k=3\): \(7542 /3 =2514\). Digits are 2,5,1,4, which includes a 1 (not allowed). Eliminate E.
---
### Step 5: Verify the valid pair
\(2475 \times 3 = 7425\), and both are valid permutations of 2,4,5,7. No other options satisfy the condition.
ANSWER: D

  - 7245 ÷ 2475 ≈ 2.92 (not an integer)
  - 7245 ÷ 2547 ≈ 2.84 (not an integer)
  - 7245 ÷ 2574 ≈ 2.81 (not an integer)
  - 7245 ÷ 2745 ≈ 2.64 (not an integer)
  - 7245 ÷ 2754 ≈ 2.63 (not an integer)
  - 7245 ÷ 4257 ≈ 1.70 (not an integer)
  - 7245 ÷ 4275 ≈ 1.69 (not an integer)
  - 7245 ÷ 4527 ≈ 1.60 (not an integer)
  - 7245 ÷ 4572 ≈ 1.58 (not an integer)
  - 7245 ÷ 4725 ≈ 1.53 (not an integer)
  - 7245 ÷ 4752 ≈ 1.52 (not an integer)
  - 7245 ÷ 5247 ≈ 1.38 (not an integer)
  - 7245 ÷ 5274 ≈ 1.37 (not an integer)
  - 7245 ÷ 5427 ≈ 1.33 (not an integer)
  - 7245 ÷ 5472 ≈ 1.32 (not an integer)
  - 7245 ÷ 5724 ≈ 1.26 (not an integer)
  - 7245 ÷ 5742 ≈ 1.26 (not an integer)
  - 7245 ÷ 7425 ≈ 0.97 (not an integer)
  - 7245 ÷ 7452 ≈ 0.97 (not an integer)
  - 7245 ÷ 7524 ≈ 0.96 (not an integer)
  - 7245 ÷ 7542 ≈ 0.96 (not an integer)

- For 7254:
  - 7254 ÷ 2457 ≈ 2.95 (not an integer)
  - 7254 ÷ 2475 ≈ 2.93 (not an integer)
  - 7254 ÷ 2547 ≈ 2.85 (not an integer)
  - 7254 ÷ 2574 ≈ 2.82 (not an integer)
  - 7254 ÷ 2745 ≈ 2.64 (not an integer)
  - 7254 ÷ 2754 ≈ 2.63 (not an integer)
  - 7254 ÷ 4257 ≈ 1.70 (not an integer)
  - 7254 ÷ 4275 ≈ 1.70 (not an integer)
  - 7254 ÷ 4527 ≈ 1.60 (not an integer)
  - 7254 ÷ 4572 ≈ 1.59 (not an integer)
  - 7254 ÷ 4725 ≈ 1.53 (not an integer)
  - 7254 ÷ 4752 ≈ 1.53 (not an integer)
  - 7254 ÷ 5247 ≈ 1.38 (not an integer)
  - 7254 ÷ 5274 ≈ 1.37 (not an integer)
  - 7254 ÷ 5427 ≈ 1.34 (not an integer)
  - 7254 ÷ 5472 ≈ 1.33 (not an integer)
  - 7254 ÷ 5724 ≈ 1.27 (not an integer)
  - 7254 ÷ 5742 ≈ 1.26 (not an integer)
  - 7254 ÷ 7425 ≈ 0.98 (not an integer)
  - 7254 ÷ 7452 ≈ 0.97 (not an integer)
  - 7254 ÷ 7524 ≈ 0.96 (not an integer)
  - 7254 ÷ 7542 ≈ 0.96 (not an integer)

- For 7425:
  - 7425 ÷ 2457 ≈ 3.02 (not an integer)
  - 7425 ÷ 2475 ≈ 3.00 (integer)
  - 7425 ÷ 2547 ≈ 2.91 (not an integer)
  - 7425 ÷ 2574 ≈ 2.88 (not an integer)
  - 7425 ÷ 2745 ≈ 2.70 (not an integer)
  - 7425 ÷ 2754 ≈ 2.69 (not an integer)
  - 7425 ÷ 4257 ≈ 1.74 (not an integer)
  - 7425 ÷ 4275 ≈ 1.74 (not an integer)
  - 7425 ÷ 4527 ≈ 1.64 (not an integer)
  - 7425 ÷ 4572 ≈ 1.62 (not an integer)
  - 7425 ÷ 4725 ≈ 1.57 (not an integer)
  - 7425 ÷ 4752 ≈ 1.56 (not an integer)
  - 7425 ÷ 5247 ≈ 1.41 (not an integer)
  - 7425 ÷ 5274 ≈ 1.41 (not an integer)
  - 7425 ÷ 5427 ≈ 1.37 (not an integer)
  - 7425 ÷ 5472 ≈ 1.36 (not an integer)
  - 7425 ÷ 5724 ≈ 1.29 (not an integer)
  - 7425 ÷ 5742 ≈ 1.29 (not an integer)
  - 7425 ÷ 7245 ≈ 1.02 (not an integer)
  - 7425 ÷ 7254 ≈ 1.02 (not an integer)
  - 7425 ÷ 7524 ≈ 0.99 (not an integer)
  - 7425 ÷ 7542 ≈ 0.98 (not an integer)

- For 7542:
  - 7542 ÷ 2457 ≈ 3.07 (not an integer)
  - 7542 ÷ 2475 ≈ 3.05 (not an integer)
  - 7542 ÷ 2547 ≈ 2.96 (not an integer)
  - 7542 ÷ 2574 ≈ 2.93 (not an integer)
  - 7542 ÷ 2745 ≈ 2.75 (not an integer)
  - 7542 ÷ 2754 ≈ 2.74 (not an integer)
  - 7542 ÷ 4257 ≈ 1.77 (not an integer)
  - 7542 ÷ 4275 ≈ 1.76 (not an integer)
  - 7542 ÷ 4527 ≈ 1.67 (not an integer)
  - 7542 ÷ 4572 ≈ 1.65 (not an integer)
  - 7542 ÷ 4725 ≈ 1.60 (not an integer)
  - 7542 ÷ 4752 ≈ 1.59 (not an integer)
  - 7542 ÷ 5247 ≈ 1.44 (not an integer)
  - 7542 ÷ 5274 ≈ 1.43 (not an integer)
  - 7542 ÷ 5427 ≈ 1.39 (not an integer)
  - 7542 ÷ 5472 ≈ 1.38 (not an integer)
  - 7542 ÷ 5724 ≈ 1.32 (not an integer)
  - 7542 ÷ 5742 ≈ 1.31 (not an integer)
  - 7542 ÷ 7245 ≈ 1.04 (not an integer)
  - 7542 ÷ 7254 ≈ 1.04 (not an integer)
  - 7542 ÷ 7425 ≈ 1.02 (not an integer)
  - 7542 ÷ 7452 ≈ 1.01 (not an integer)

**Step 5: Identify the correct answer**
From the divisions, we see that 7425 ÷ 2475 = 3, which is an integer. Therefore, 7425 is a multiple of 2475.

**Step 6: Rule out the wrong choices**
- A. 5724: Not a multiple of any other number in the list.
- B. 7245: Not a multiple of any other number in the list.
- C. 7254: Not a multiple of any other number in the list.
- D. 7425: Is a multiple of 2475.
- E. 7542: Not a multiple of any other number in the list.

**Final answer**
ANSWER: D

other permutation.
- Check if any other permutation is divisible by 7425.
- 7425 is large, so unlikely to be a divisor of another permutation.
- Check divisibility of 7425 by other permutations:
  - 7425 ÷ 2457 ≈ 3.02 (close to 3, but not exact)
  - 7425 ÷ 2547 ≈ 2.91 (not an integer)
  - No obvious divisors found.
- **Conclusion:** Not the correct choice.

#### **Choice E: 7542**
- Check if 7542 is divisible by any other permutation.
- Check if any other permutation is divisible by 7542.
- 7542 is large, so unlikely to be a divisor of another permutation.
- Check divisibility of 7542 by other permutations:
  - 7542 ÷ 2457 ≈ 3.07 (not an integer)
  - 7542 ÷ 2547 ≈ 2.96 (not an integer)
  - 7542 ÷ 3721 ≈ 2.03 (not an integer, and 3721 is not a permutation of 2,4,5,7)
  - Wait, let's check if 7542 is a multiple of another permutation:
    - 7542 ÷ 2514 ≈ 2.999 (close to 3, but 2514 is not a permutation of 2,4,5,7)
    - 7542 ÷ 3771 = 2 (but 3771 is not a permutation of 2,4,5,7)
    - Hmm, this seems not promising.
- **Wait, let's try the other way: is 7542 a multiple of another permutation?**
  - Let's list some permutations and see if 7542 is divisible by them:
    - 7542 ÷ 2457 ≈ 3.07 (not an integer)
    - 7542 ÷ 2547 ≈ 2.96 (not an integer)
    - 7542 ÷ 2745 ≈ 2.75 (not an integer)
    - 7542 ÷ 2754 ≈ 2.74 (not an integer)
    - 7542 ÷ 3721 ≈ 2.03 (not an integer)
  - **But wait, let's check if another permutation is a multiple of 7542:**
    - The largest permutation is 7542, so no permutation is larger than 7542. Thus, 7542 cannot be a divisor of another permutation.
- **Wait, perhaps I made a mistake. Let's try a different approach.**

---

### **Step 5: Alternative Approach**
Instead of checking each choice, let's find all 24 permutations and look for a pair where one is a multiple of the other.

However, since listing all 24 is tedious, let's look for a pair where one number is double or triple another number.

- The digits are 2, 4, 5, 7. The sum of digits is 2 + 4 + 5 + 7 = 18, so all permutations are divisible by 9.
- Since all numbers are divisible by 9, it's possible that one is a multiple of another.

Let's look for numbers where one is double another:
- The smallest permutation is 2457, and the largest is 7542.
- Double of 2457 is 4914 (not a permutation of 2,4,5,7).
- Double of 2475 is 4950 (not a permutation).
- Double of 2547 is 5094 (not a permutation).
- Double of 2574 is 5148 (not a permutation).
- Double of 2745 is 5490 (not a permutation).
- Double of 2754 is 5508 (not a permutation).
- Double of 4257 is 8514 (not a permutation).
- Double of 4275 is 8550 (not a permutation).
- Double of 4527 is 9054 (not a permutation).
- Double of 4572 is 9144 (not a permutation).
- Double of 4725 is 9450 (not a permutation).
- Double of 4752 is 9504 (not a permutation).
- Double of 5247 is 10494 (too large).
- It seems no permutation is double another.

Now, let's check for triple:
- Triple of 2457 is 7371 (not a permutation).
- Triple of 2475 is 7425 (this is a permutation of 2,4,5,7!).
- Let's verify: 2475 * 3 = 7425.
- Both 2475 and 7425 are permutations of 2,4,5,7.
- Check if 2475 is in the list of permutations: Yes, it is (2475).
- Check if 7425 is in the list of permutations: Yes, it is (Choice D).

Thus, 7425 is a multiple of 2475, and both are permutations of the digits 2,4,5,7.

---

### **Step 6: Verify uniqueness**
The problem states that only one such pair exists. We found that 7425 is a multiple of 2475. To ensure uniqueness, let's check if there are other such pairs:
- Check if any other permutation is triple another:
  - Triple of 2547 is 7641 (not a permutation).
  - Triple of 2574 is 7722 (not a permutation).
  - Triple of 2745 is 8235 (not a permutation).
  - Triple of 2754 is 8262 (not a permutation).
  - Triple of 4257 is 12771 (too large).
- No other such pairs found.

---

### **Step 7: Match with choices**
The pair we found is 2475 and 7425. Among the choices, 7425 is Choice D.

---

**ANSWER: D**

I need to find the 98th term of the sequence beginning 98, 49, ... by following these rules:
- If n < 10: multiply by 9
- If n ≥ 10 and even: divide by 2
- If n ≥ 10 and odd: subtract 5

**Step 1: Build the sequence from the start**

Term 1: 98 (given)
Term 2: 49 (given)

Let me continue:

Term 3: 49 is odd and > 9 → 49 - 5 = 44
Term 4: 44 is even and > 9 → 44 ÷ 2 = 22
Term 5: 22 is even and > 9 → 22 ÷ 2 = 11
Term 6: 11 is odd and > 9 → 11 - 5 = 6
Term 7: 6 is < 10 → 6 × 9 = 54
Term 8: 54 is even and > 9 → 54 ÷ 2 = 27
Term 9: 27 is odd and > 9 → 27 - 5 = 22
Term 10: 22 is even and > 9 → 22 ÷ 2 = 11
Term 11: 11 is odd and > 9 → 11 - 5 = 6
Term 12: 6 is < 10 → 6 × 9 = 54
Term 13: 54 is even and > 9 → 54 ÷ 2 = 27

**Step 2: Identify the cycle**

Looking at the terms:
- Terms 1-6: 98, 49, 44, 22, 11, 6
- Term 7: 54
- Term 8: 27
- Term 9: 22
- Term 10: 11
- Term 11: 6
- Term 12: 54
- Term 13: 27

Starting from term 7, I see a repeating pattern: 54, 27, 22, 11, 6, 54, 27, 22, 11, 6, ...

The cycle is: **54 → 27 → 22 → 11 → 6** (length 5)

**Step 3: Find where term 98 falls**

Terms 1-6 are the initial sequence before the cycle starts.
Starting from term 7, we have a repeating cycle of length 5.

For term n where n ≥ 7:
- Position in cycle = (n - 7) mod 5
- If remainder is 0, it's the 5th element: 6
- If remainder is 1, it's the 1st element: 54
- If remainder is 2, it's the 2nd element: 27
- If remainder is 3, it's the 3rd element: 22
- If remainder is 4, it's the 4th element: 11

For term 98:
98 - 7 = 91
91 ÷ 5 = 18 remainder 1

So the remainder is 1, which means term 98 is at position 1 of the cycle.

Position 1 of the cycle is: **54**

**Step 4: Verify with another approach**

Term 7 = 54
Term 12 = 54 (after 5 terms)
Term 17 = 54 (after 5 terms)
...
Term 7 + 5k = 54

We need 7 + 5k = 98
5k = 91
k = 18.2

So term 98 is not exactly at a start of a cycle. Let me recalculate:

For term 98, I need (98 - 7) mod 5 = 91 mod 5 = 1

The cycle starting at term 7: 54(pos 1), 27(pos 2), 22(pos 3), 11(pos 4), 6(pos 5), 54(pos 1), ...

Position 1 (remainder 1) = 54

ANSWER: E

1. Restate: Start sequence with 98, 49, ... apply rules. Find 98th term.

2. Build sequence:
- Term 1: 98 (even >9 → /2 = 49)
- Term 2: 49 (odd >9 → -5 = 44)
- Term 3: 44 (even >9 → /2 = 22)
- Term 4: 22 (even >9 → /2 = 11)
- Term 5: 11 (odd >9 → -5 = 6)
- Term 6: 6 (<10 → ×9 = 54)
- Term 7: 54 (even >9 → /2 = 27)
- Term 8: 27 (odd >9 → -5 = 22)
- Term 9: 22 → 11
- Term 10: 11 → 6
- Term 11: 6 → 54
- Term 12: 54 → 27
- Term 13: 27 → 22

3. Detect cycle: From term 3 onward: 22, 11, 6, 54, 27, then repeats 22... 
Terms 3,4,5,6,7 = 22,11,6,54,27. Term 8 = 22 again. So cycle length 5 starting at term 3.

4. Find term 98: For n≥3, position in cycle = (n-3) mod 5.
n=98: (98-3) = 95, 95 mod 5 = 0 → first element of cycle = 22.

5. Check rule-out: cycle positions: 0→22, 1→11, 2→6, 3→54, 4→27. So 22.

6. Eliminate others: 11 (pos1), 6(pos2), 54(pos3), 27(pos4) — all wrong positions.

ANSWER: C

## Step-by-Step Solution

### Step 1: Restate the Problem
Starting with 98, 49, ..., I need to find the 98th term using these rules:
- If the integer < 10: multiply by 9
- If even and > 9: divide by 2
- If odd and > 9: subtract 5

### Step 2: Generate the Sequence
Let me build the sequence and look for a cycle.

| Term # | Value | Rule Applied |
|--------|-------|--------------|
| 1 | 98 | even > 9: ÷2 |
| 2 | 49 | odd > 9: −5 |
| 3 | 44 | even > 9: ÷2 |
| 4 | 22 | even > 9: ÷2 |
| 5 | 11 | odd > 9: −5 |
| 6 | 6 | < 10: ×9 |
| 7 | 54 | even > 9: ÷2 |
| 8 | 27 | odd > 9: −5 |
| 9 | 22 | even > 9: ÷2 |
| 10 | 11 | odd > 9: −5 |
| 11 | 6 | < 10: ×9 |
| 12 | 54 | even > 9: ÷2 |
| 13 | 27 | ... |

### Step 3: Identify the Cycle
Starting from term 4: **22, 11, 6, 54, 27, 22, 11, 6, 54, 27, ...**

The cycle is: **22 → 11 → 6 → 54 → 27 → 22 → ...** with **period 5**, beginning at term 4.

### Step 4: Find the 98th Term
Terms 4, 9, 14, 19, ... (i.e., terms of the form 4 + 5k) all equal **22**.

I need to find where term 98 falls in the cycle. Subtract term 4's position:
$$98 - 4 = 94$$

Divide 94 by the cycle length 5:
$$94 = 5 \times 18 + 4$$

So term 98 is 4 positions after a "22" in the cycle.

Cycle starting at position 0: 22(+0), 11(+1), 6(+2), 54(+3), 27(+4)

Position +4 in the cycle = **27**

### Step 5: Verify
- Term 4: 22 (offset 0)
- Term 8: 27 (offset 4) ✓ — matches our table above
- Term 98: offset = (98−4) mod 5 = 94 mod 5 = 4 → **27** ✓

### Step 6: Rule Out Wrong Answers
- A (6): offset 2 — would be term 6, 11, 16, ...
- B (11): offset 1 — would be term 5, 10, 15, ...
- C (22): offset 0 — would be term 4, 9, 14, ...
- **D (27): offset 4 — term 8, 13, 18, ..., 98 ✓**
- E (54): offset 3 — would be term 7, 12, 17, ...

ANSWER: D

1. **Restate the problem.**  
   We start with the sequence:
   - 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.

   The sequence begins:
   \[
   98,\ 49,\ \dots
   \]
   We need the **98th term**.

2. **Generate the pattern.**  
   Start with 98:
   - \(98\) is even and greater than 9, so next is \(98 \div 2 = 49\).
   - \(49\) is odd and greater than 9, so next is \(49 - 5 = 44\).
   - \(44\) is even and greater than 9, so next is \(44 \div 2 = 22\).
   - \(22\) is even and greater than 9, so next is \(22 \div 2 = 11\).
   - \(11\) is odd and greater than 9, so next is \(11 - 5 = 6\).
   - \(6\) is less than 10, so next is \(6 \cdot 9 = 54\).
   - \(54\) is even and greater than 9, so next is \(54 \div 2 = 27\).
   - \(27\) is odd and greater than 9, so next is \(27 - 5 = 22\).

3. **Look for a repeating cycle.**  
   From \(22\) onward:
   \[
   22,\ 11,\ 6,\ 54,\ 27,\ 22,\dots
   \]
   So the cycle is:
   \[
   22,\ 11,\ 6,\ 54,\ 27
   \]
   This cycle has length 5.

4. **Find where the 98th term falls in the cycle.**  
   Since the 3rd term is \(44\), let’s list terms from the 3rd onward:
   \[
   3:44,\ 4:22,\ 5:11,\ 6:6,\ 7:54,\ 8:27,\ 9:22,\dots
   \]
   After the 3rd term, the 5-term cycle repeats:
   \[
   44,\ 22,\ 11,\ 6,\ 54,\ 27
   \]
   Better to use the cycle starting at term 4:
   \[
   4:22,\ 5:11,\ 6:6,\ 7:54,\ 8:27
   \]
   Then:
   - term 9 is \(22\)
   - term 10 is \(11\), etc.

   The cycle length is 5, so we compute:
   \[
   98 - 4 = 94
   \]
   and
   \[
   94 \div 5 = 18 \text{ remainder } 4.
   \]
   That means the 98th term is the 4th term in the cycle \(22, 11, 6, 54, 27\), which is:
   \[
   54
   \]

5. **Check against the answer choices.**  
   The choices are:
   - A. 6
   - B. 11
   - C. 22
   - D. 27
   - E. 54

   Our result is \(54\), which matches **E**.

ANSWER: E

An elegant way to solve this problem is to generate the first few terms of the sequence, identify a repeating cycle, and then use the cycle to find the 98th term.

### Step 1: Restate the rules and the goal
We are given a sequence starting with $a_1 = 98$ and $a_2 = 49$. The rules to find the next term $a_{n+1}$ from $a_n$ are:
1. If $a_n < 10$, then $a_{n+1} = 9 \times a_n$.
2. If $a_n > 9$ and $a_n$ is even, then $a_{n+1} = a_n / 2$.
3. If $a_n > 9$ and $a_n$ is odd, then $a_{n+1} = a_n - 5$.

We need to find the 98th term, $a_{98}$.

### Step 2: Compute the first few terms
Let's calculate the terms starting from $a_1$:
* $a_1 = 98$ (given)
* $a_2 = 49$ (since $98 > 9$ and is even, $98 / 2 = 49$)
* $a_3 = 49 - 5 = 44$ (since $49 > 9$ and is odd)
* $a_4 = 44 / 2 = 22$ (since $44 > 9$ and is even)
* $a_5 = 22 / 2 = 11$ (since $22 > 9$ and is even)
* $a_6 = 11 - 5 = 6$ (since $11 > 9$ and is odd)
* $a_7 = 6 \times 9 = 54$ (since $6 < 10$)
* $a_8 = 54 / 2 = 27$ (since $54 > 9$ and is even)
* $a_9 = 27 - 5 = 22$ (since $27 > 9$ and is odd)
* $a_{10} = 22 / 2 = 11$

### Step 3: Identify the repeating cycle
We notice that $a_9 = 22$, which is the same as $a_4 = 22$. Since the rules are deterministic, the sequence will now repeat the same values in a cycle.
The repeating cycle is:
$$\mathbf{22, 11, 6, 54, 27}$$

This cycle has a length of $5$ terms:
* $a_4 = 22$
* $a_5 = 11$
* $a_6 = 6$
* $a_7 = 54$
* $a_8 = 27$

And then it repeats:
* $a_9 = 22$
* $a_{10} = 11$
* $a_{11} = 6$
* $a_{12} = 54$
* $a_{13} = 27$

### Step 4: Compute the 98th term
Since the cycle starts at $a_4$ and repeats every 5 terms, we can find the value of $a_{98}$ by looking at the index relative to the start of the cycle:
$$98 - 4 = 94$$

Now, we divide $94$ by the cycle length $5$:
$$94 \div 5 = 18 \text{ with a remainder of } 4$$

This means that after $18$ full cycles starting from $a_4$, we go $4$ more steps into the next cycle:
* $0$ steps after $a_4$: $a_{4 + 5(18)} = a_{94} = 22$
* $1$ step after: $a_{95} = 11$
* $2$ steps after: $a_{96} = 6$
* $3$ steps after: $a_{97} = 54$
* $4$ steps after: $a_{98} = 27$

Thus, the 98th term is $27$.

### Step 5: Match with the choices
* A. 6 (This is $a_{96}$)
* B. 11 (This is $a_{95}$)
* C. 22 (This is $a_{94}$)
* D. 27 (This is $a_{98}$)
* E. 54 (This is $a_{97}$)

The correct choice is D.

ANSWER: D

The sequence begins with \(a_1 = 98\) and \(a_2 = 49\). The given rules are applied iteratively to generate subsequent terms as follows:

- \(a_3 = 49 - 5 = 44\)
- \(a_4 = 44 / 2 = 22\)
- \(a_5 = 22 / 2 = 11\)
- \(a_6 = 11 - 5 = 6\)
- \(a_7 = 6 \times 9 = 54\)
- \(a_8 = 54 / 2 = 27\)
- \(a_9 = 27 - 5 = 22\)

From \(a_4 = 22\) onward, the sequence enters the repeating cycle \(22, 11, 6, 54, 27\) of length 5. To confirm the cycle:

- \(22 / 2 = 11\)
- \(11 - 5 = 6\)
- \(6 \times 9 = 54\)
- \(54 / 2 = 27\)
- \(27 - 5 = 22\)

and the pattern repeats. Thus, for all \(n \geq 4\), the terms satisfy the periodic relation with period 5. The position within the cycle is given by the remainder \(r = (n-4) \pmod{5}\), which maps to the cycle values as follows:

- \(r = 0 \to 22\)
- \(r = 1 \to 11\)
- \(r = 2 \to 6\)
- \(r = 3 \to 54\)
- \(r = 4 \to 27\)

For \(n = 98\), compute \(r = (98-4) \pmod{5} = 94 \pmod{5}\). Since \(94 = 5 \times 18 + 4\), it follows that \(r = 4\), so \(a_{98} = 27\).

This rules out the other choices, as none of \(6, 11, 22, 54\) occurs at a position congruent to 4 modulo 5 in the cycle.

ANSWER: D

1. The problem asks 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. To find the next term, we apply the rules: since 98 is even and greater than 9, we divide it by 2 to get 49.

3. Now we are at 49. Since 49 is odd and greater than 9, we subtract 5 to get 44.

4. Continuing this process:
   - 44 is even and greater than 9, so we divide by 2 to get 22.
   - 22 is even and greater than 9, so we divide 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 by 9 to get 54.
   - 54 is even and greater than 9, so we divide 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, and 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, ... after the term 44.

6. Let's list the terms with their positions:
   - 1st term: 98
   - 2nd term: 49
   - 3rd term: 44
   - 4th term: 22
   - 5th term: 11
   - 6th term: 6
   - 7th term: 54
   - 8th term: 27
   - 9th term: 22
   - The loop (22, 11, 6, 54, 27) starts from the 4th term and has a length of 5.

7. To find the 98th term, we first note that the sequence loops every 5 terms starting from the 4th term. 
   - The terms from the 4th onwards follow the loop: 22, 11, 6, 54, 27.

8. Since the loop starts at the 4th term, we need to find where the 98th term falls within this loop.
   - The position within the loop is given by (98 - 3) mod 5, because the loop starts from the 4th term. So, (98 - 3) = 95, and 95 mod 5 = 0.

9. Since 95 mod 5 = 0, it means the 98th term is the last term of the loop, which is 27.

10. Checking the given choices:
    - A. 6: This is the 6th term in the sequence, not the 98th.
    - B. 11: This is the 5th term, not the 98th.
    - C. 22: This is the 4th and 9th term, but the pattern repeats every 5 terms, so it's not the 98th.
    - D. 27: This matches our calculation for the 98th term.
    - E. 54: This is the 7th term, not the 98th.

ANSWER: D

1. **Understand the rules:**  
   - If the number 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.

2. **List the first several terms starting from 98:**  
   - Term 1: 98  
   - Term 2: 98 is even >9 → 98 ÷ 2 = 49  
   - Term 3: 49 is odd >9 → 49 − 5 = 44  
   - Term 4: 44 is even >9 → 44 ÷ 2 = 22  
   - Term 5: 22 is even >9 → 22 ÷ 2 = 11  
   - Term 6: 11 is odd >9 → 11 − 5 = 6  
   - Term 7: 6 < 10 → 6 × 9 = 54  
   - Term 8: 54 is even >9 → 54 ÷ 2 = 27  
   - Term 9: 27 is odd >9 → 27 − 5 = 22  
   - Term 10: 22 ÷ 2 = 11  
   - Term 11: 11 − 5 = 6  
   - Term 12: 6 × 9 = 54  
   - Term 13: 54 ÷ 2 = 27  
   - Term 14: 27 − 5 = 22

3. **Identify the cycle:**  
   - From term 4 onward, the values repeat in a cycle of length 5:  
     **22, 11, 6, 54, 27**  
   - Term 4 → 22 (1st in cycle)  
   - Term 5 → 11 (2nd)  
   - Term 6 → 6 (3rd)  
   - Term 7 → 54 (4th)  
   - Term 8 → 27 (5th)  
   - Term 9 → 22 (1st) and so on.

4. **Find the position of the 98th term in the cycle:**  
   - The cycle begins at term 4.  
   - Subtract 3 to shift the index: position in cycle = (98 − 3) = 95.  
   - Divide by cycle length 5: 95 ÷ 5 = 19 remainder 0.  
   - A remainder of 0 means it is the last (5th) element in the cycle.  
   - The 5th element is 27.  
   - (Double-check: term 8, 13, 18, ..., 98 are all 27 because 98−8=90 is a multiple of 5.)

5. **Match with the choices:**  
   - 27 corresponds to option D.  
   - The other options are parts of the cycle but not the 98th term: 6 (A), 11 (B), 22 (C), 54 (E) appear earlier or at different positions.

ANSWER: D

# Finding the 98th Term of Terri's Sequence

## Step 1: Restate the Problem
We start with 98 and apply these rules to generate each subsequent term:
- **Rule 1:** If the number < 10, multiply by 9.
- **Rule 2:** If the number is even and > 9, divide by 2.
- **Rule 3:** If the number is odd and > 9, subtract 5.

We need to find the 98th term.

## Step 2: Compute Terms Until a Cycle Appears

| Term | Value | Rule Applied |
|------|-------|-------------|
| 1 | 98 | even, >9 → ÷2 |
| 2 | 49 | odd, >9 → −5 |
| 3 | 44 | even, >9 → ÷2 |
| 4 | **22** | even, >9 → ÷2 |
| 5 | **11** | odd, >9 → −5 |
| 6 | **6** | <10 → ×9 |
| 7 | **54** | even, >9 → ÷2 |
| 8 | **27** | odd, >9 → −5 |
| 9 | **22** | even, >9 → ÷2 |
| 10 | 11 | (repeats) |

## Step 3: Identify the Cycle
Starting at **Term 4**, the sequence enters a repeating cycle of length **5**:
$$22, 11, 6, 54, 27, 22, 11, 6, 54, 27, \ldots$$

## Step 4: Find the 98th Term Using the Cycle
For any term $n \geq 4$, its position in the cycle is $(n - 4) \mod 5$:

$$98 - 4 = 94$$
$$94 \div 5 = 18 \text{ remainder } 4$$
$$(98 - 4) \mod 5 = 4$$

Mapping remainders to cycle values:
- Remainder 0 → 22
- Remainder 1 → 11
- Remainder 2 → 6
- Remainder 3 → 54
- **Remainder 4 → 27**

## Step 5: Rule Out Wrong Choices
- **A (6):** Would require remainder 2, but 94 mod 5 = 4, not 2.
- **B (11):** Would require remainder 1, but 94 mod 5 = 4, not 1.
- **C (22):** Would require remainder 0, but 94 mod 5 = 4, not 0.
- **E (54):** Would require remainder 3, but 94 mod 5 = 4, not 3.

Only **27** matches remainder 4.

ANSWER: D

**Step 1: Understand the problem.**  
We must find the 98th term of a sequence that begins 98, 49. The rule for producing the next term is:
- less than 10 → multiply by 9;
- even and greater than 9 → divide by 2;
- odd and greater than 9 → subtract 5.

**Step 2: Compute the first several terms explicitly.**  
- \(a_1 = 98\) (given)  
- \(a_2\): 98 is even and \(>9\) → \(98/2 = 49\)  
- \(a_3\): 49 is odd and \(>9\) → \(49-5 = 44\)  
- \(a_4\): 44 is even and \(>9\) → \(44/2 = 22\)  
- \(a_5\): 22 is even and \(>9\) → \(22/2 = 11\)  
- \(a_6\): 11 is odd and \(>9\) → \(11-5 = 6\)  
- \(a_7\): \(6<10\) → \(6\times 9 = 54\)  
- \(a_8\): 54 is even and \(>9\) → \(54/2 = 27\)  
- \(a_9\): 27 is odd and \(>9\) → \(27-5 = 22\)

**Step 3: Identify the repeating cycle.**  
Notice that \(a_9 = 22 = a_4\). Because each term depends only on the previous one, the sequence must now repeat indefinitely. The 5-term cycle is:

\[
22 \xrightarrow{/2} 11 \xrightarrow{-5} 6 \xrightarrow{\times 9} 54 \xrightarrow{/2} 27 \xrightarrow{-5} 22
\]

So for all \(n \ge 4\), the values repeat every 5 terms:
\[
(a_4, a_5, a_6, a_7, a_8) = (22, 11, 6, 54, 27)
\]

**Step 4: Locate the 98th term inside the cycle.**  
The cycle starts at term 4. To find where term 98 falls, compute the offset from term 4:
\[
98 - 4 = 94
\]
Now reduce modulo 5:
\[
94 \div 5 = 18\text{ remainder }4 \quad\Rightarrow\quad 94 \equiv 4 \pmod{5}
\]

An offset of 4 means term 98 corresponds to the same position in the cycle as term \(4+4 = 8\).

**Step 5: Read off the value.**  
From Step 2, \(a_8 = 27\). Equivalently, counting 4 steps forward in the cycle from 22:
- offset 0: 22
- offset 1: 11
- offset 2: 6
- offset 3: 54
- offset 4: 27

Hence the 98th term is **27**.

**Step 6: Rule out the wrong choices.**  
Since \(98 \equiv 3 \pmod{5}\), and for \(n\ge 4\) we have:
- \(n\equiv 4\pmod 5 \to 22\) (Choice C)
- \(n\equiv 0\pmod 5 \to 11\) (Choice B)
- \(n\equiv 1\pmod 5 \to 6\) (Choice A)
- \(n\equiv 2\pmod 5 \to 54\) (Choice E)
- \(n\equiv 3\pmod 5 \to 27\) (Choice D)

Because \(98 \equiv 3 \pmod 5\), only Choice D matches.

ANSWER: D

Here is the step-by-step solution:

**1. Understand the Rules and Goal**
We are given a sequence defined by specific transformation rules based on the value of the current integer:
*   **Rule 1:** If $x < 10$, the next term is $9x$.
*   **Rule 2:** If $x \ge 10$ and $x$ is even, the next term is $x/2$.
*   **Rule 3:** If $x \ge 10$ and $x$ is odd, the next term is $x-5$.

The sequence begins with 98. We need to find the value of the 98th term ($a_{98}$).

**2. Generate the Sequence Terms**
Let's compute the terms sequentially until we detect a repeating pattern (a cycle).

*   **Term 1:** 98
    *   Condition: $\ge 10$ and even. Apply Rule 2: $98 / 2 = 49$.
*   **Term 2:** 49
    *   Condition: $\ge 10$ and odd. Apply Rule 3: $49 - 5 = 44$.
*   **Term 3:** 44
    *   Condition: $\ge 10$ and even. Apply Rule 2: $44 / 2 = 22$.
*   **Term 4:** 22
    *   Condition: $\ge 10$ and even. Apply Rule 2: $22 / 2 = 11$.
*   **Term 5:** 11
    *   Condition: $\ge 10$ and odd. Apply Rule 3: $11 - 5 = 6$.
*   **Term 6:** 6
    *   Condition: $< 10$. Apply Rule 1: $6 \times 9 = 54$.
*   **Term 7:** 54
    *   Condition: $\ge 10$ and even. Apply Rule 2: $54 / 2 = 27$.
*   **Term 8:** 27
    *   Condition: $\ge 10$ and odd. Apply Rule 3: $27 - 5 = 22$.
*   **Term 9:** 22
    *   We have returned to 22, which was **Term 4**.

**3. Identify the Cycle**
The sequence from Term 4 onwards enters a repeating loop.
*   **Pre-cycle terms:**
    *   $a_1 = 98$
    *   $a_2 = 49$
    *   $a_3 = 44$
*   **Cyclic part (starts at Term 4):**
    The cycle consists of the numbers: **22, 11, 6, 54, 27**.
    The length of this cycle is **5**.

    Mapping the cycle positions to remainders (letting $k$ be the index $n$):
    *   $n=4$: 22 (Remainder 0 when offset by 4)
    *   $n=5$: 11 (Remainder 1)
    *   $n=6$: 6 (Remainder 2)
    *   $n=7$: 54 (Remainder 3)
    *   $n=8$: 27 (Remainder 4)
    *   $n=9$: 22 (Remainder 0) ...

**4. Compute the 98th Term**
We want to find $a_{98}$.
Since the cycle starts at index $n=4$, we can determine the position of the 98th term within the cycle using modular arithmetic.
We calculate how far index 98 is from the start of the cycle (index 4):
$$ \text{Offset} = 98 - 4 = 94 $$

Now, we find the remainder when this offset is divided by the cycle length (5):
$$ 94 \div 5 = 18 \text{ with a remainder of } 4 $$

A remainder of 4 means that $a_{98}$ corresponds to the **5th element** in our cycle list (using 0-based indexing for the remainder: 0$\to$1st, 1$\to$2nd, 2$\to$3rd, 3$\to$4th, 4$\to$5th).
The cycle list is: {22, 11, 6, 54, **27**}.
The 5th element is **27**.

**5. Evaluate Choices**
*   A. 6: This is the 3rd item in the cycle (corresponds to remainder 2). Incorrect.
*   B. 11: This is the 2nd item in the cycle (corresponds to remainder 1). Incorrect.
*   C. 22: This is the 1st item in the cycle (corresponds to remainder 0). Incorrect.
*   D. 27: This matches our calculated result for the 98th term. Correct.
*   E. 54: This is the 4th item in the cycle (corresponds to remainder 3). Incorrect.

ANSWER: D

**Step 1 – Understand the rules**

The sequence \(a_1,a_2,\dots\) is built by the rule  

\[
f(n)=\begin{cases}
9n &\text{if } n<10,\\[2mm]
\displaystyle\frac{n}{2} &\text{if } n\ge 10\text{ and }n\text{ is even},\\[2mm]
n-5 &\text{if } n\ge 10\text{ and }n\text{ is odd}.
\end{cases}
\]

We are told the sequence begins \(a_1=98,\;a_2=49,\dots\).

--------------------------------------------------------------------

**Step 2 – Generate the first few terms**

\(a_1=98\) (even, \(>9\))  
\(\displaystyle a_2=\frac{98}{2}=49\).

\(a_2=49\) (odd, \(>9\))  
\(\displaystyle a_3=49-5=44\).

\(a_3=44\) (even, \(>9\))  
\(\displaystyle a_4=\frac{44}{2}=22\).

\(a_4=22\) (even, \(>9\))  
\(\displaystyle a_5=\frac{22}{2}=11\).

\(a_5=11\) (odd, \(>9\))  
\(\displaystyle a_6=11-5=6\).

Now \(a_6=6\) is **< 10**, so we apply the first rule:

\[
a_7 = 9\cdot6 = 54.
\]

--------------------------------------------------------------------

**Step 3 – Discover the cycle**

Continue with the new rules:

\[
\begin{aligned}
a_8 &= \frac{54}{2}=27 \quad(\text{even, } >9)\\
a_9 &= 27-5=22 \quad(\text{odd, } >9)\\
a_{10}&= \frac{22}{2}=11 \quad(\text{even, } >9)\\
a_{11}&= 11-5=6 \quad(\text{odd, } >9).
\end{aligned}
\]

We are back at **6**, so the sequence will now repeat the five‑term cycle

\[
6,\;54,\;27,\;22,\;11,\;6,\;54,\dots
\]

--------------------------------------------------------------------

**Step 4 – Locate the 98‑th term inside the cycle**

The first term of the cycle is \(a_6=6\).  
For any term \(a_n\) with \(n\ge 6\),

\[
\text{position in cycle}= (n-6)\pmod{5}.
\]

Compute for \(n=98\):

\[
98-6 = 92,\qquad 92\div5 = 18\text{ remainder }2.
\]

Thus the position is \(2\) (counting \(0,1,2,3,4\) from the start of the cycle).  
The mapping inside the cycle is

\[
\begin{array}{c|c}
\text{position} & \text{value}\\\hline
0 & 6\\
1 & 54\\
2 & 27\\
3 & 22\\
4 & 11
\end{array}
\]

Position \(2\) corresponds to **27**.

--------------------------------------------------------------------

**Step 5 – Select the correct answer**

The 98‑th term is \(27\), which matches choice **D**.

--------------------------------------------------------------------

**Final answer**

ANSWER: D

1. **Restate what is being asked**:
   - We are given a sequence - building rule. Starting with the first term $a_1 = 98$, we need to apply the given rules to generate the sequence and find the 98th term. The rules are:
     - If $n\lt10$, then $a_{k + 1}=9n$ (where $n=a_k$).
     - If $n$ is even and $n > 10$, then $a_{k + 1}=\frac{n}{2}$.
     - If $n$ is odd and $n>10$, then $a_{k + 1}=n - 5$.
2. **Generate the sequence step - by - step**:
   - Given $a_1 = 98$. Since $98>10$ and is even, we use the rule $a_{k + 1}=\frac{n}{2}$. So, $a_2=\frac{98}{2}=49$.
   - Since $49>10$ and is odd, we use the rule $a_{k + 1}=n - 5$. So, $a_3=49 - 5 = 44$.
   - Since $44>10$ and is even, we use the rule $a_{k + 1}=\frac{n}{2}$. So, $a_4=\frac{44}{2}=22$.
   - Since $22>10$ and is even, we use the rule $a_{k + 1}=\frac{n}{2}$. So, $a_5=\frac{22}{2}=11$.
   - Since $11>10$ and is odd, we use the rule $a_{k + 1}=n - 5$. So, $a_6=11 - 5 = 6$.
   - Since $6\lt10$, we use the rule $a_{k + 1}=9n$. So, $a_7=9\times6 = 54$.
   - Since $54>10$ and is even, we use the rule $a_{k + 1}=\frac{n}{2}$. So, $a_8=\frac{54}{2}=27$.
   - Since $27>10$ and is odd, we use the rule $a_{k + 1}=n - 5$. So, $a_9=27 - 5 = 22$.
3. **Identify the cycle**:
   - We have found that $a_4 = 22$ and $a_9 = 22$. This means that the sequence is periodic after the 4th term. The cycle starts from the 4th term and the cycle length $T=9 - 4+1 = 6$.
4. **Adjust the term number to account for the non - cyclic part**:
   - We want to find the 98th term. The first 3 terms ($a_1,a_2,a_3$) are non - cyclic. So, we consider the position of the term within the cycle. We first subtract 3 from 98, getting $98-3 = 95$.
5. **Find the position within the cycle**:
   - Divide 95 by the cycle length 6: $95\div6 = 15\cdots\cdots5$.
   - This means that after 15 full cycles, we are at the 5th term of the next cycle.
6. **Determine the 98th term based on the cycle**:
   - The cycle is $\{22,11,6,54,27,22\}$. The 5th term of the cycle is 27.
7. **Rule out the wrong choices**:
   - Option A: 6 is the 6th term of the cycle, not the 98th term.
   - Option B: 11 is the 2nd term of the cycle, not the 98th term.
   - Option C: 22 is the 1st and 6th term of the cycle, not the 98th term.
   - Option E: 54 is the 4th term of the cycle, not the 98th term.

ANSWER: D

### Step 1: Restate the problem
We need to find the 98th term of a sequence starting with \(a_1=98\), \(a_2=49\), generated by these rules:
1. If the term is <10, multiply by 9
2. If the term is even and >9, divide by 2
3. If the term is odd and >9, subtract 5
The answer choices are A.6, B.11, C.22, D.27, E.54.

---
### Step 2: List sequence terms to identify the repeating cycle
Generate the sequence starting from the first term:
- \(a_1=98\), \(a_2=49\) (given)
- \(a_3=49-5=44\) (49 is odd >9)
- \(a_4=44/2=22\) (44 is even >9)
- \(a_5=22/2=11\) (22 is even >9)
- \(a_6=11-5=6\) (11 is odd >9)
- \(a_7=6*9=54\) (6 <10)
- \(a_8=54/2=27\) (54 is even >9)
- \(a_9=27-5=22\) (27 is odd >9)

From \(a_4\) onward, the sequence repeats the 5-term cycle: **[22, 11, 6, 54, 27]**. The cycle starts at index 4, and has length 5.

---
### Step3: Calculate the position of the 98th term in the cycle
The first term of the cycle is at index 4. For any index \(n \geq4\), the offset into the cycle is \((n-4)\). For \(n=98\):
- Offset = \(98-4=94\)
- \(94 \mod 5 = 4\) (since 5*18=90, 94-90=4)

This offset maps to the 5th element of the cycle (index 4 in 0-indexed cycle terms), which is 27.

---
### Step4: Eliminate incorrect options
- A.6: Occurs at indices 6,11,16... (never 98)
- B.11: Occurs at indices5,10,15... (never98)
- C.22: Occurs at indices4,9,14... (never98)
- E.54: Occurs at indices7,12,17... (never98)

Only D.27 is the correct 98th term.

ANSWER: D

### Step 1: Restate the problem and rules
We need to find the 98th term of a sequence starting with 98, 49, ..., where each term is generated by:
1.  If the term is <10: multiply by 9.
2.  If the term is even and >9: divide by 2.
3.  If the term is odd and >9: subtract 5.
---
### Step 2: Generate initial terms to identify cycles
We list terms sequentially, starting with term 1 = 98, term 2 = 49 (given):
- Term 1: 98 (even >9) → Term 2 = 98/2 = 49
- Term 2: 49 (odd >9) → Term3 = 49-5 = 44
- Term3:44 (even>9) → Term4=44/2=22
- Term4:22 (even>9) → Term5=22/2=11
- Term5:11 (odd>9) → Term6=11-5=6
- Term6:6 (<10) → Term7=6*9=54
- Term7:54 (even>9) → Term8=54/2=27
- Term8:27 (odd>9) → Term9=27-5=22
---
### Step 3: Identify the cycle
Term9 = 22, which is identical to Term4. This means the sequence repeats from Term4 onward. The repeating cycle is:
`[22, 11, 6, 54, 27]`, which has a **cycle length of 5**.
The first 3 terms (98, 49, 44) are unique pre-cycle terms.
---
### Step 4: Calculate the position of the 98th term in the cycle
Terms 1–3 are pre-cycle. For any term n ≥ 4, its position in the 5-term cycle is given by:
`cycle_index = (n - 3) mod 5`
- If `cycle_index = 0`, it corresponds to the 5th element of the cycle (since mod 5 returns 0 for multiples of 5).
For n=98:
1.  Subtract the 3 pre-cycle terms: 98 - 3 = 95 (this is the number of terms into the cycle).
2.  Compute 95 mod 5: 95 ÷5 = 19 with remainder 0.
---
### Step 5: Map the cycle index to the term value
The cycle is [22 (1st), 11 (2nd), 6 (3rd), 54 (4th), 27 (5th)]. A remainder of 0 means the 5th element of the cycle, which is 27.
---
### Step 6: Verify consistency
We can confirm with earlier terms:
- Term8: (8-3)=5 mod5=0 → 5th cycle element =27 (correct).
- Term13: (13-3)=10 mod5=0 →27 (correct).
- Term98 follows the same pattern, so it is 27.
ANSWER: D

1. **Identify the sequence rules**:  The sequence rules are as follows:
- If the integer is less than 10, multiply it by 9.
- If the integer is even and greater than 9, divide it by 2.
- If the integer is odd and greater than 9, subtract 5.

2. **Determine the first few terms of the sequence**:  Starting with the first term 98:
- 98 is even and greater than 9, so divide by 2: \( 98 / 2 = 49 \).
- 49 is odd and greater than 9, so subtract 5: \( 49 - 5 = 44 \).
- 44 is even and greater than 9, so divide by 2: \( 44 / 2 = 22 \).
- 22 is even and greater than 9, so divide by 2: \( 22 / 2 = 11 \).
- 11 is odd and greater than 9, so subtract 5: \( 11 - 5 = 6 \).
- 6 is less than 10, so multiply by 9: \( 6 \times 9 = 54 \).
- 54 is even and greater than 9, so divide by 2: \( 54 / 2 = 27 \).
- 27 is odd and greater than 9, so subtract 5: \( 27 - 5 = 22 \).

3. **Identify the repeating cycle**:  From the above calculations, the sequence begins to repeat after reaching 22:
- 22 -> 11 -> 6 -> 54 -> 27 -> 22.

4. **Determine the length of the repeating cycle**:  The repeating cycle is 22, 11, 6, 54, 27. This cycle has a length of 5.

5. **Find the position within the cycle for the 98th term**:  To find the 98th term, we need to determine its position within the repeating cycle:
- Calculate \( 98 \mod 5 \): \( 98 \div 5 = 19 \) remainder \( 3 \).
- Thus, the 98th term corresponds to the 3rd term in the cycle.

6. **Identify the 3rd term in the cycle**:  The cycle is 22, 11, 6, 54, 27. The 3rd term is 6.

7. **Conclude the correct answer**:  The 98th term of the sequence is 6.

ANSWER: A

**Problem Restatement:**
We need to find the 98th term of a sequence generated by the following rules:
1. If the current term is less than 10, multiply it by 9.
2. If the current term is even and greater than 9, divide it by 2.
3. If the current term is odd and greater than 9, subtract 5.

The sequence starts with 98, 49, ...

---

### Step 1: Understand the Rules and Initial Terms
- The first term is **98**.
- Since 98 is even and greater than 9, the second term is 98 / 2 = **49**.
- Now, 49 is odd and greater than 9, so the third term is 49 - 5 = **44**.
- Continue applying the rules to generate more terms until a pattern or cycle is detected.

---

### Step 2: Generate the Sequence Until a Cycle is Found
Let's compute the sequence step-by-step:

1. Term 1: 98 (even, >9) → 98 / 2 = **49**
2. Term 2: 49 (odd, >9) → 49 - 5 = **44**
3. Term 3: 44 (even, >9) → 44 / 2 = **22**
4. Term 4: 22 (even, >9) → 22 / 2 = **11**
5. Term 5: 11 (odd, >9) → 11 - 5 = **6**
6. Term 6: 6 (<10) → 6 * 9 = **54**
7. Term 7: 54 (even, >9) → 54 / 2 = **27**
8. Term 8: 27 (odd, >9) → 27 - 5 = **22**
9. Term 9: 22 (even, >9) → 22 / 2 = **11**
10. Term 10: 11 (odd, >9) → 11 - 5 = **6**
11. Term 11: 6 (<10) → 6 * 9 = **54**
12. Term 12: 54 (even, >9) → 54 / 2 = **27**

At this point, we observe a cycle starting from Term 7:
**22, 11, 6, 54, 27, 22, 11, 6, 54, 27, ...**

The cycle length is **5** (22, 11, 6, 54, 27).

---

### Step 3: Determine the Position in the Cycle for the 98th Term
- The cycle starts at Term 7.
- To find the position of the 98th term in the cycle, compute:
  (98 - 6) mod 5 = 92 mod 5 = **2** (since 92 / 5 = 18 with remainder 2).

This means the 98th term corresponds to the **2nd term in the cycle**.

The cycle is:
1. 22
2. 11
3. 6
4. 54
5. 27

Thus, the 98th term is **11**.

---

### Step 4: Rule Out Incorrect Choices
- A. 6 → This is the 3rd term in the cycle, not the 2nd.
- B. 11 → Matches our calculation.
- C. 22 → This is the 1st term in the cycle.
- D. 27 → This is the 5th term in the cycle.
- E. 54 → This is the 4th term in the cycle.

---

**ANSWER: B**

I need to find all possible values of n such that when n is added to {3, 6, 9, 10}, the mean equals the median.

**Step 1: Understand what we need**
- Original set: {3, 6, 9, 10}
- Add n to get five numbers
- Find all n where: mean = median

**Step 2: Set up the mean**
Mean of five numbers = (3 + 6 + 9 + 10 + n)/5 = (28 + n)/5

**Step 3: Determine possible medians**
With five numbers, the median is the middle (3rd) value when arranged in order.

The median depends on where n falls in the ordered list. Let me consider cases:

**Case 1: n ≤ 3**
Ordered list: {n, 3, 6, 9, 10}
Median = 6
Mean = Median: (28 + n)/5 = 6
28 + n = 30
n = 2 ✓ (satisfies n ≤ 3)

**Case 2: 3 < n ≤ 6**
Ordered list: {3, n, 6, 9, 10}
Median = 6
Mean = Median: (28 + n)/5 = 6
n = 2 ✗ (doesn't satisfy 3 < n ≤ 6)

**Case 3: 6 < n ≤ 9**
Ordered list: {3, 6, n, 9, 10}
Median = n
Mean = Median: (28 + n)/5 = n
28 + n = 5n
28 = 4n
n = 7 ✓ (satisfies 6 < n ≤ 9)

**Case 4: 9 < n ≤ 10**
Ordered list: {3, 6, 9, n, 10}
Median = 9
Mean = Median: (28 + n)/5 = 9
28 + n = 45
n = 17 ✗ (doesn't satisfy 9 < n ≤ 10)

**Case 5: n > 10**
Ordered list: {3, 6, 9, 10, n}
Median = 9
Mean = Median: (28 + n)/5 = 9
n = 17 ✓ (satisfies n > 10)

**Step 4: Verify the solutions**

For n = 2:
- Set: {2, 3, 6, 9, 10}
- Mean: (2 + 3 + 6 + 9 + 10)/5 = 30/5 = 6
- Median: 6 ✓

For n = 7:
- Set: {3, 6, 7, 9, 10}
- Mean: (3 + 6 + 7 + 9 + 10)/5 = 35/5 = 7
- Median: 7 ✓

For n = 17:
- Set: {3, 6, 9, 10, 17}
- Mean: (3 + 6 + 9 + 10 + 17)/5 = 45/5 = 9
- Median: 9 ✓

**Step 5: Count solutions**
There are exactly 3 possible values of n: {2, 7, 17}

ANSWER: C

1. **What's asked:** Add n to {3,6,9,10} making mean = median of the 5 numbers. Count possible n values.

2. **Sum and mean:** Sum = 3+6+9+10 = 28. With n: total = 28+n. Mean = (28+n)/5.

3. **Median depends on where n falls.** Sort the five numbers; median is the 3rd.

**Case 1: n ≤ 6** (n among smallest). Sorted, the middle value (3rd) is... let's check.
- If n ≤ 3: order n,3,6,9,10 → median = 6.
- If 3≤n≤6: order 3,n,6,9,10 → median = 6.
So median = 6 when n ≤ 6.

Set (28+n)/5 = 6 → 28+n = 30 → n = 2. Check n=2 ≤ 6 ✓. Valid.

**Case 2: 6 ≤ n ≤ 9** (n is middle). Order 3,6,n,9,10 → median = n.
Set (28+n)/5 = n → 28+n = 5n → 28 = 4n → n = 7. Check 6≤7≤9 ✓. Valid.

**Case 3: n ≥ 9** (n among largest). Median = 9.
- If 9≤n: order 3,6,9,n,10 or 3,6,9,10,n → 3rd is 9.
Set (28+n)/5 = 9 → 28+n = 45 → n = 17. Check 17 ≥ 9 ✓. Valid.

4. **Count:** n = 2, 7, 17 → three values.

5. **Rule out others:** Each case gives exactly one valid solution within its range, totaling 3. So not 1, 2, 4, or more than 4.

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 Cases for the Median

The median of 5 numbers is the 3rd value when sorted. I'll consider cases based on where n falls.

**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: n = 2**

**Case 2: 3 < n ≤ 6**
Sorted order: 3, n, 6, 9, 10 → Median = 6
Set mean = median: (28 + n)/5 = 6 → n = 2
Check: n = 2 is NOT in (3, 6] ✗ **No solution in this case**

**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: n = 7 and 6 < 7 ≤ 9 ✓ **Valid: n = 7**

**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] ✗ **No solution in this case**

**Case 5: n > 10**
Sorted order: 3, 6, 9, 10, n → Median = 9
Set mean = median: (28 + n)/5 = 9 → n = 17
Check: n = 17 > 10 ✓ **Valid: n = 17**

## Step 3: Count Valid Values

The valid values of n are: **2, 7, 17** — that's **3 values**.

## Step 4: Rule Out Wrong Choices

- A (1): Wrong, there are 3 solutions
- B (2): Wrong, there are 3 solutions
- **C (3): Correct**
- D (4): Wrong, we checked all cases exhaustively
- E (more than 4): Wrong

ANSWER: C

1. **Restate the problem:**  
   We have the set \(\{3,6,9,10\}\). We add a fifth number \(n\). We want the **mean** of the five numbers to equal the **median** of the five numbers. We must find how many possible values of \(n\) there are.

2. **Find the mean condition:**  
   The sum of the original four numbers is
   \[
   3+6+9+10=28.
   \]
   So the mean of the five numbers is
   \[
   \frac{28+n}{5}.
   \]

3. **Figure out the median depending on where \(n\) falls:**  
   With 5 numbers, the median is the **middle number** after sorting.

   Since the original numbers are \(3<6<9<10\), the median depends on where \(n\) fits:
   - If \(n \le 6\), the sorted list is \(\{3,n,6,9,10\}\) or \(\{3,6, n, 9,10\}\) depending on \(n\), but the median may be 6.
   - If \(6 \le n \le 9\), then the median is \(n\).
   - If \(n \ge 9\), the median may be 9.

   We test these cases carefully.

4. **Case 1: \(n \le 6\)**  
   Then 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 both situations, the median is **6**.

   Set mean = median:
   \[
   \frac{28+n}{5}=6
   \]
   \[
   28+n=30
   \]
   \[
   n=2.
   \]
   This fits \(n\le 6\), so it works.

5. **Case 2: \(6 \le n \le 9\)**  
   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 fits \(6\le n\le 9\), so it works.

6. **Case 3: \(n \ge 9\)**  
   Then the sorted list is \(3,6,9,n,10\) if \(9\le n\le 10\), or \(3,6,9,10,n\) if \(n\ge 10\).  
   In both situations, the median is **9**.

   Set mean = median:
   \[
   \frac{28+n}{5}=9
   \]
   \[
   28+n=45
   \]
   \[
   n=17.
   \]
   This fits \(n\ge 9\), so it works.

7. **Count the solutions:**  
   The possible values of \(n\) are:
   \[
   2,\ 7,\ 17
   \]
   so there are **3** possible values.

8. **Choose the correct answer:**  
   The correct choice is **C**.

ANSWER: C

An elegant way to solve this problem is to analyze the possible values of the median of the five-number set $\{3, 6, 9, 10, n\}$ based on the position of $n$, and then set the mean equal to that median.

### **Step 1: Restate the Goal**
We need to find the number of possible values of $n$ such that the mean of the set $\{3, 6, 9, 10, n\}$ is equal to its median.

---

### **Step 2: Set up the Math for Mean**
The mean of the five numbers is the sum of the numbers divided by 5:
$$\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$$

---

### **Step 3: Analyze the Median by Cases**
The median of the five numbers depends on where $n$ lies relative to the sorted list of the other four numbers: $3, 6, 9, 10$.

#### **Case 1: $n \le 6$**
If $n \le 6$, the ordered set is either $\{n, 3, 6, 9, 10\}$ (if $n \le 3$) or $\{3, n, 6, 9, 10\}$ (if $3 < n \le 6$). In both cases, the middle (3rd) number is $6$.
* Therefore, the median is $6$.
* Setting the mean equal to the median:
  $$\frac{28 + n}{5} = 6$$
  $$28 + n = 30 \implies n = 2$$
* Since $2 \le 6$, this is a valid solution.

#### **Case 2: $6 < n \le 9$**
If $6 < n \le 9$, the ordered set is $\{3, 6, n, 9, 10\}$. The middle (3rd) number is $n$.
* Therefore, the median is $n$.
* Setting the mean equal to the median:
  $$\frac{28 + n}{5} = n$$
  $$28 + n = 5n$$
  $$4n = 28 \implies n = 7$$
* Since $6 < 7 \le 9$, this is a valid solution.

#### **Case 3: $n > 9$**
If $n > 9$, the ordered set is either $\{3, 6, 9, n, 10\}$ (if $9 < n \le 10$) or $\{3, 6, 9, 10, n\}$ (if $n > 10$). In both cases, the middle (3rd) number is $9$.
* Therefore, the median is $9$.
* Setting the mean equal to the median:
  $$\frac{28 + n}{5} = 9$$
  $$28 + n = 45 \implies n = 17$$
* Since $17 > 9$, this is a valid solution.

---

### **Step 4: Count the Solutions and Rule Out Wrong Choices**
We have found exactly three valid values for $n$:
1. $n = 2$ (giving the set $\{2, 3, 6, 9, 10\}$ with mean = median = $6$)
2. $n = 7$ (giving the set $\{3, 6, 7, 9, 10\}$ with mean = median = $7$)
3. $n = 17$ (giving the set $\{3, 6, 9, 10, 17\}$ with mean = median = $9$)

Thus, there are exactly 3 possible values of $n$.
* Choice A (1) is incorrect because we found more than 1 value.
* Choice B (2) is incorrect because we found 3 values.
* Choice D (4) and Choice E (more than 4) are incorrect because our exhaustive case analysis shows no other solutions are possible.

This matches Choice C.

ANSWER: C

The original set is \(\{3, 6, 9, 10\}\), with sum \(28\). After adding \(n\), the new sum is \(28 + n\), so the mean is \((28 + n)/5\).

For the mean to equal the median, it is required that \(n = 5m - 28\), where \(m\) denotes the median of the resulting five-element set. The value of \(m\) is the third element in the ordered list, which depends on the position of \(n\) relative to the fixed elements \(3, 6, 9, 10\). These positions are partitioned into the five exhaustive intervals below, and the resulting candidate value of \(n\) is checked for membership in each interval.

- If \(n \leq 3\), the ordered list is \(n, 3, 6, 9, 10\), so \(m = 6\). Then \(n = 5 \cdot 6 - 28 = 2\). The candidate satisfies \(2 \leq 3\), and is therefore valid.
- If \(3 < n \leq 6\), the ordered list is \(3, n, 6, 9, 10\), so \(m = 6\). Then \(n = 2\), which lies outside \((3, 6]\) and yields no solution.
- If \(6 < n \leq 9\), the ordered list is \(3, 6, n, 9, 10\), so \(m = n\). Then \(n = 5n - 28\), or \(n = 7\). The candidate satisfies \(6 < 7 \leq 9\), and is therefore valid.
- If \(9 < n \leq 10\), the ordered list is \(3, 6, 9, n, 10\), so \(m = 9\). Then \(n = 5 \cdot 9 - 28 = 17\), which lies outside \((9, 10]\) and yields no solution.
- If \(n > 10\), the ordered list is \(3, 6, 9, 10, n\), so \(m = 9\). Then \(n = 17\). The candidate satisfies \(17 > 10\), and is therefore valid.

Boundary points where \(n\) equals one of the original elements were checked explicitly and produce no additional solutions. No other values of \(m\) are possible: the median cannot be \(3\) (at most two elements can be \(\leq 3\)) and cannot be \(10\) (at most two elements can be \(\geq 10\)).

Thus exactly three values of \(n\) work: \(2, 7, 17\).

ANSWER: C

1. **Understanding the Problem**: We are given a set of four numbers: {3, 6, 9, 10} and need to find a fifth number, n, such that when added to this set, the mean (average) of the resulting set of five numbers equals its median (the middle value when the numbers are arranged in ascending order).

2. **Setting Up the Math**: First, let's understand what the mean and median of the new set will be. The mean is the sum of all numbers divided by the count of numbers. So, the mean = (3 + 6 + 9 + 10 + n) / 5 = (28 + n) / 5.

3. **Calculating the Median**: To find the median of the set {3, 6, 9, 10, n}, we need to consider the possible positions of n in the ordered set. The set can be ordered in different ways based on the value of n:
   - If n ≤ 6, the ordered set is {n, 3, 6, 9, 10} and the median is 6.
   - If 6 < n ≤ 9, the ordered set is {3, 6, n, 9, 10} and the median is n.
   - If n > 9, the ordered set is {3, 6, 9, 10, n} and the median is 9.

4. **Equating Mean and Median**: We need to equate the mean to the median for each case and solve for n.
   - For n ≤ 6, mean = median gives (28 + n) / 5 = 6. So, 28 + n = 30, hence n = 2. Since n = 2 is less than 6, it's a valid solution.
   - For 6 < n ≤ 9, mean = median gives (28 + n) / 5 = n. So, 28 + n = 5n, hence 4n = 28, giving n = 7. Since 6 < 7 ≤ 9, it's a valid solution.
   - For n > 9, mean = median gives (28 + n) / 5 = 9. So, 28 + n = 45, hence n = 17. Since n = 17 is greater than 9, it's a valid solution.

5. **Checking the Validity of Solutions**: We have found three valid values of n (2, 7, and 17) that satisfy the condition that the mean equals the median.

6. **Ruling Out Wrong Choices**: 
   - A. 1 is incorrect because we found more than one value of n.
   - B. 2 is incorrect because we found three values, not just two.
   - D. 4 is incorrect because we did not find four values.
   - E. more than 4 is incorrect because we found exactly three values.

7. **Conclusion**: Since we have exactly three valid values of n, the correct answer is the choice that corresponds to this count.

ANSWER: C

Let’s go step-by-step.

---

**Step 1 – Restate the problem**  
We have a set: \(\{3, 6, 9, 10\}\).  
A fifth number \(n\) is added.  
We want the **mean** of all five numbers to equal the **median**.  
We must find how many possible values of \(n\) satisfy this.

---

**Step 2 – Write the condition for the mean**  
The sum of the original set is:  
\[
3 + 6 + 9 + 10 = 28
\]  
The sum after adding \(n\) is \(28 + n\), and the mean is:
\[
\frac{28 + n}{5}
\]

---

**Step 3 – Consider possible orderings for the median**  
The median of five numbers is the **third** when arranged in increasing order. The original numbers sorted: \(3, 6, 9, 10\).

Depending on \(n\), it could be:

- smallest (less than 3)
- between 3 and 6
- between 6 and 9
- between 9 and 10
- largest (greater than 10)

Each case gives a different median, so let’s check them.

---

**Step 4 – Case 1: \(n \le 3\)**  
Order: \(n, 3, 6, 9, 10\)  
Median = 6.  
Set mean = median:  
\[
\frac{28 + n}{5} = 6
\]  
\[
28 + n = 30
\]  
\[
n = 2
\]  
Check ordering: \(n = 2 \le 3\) — valid.  
So \(n = 2\) works.

---

**Step 5 – Case 2: \(3 < n \le 6\)**  
Order: \(3, n, 6, 9, 10\) or \(3, 6, n, 9, 10\) — in either possibility, the third number is 6.  
Median = 6.  
We get \(n = 2\) again from the same equation, but \(n \le 3\) was required — \(n=2\) does not fit here. So no solution in this case.

---

**Step 6 – Case 3: \(6 < n \le 9\)**  
Order: \(3, 6, n, 9, 10\) — the third number is \(n\).  
Median = \(n\).  
Equation:
\[
\frac{28 + n}{5} = n
\]  
\[
28 + n = 5n
\]  
\[
28 = 4n
\]  
\[
n = 7
\]  
Check: \(6 < 7 \le 9\) — valid.  
So \(n = 7\) works.

---

**Step 7 – Case 4: \(9 < n \le 10\)**  
Order: \(3, 6, 9, n, 10\) — the third number is 9.  
Median = 9.  
Equation:
\[
\frac{28 + n}{5} = 9
\]  
\[
28 + n = 45
\]  
\[
n = 17
\]  
But \(n\) must be \(\le 10\) here — \(17\) is outside range, so no solution.

---

**Step 8 – Case 5: \(n > 10\)**  
Order: \(3, 6, 9, 10, n\) — median is 9.  
Same equation, gives \(n = 17\).  
Now check: \(17 > 10\) — valid.  
So \(n = 17\) works.

---

**Step 9 – Summary of valid \(n\)**  
We found:  
- \(n = 2\) (Case 1)  
- \(n = 7\) (Case 3)  
- \(n = 17\) (Case 5)  

That is **3** possible values.

---

**Step 10 – Eliminate wrong choices**  
- A (1): too few  
- B (2): too few  
- C (3): matches  
- D (4): too many  
- E (>4): too many

---

ANSWER: C

# Finding the Number of Possible Values of n

## Step 1: Restate the Problem
We have the set {3, 6, 9, 10} and add a fifth number *n*. We need the **mean** of the five numbers to equal the **median**. We must find how many values of *n* satisfy this.

## Step 2: Express the Mean
The mean of the five numbers is:
$$\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$$

## Step 3: Determine the Median by Cases
The median of 5 numbers is the **3rd value** when sorted. The original sorted values are 3, 6, 9, 10. The position of *n* changes the median, so I consider all cases:

### Case 1: n ≤ 3
Sorted order: **n, 3, 6, 9, 10** → Median = 6
$$\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2$$
Check: Is 2 ≤ 3? **Yes** ✓ → **n = 2 is valid**

### Case 2: 3 < n ≤ 6
Sorted order: **3, n, 6, 9, 10** → Median = 6
$$\frac{28 + n}{5} = 6 \implies n = 2$$
Check: Is 3 < 2 ≤ 6? **No** ✗ → Not valid in this case

### Case 3: 6 < n ≤ 9
Sorted order: **3, 6, n, 9, 10** → Median = n
$$\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 28 = 4n \implies n = 7$$
Check: Is 6 < 7 ≤ 9? **Yes** ✓ → **n = 7 is valid**

### Case 4: 9 < n ≤ 10
Sorted order: **3, 6, 9, n, 10** → Median = 9
$$\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17$$
Check: Is 9 < 17 ≤ 10? **No** ✗ → Not valid in this case

### Case 5: n > 10
Sorted order: **3, 6, 9, 10, n** → Median = 9
$$\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17$$
Check: Is 17 > 10? **Yes** ✓ → **n = 17 is valid**

## Step 4: Verify All Solutions
| n | Sorted Set | Mean | Median | Match? |
|---|-----------|------|--------|--------|
| 2 | {2, 3, 6, 9, 10} | 30/5 = 6 | 6 | ✓ |
| 7 | {3, 6, 7, 9, 10} | 35/5 = 7 | 7 | ✓ |
| 17 | {3, 6, 9, 10, 17} | 45/5 = 9 | 9 | ✓ |

## Step 5: Rule Out Other Choices
- **A (1)** and **B (2)**: Too few — we found 3 valid values.
- **D (4)** and **E (more than 4)**: Too many — we exhaustively checked all 5 possible ordering cases and only 3 produced valid solutions.

There are exactly **3** possible values of n: {2, 7, 17}.

ANSWER: C

**Step 1: Restate the problem.**  
We start with the four numbers \(3, 6, 9, 10\). We add a fifth real number \(n\) so that the mean of the five numbers equals the median of the five numbers. We need to find how many different values of \(n\) make this true.

**Step 2: Express the mean algebraically.**  
The sum of the original four numbers is \(3+6+9+10=28\).  
With the added number \(n\), the mean is  
\[
\text{Mean}=\frac{28+n}{5}.
\]

**Step 3: Express the median as a piecewise function of \(n\).**  
Because a set of five numbers has an odd count, its median is the third number after the numbers are arranged in order. The value of the median depends on where \(n\) falls relative to \(3,6,9,10\).

- **If \(n\le 6\):** however \(n\) is placed on the left, the ordered list looks like \(\dots, 6, \dots\). The third entry is \(6\).  
  Median \(=6\).

- **If \(6\le n\le 9\):** the ordered list is \(3,6,n,9,10\). The third entry is \(n\) itself.  
  Median \(=n\).

- **If \(n\ge 9\):** the ordered list is \(3,6,9,\dots\). The third entry is \(9\).  
  Median \(=9\).

(At the endpoints \(n=6\) and \(n=9\) the formulas agree, so the description is consistent.)

**Step 4: Set mean equal to median in each region and solve.**

- **Region A: \(n\le 6\)**  
  \[
  \frac{28+n}{5}=6 \implies 28+n=30 \implies n=2.
  \]
  Since \(2\le 6\), this solution is valid.

- **Region B: \(6\le n\le 9\)**  
  \[
  \frac{28+n}{5}=n \implies 28+n=5n \implies 28=4n \implies n=7.
  \]
  Since \(6\le 7\le 9\), this solution is valid.

- **Region C: \(n\ge 9\)**  
  \[
  \frac{28+n}{5}=9 \implies 28+n=45 \implies n=17.
  \]
  Since \(17\ge 9\), this solution is valid.

**Step 5: Check the boundary values to make sure no extra solutions are missed.**  
- At \(n=6\): Mean \(=\frac{34}{5}=6.8\), Median \(=6\). Not equal.  
- At \(n=9\): Mean \(=\frac{37}{5}=7.4\), Median \(=9\). Not equal.  
- At \(n=3\) or \(n=10\): equality also fails. Thus no additional solutions occur at the boundaries.

**Step 6: List all valid values of \(n\).**  
The only values that work are \(n=2\), \(n=7\), and \(n=17\). That gives **exactly three** possible values.

**Step 7: Rule out the wrong answer choices.**  
- Choice A (1) is wrong because we found three solutions, not one.  
- Choice B (2) is wrong because we found more than two solutions.  
- Choice D (4) is wrong because we did not find four distinct solutions.  
- Choice E (more than 4) is wrong because the three linear equations in the three regions produced only one solution each, giving a total of three.  

Therefore the correct choice is C.

ANSWER: C

Here is the step-by-step solution:

**1. Understand the Goal**
We need to find how many possible values exist for a number $n$ such that when added to the set $\{3, 6, 9, 10\}$, the **mean** (average) of the resulting set of 5 numbers equals the **median** (middle number).

**2. Formulate the Mean**
Let the new set be $S' = \{3, 6, 9, 10, n\}$.
The sum of the original four numbers is $3 + 6 + 9 + 10 = 28$.
The sum of the new set of five numbers is $28 + n$.
The formula for the **Mean** is:
$$Mean = \frac{\text{Sum}}{\text{Count}} = \frac{28 + n}{5}$$

**3. Determine Possible Medians**
The median of a set with 5 elements is the 3rd element when the numbers are arranged in increasing order. The value of the median depends on where $n$ falls relative to the existing numbers $\{3, 6, 9, 10\}$. We must consider different intervals for $n$:

*   **Case 1: $n \leq 6$**
    If $n$ is small (less than or equal to 6), the ordered list looks like $\{n, 3, 6, 9, 10\}$ or $\{3, n, 6, 9, 10\}$ or $\{3, 6, n, ...\}$ (if $n=6$, it doesn't change the middle spot being occupied by 6). In all these sub-cases, the 3rd number is **6**.
    
    *Equation:* $\frac{28 + n}{5} = 6$
    *Solution:* $28 + n = 30 \Rightarrow n = 2$
    *Check:* Is $2 \leq 6$? **Yes.** So, $n=2$ is a valid solution.

*   **Case 2: $6 < n \leq 9$**
    If $n$ is between 6 and 9, the ordered list is $\{3, 6, n, 9, 10\}$. Here, $n$ is the 3rd number. Thus, the **Median is $n$**.
    
    *Equation:* $\frac{28 + n}{5} = n$
    *Solution:* $28 + n = 5n \Rightarrow 28 = 4n \Rightarrow n = 7$
    *Check:* Is $6 < 7 \leq 9$? **Yes.** So, $n=7$ is a valid solution.

*   **Case 3: $n \geq 9$**
    If $n$ is large (greater than or equal to 9), the ordered list looks like $\{3, 6, 9, n, 10\}$ or $\{3, 6, 9, 10, n\}$. In these scenarios, the 3rd number is **9**.
    
    *Equation:* $\frac{28 + n}{5} = 9$
    *Solution:* $28 + n = 45 \Rightarrow n = 17$
    *Check:* Is $17 \geq 9$? **Yes.** So, $n=17$ is a valid solution.

**4. Count the Solutions**
We found three distinct values for $n$ that satisfy the conditions:
1.  $n = 2$
2.  $n = 7$
3.  $n = 17$

There are exactly **3** possible values.

**5. Evaluate Choices**
A. 1 (Incorrect)
B. 2 (Incorrect)
C. 3 (Correct)
D. 4 (Incorrect)
E. more than 4 (Incorrect)

ANSWER: C

**1. Restate the problem**

We have the four numbers \(\{3,6,9,10\}\).  
A fifth number \(n\) is added, making a set of five numbers.  
We want the **mean** of the five numbers to equal the **median** of the five numbers.  
How many different values can \(n\) take?

**2. Find the median as a function of \(n\)**  

Sorted order of the four original numbers:  

\[
3\le 6\le 9\le 10 .
\]

When we insert \(n\) there are three possibilities for where it falls:

| Range of \(n\) | Ordered list (5 numbers) | Median (3rd number) |
|----------------|--------------------------|--------------------|
| \(n\le 6\)     | \(n,3,6,9,10\)           | \(6\) |
| \(6<n\le 9\)   | \(3,6,n,9,10\)           | \(n\) |
| \(n>9\)        | \(3,6,9,10,n\)           | \(9\) |

(At the boundaries \(n=6\) and \(n=9\) the median is still 6 and 9 respectively, so the piecewise definition above is correct.)

**3. Write the mean**

\[
\text{Mean}= \frac{3+6+9+10+n}{5}= \frac{28+n}{5}.
\]

**4. Set mean = median in each case**

*Case 1: \(n\le 6\)*  
\[
\frac{28+n}{5}=6\quad\Longrightarrow\quad28+n=30\quad\Longrightarrow\quad n=2.
\]
Check: \(2\le6\) ✔︎ → valid.

*Case 2: \(6<n\le 9\)*  
\[
\frac{28+n}{5}=n\quad\Longrightarrow\quad28+n=5n\quad\Longrightarrow\quad28=4n\quad\Longrightarrow\quad n=7.
\]
Check: \(6<7\le9\) ✔︎ → valid.

*Case 3: \(n>9\)*  
\[
\frac{28+n}{5}=9\quad\Longrightarrow\quad28+n=45\quad\Longrightarrow\quad n=17.
\]
Check: \(17>9\) ✔︎ → valid.

**5. Verify that no other \(n\) satisfy the condition**

The median is forced to be one of the three numbers (6, n, 9) depending on the region, and each region gives a single linear equation. Solving each yields exactly one candidate, and each candidate indeed lies in the region that produced it. No other values work.

**6. Count the possible values**

We obtained three distinct values:  

\[
n = 2,\;7,\;17 .
\]

Hence there are **3** possible values of \(n\).

**7. Match to answer choices**

Choice C says “3”.  

**Answer:**  

ANSWER: C

1. First, understand what is being asked:
   - We have a set of four numbers $\{3,6,9,10\}$, and we are adding a fifth number $n$ to it. We need to find the number of possible values of $n$ such that the mean of the new - set of five numbers is equal to its median.
2. Recall the formulas for mean and median:
   - The mean of a set of numbers $x_1,x_2,\cdots,x_k$ is $\bar{x}=\frac{x_1 + x_2+\cdots+x_k}{k}$. For the set $\{3,6,9,10,n\}$, the mean is $\frac{3 + 6+9 + 10 + n}{5}=\frac{28 + n}{5}$.
   - The median of a set of five numbers (an odd - numbered set) is the middle number when the numbers are arranged in ascending order.
3. Consider different cases based on the position of $n$ when the numbers are arranged in ascending order:
   - **Case 1: $n\leqslant6$**
     - Arrange the numbers in ascending order: $\{n,3,6,9,10\}$. The median is $6$.
     - Set the mean equal to the median: $\frac{28 + n}{5}=6$.
     - Multiply both sides of the equation by $5$: $28 + n = 30$.
     - Subtract $28$ from both sides: $n = 2$. Since $2\leqslant6$, this is a valid solution.
   - **Case 2: $6\lt n\lt9$**
     - Arrange the numbers in ascending order: $\{3,6,n,9,10\}$. The median is $n$.
     - Set the mean equal to the median: $\frac{28 + n}{5}=n$.
     - Multiply both sides by $5$: $28 + n = 5n$.
     - Subtract $n$ from both sides: $28 = 4n$.
     - Divide both sides by $4$: $n = 7$. Since $6\lt7\lt9$, this is a valid solution.
   - **Case 3: $n\geqslant9$**
     - Arrange the numbers in ascending order: $\{3,6,9,n,10\}$ or $\{3,6,9,10,n\}$. The median is $9$.
     - Set the mean equal to the median: $\frac{28 + n}{5}=9$.
     - Multiply both sides by $5$: $28 + n = 45$.
     - Subtract $28$ from both sides: $n = 17$. Since $17\geqslant9$, this is a valid solution.


4. Rule out the wrong choices:
   - We have found three possible values of $n$ ($n = 2$, $n = 7$, and $n = 17$). So, the number of possible values of $n$ is $3$. Choices A (1), B (2), D (4), and E (more than 4) are incorrect.

ANSWER: C

### Step 1: Restate the Problem
We add a fifth number $n$ to the set $\{3,6,9,10\}$ to create a 5-number set. The condition is that the mean of this new set equals its median. We need to count how many distinct valid values of $n$ exist.
---
### Step 2: Calculate the Mean of the 5-number set
The sum of the original set is $3+6+9+10 = 28$. Adding $n$, the total sum is $28 + n$,  the mean is $\frac{28+n}{5}$ for any value of $n$.
---
### Step 3: Define the Median of the 5-number set
For an odd number of sorted elements, the median is the middle (3rd) element in the sorted 5-number list. The median depends on where $n$ falls relative to the original sorted set $\{3,6,9,10\}$, creating three mutually exclusive, exhaustive cases:
1.  If $n \leq 6$: The sorted set will always have 6 as the 3rd element (median = 6), regardless of whether $n$ is less than 3 or between 3 and 6.
2.  If $6 < n \leq 9$: The sorted set will have $n$ as the 3rd element (median = $n$).
3.  If $n > 9$: The sorted set will always have 9 as the 3rd element (median =9), regardless of whether $n$ is between 9 and 10 or greater than 10.
No other medians are possible: the median cannot be 3 (only one element ≤3 exists, even after adding $n$, we cannot have three elements ≤3 to make 3 the middle element) and cannot be 10 (only one element ≥10 exists, even after adding $n$, we cannot have three elements ≥10 to make 10 the middle element).
---
### Step 4: Solve for $n$ in each case
For each case, set the mean equal to the median, solve for $n$, and verify the solution falls within the case's interval for $n$:
1.  **Case 1 (median=6, $n\leq6$):**
    $\frac{28+n}{5}=6 \implies 28+n=30 \implies n=2$. $2\leq6$, so this is valid.
2.  **Case 2 (median=$n$, $6<n\leq9$):**
    $\frac{28+n}{5}=n \implies 28+n=5n \implies 4n=28 \implies n=7$. $6<7\leq9$,  this is valid.
3.  **Case3 (median=9, $n>9$):**
    $\frac{28+n}{5}=9 \implies 28+n=45 \implies n=17$. $17>9$,  this is valid.
---
### Step 5: Count valid solutions and eliminate incorrect choices
The only valid values of $n$ are 2, 7, and 17: exactly 3 values. This rules out options A (1), B(2), D(4), and E(more than 4).
ANSWER: C

### Step 1: Restate the problem and define key terms
We start with the set {3, 6, 9, 10} and add a number \( n \) to make a 5-element set. We need the **mean** (average) of the 5-number set to equal its **median** (the middle value when sorted, which is the 3rd element for 5 numbers). We count how many valid \( n \) satisfy this.
---
### Step 2: Calculate the mean of the 5-number set
The sum of the original 4 numbers is \( 3 + 6 + 9 + 10 = 28 \). The sum of the 5-number set is \( 28 + n \), so the mean is:
\[
\text{Mean} = \frac{28 + n}{5}
\]
---
### Step 3: Analyze the median based on the position of \( n \)
When sorted, the original set is [3, 6, 9, 10]. The median depends on where \( n \) falls in the sorted order, so we split into cases:
1.  **Case 1: \( n \leq 3 \)**  
    Sorted set: \([n, 3, 6, 9, 10]\). Median = 6 (3rd element).
2.  **Case 2: \( 3 < n \leq 6 \)**  
    Sorted set: \([3, n, 6, 9, 10]\). Median = 6 (3rd element).
3.  **Case 3: \( 6 < n \leq 9 \)**  
    Sorted set: \([3, 6, n, 9, 10]\). Median = \( n \) (3rd element).
4.  **Case 4: \( 9 < n \leq 10 \)**  
    Sorted set: \([3, 6, 9, n, 10]\). Median =9 (3rd element).
5.  **Case 5: \( n > 10 \)**  
    Sorted set: \([3, 6, 9, 10, n]\). Median =9 (3rd element).
---
### Step 4: Solve for \( n \) in each case, and validate against the case range
For each case, set mean = median, solve for \( n \), and check if \( n \) fits the case's range:
1.  **Cases 1 & 2 (median = 6):**  
    Set \( \frac{28 + n}{5} = 6 \). Multiply by 5: \( 28 + n = 30 \), so \( n = 2 \).  
    Check: \( 2 \leq 3 \), which fits Cases 1/2. Valid.
2.  **Case 3 (median = \( n \)):**  
    Set \( \frac{28 + n}{5} = n \). Multiply by 5: \( 28 + n = 5n \),  \( 4n =28 \),  \( n=7 \).  
    Check: \( 6 <7 \leq9 \), which fits Case3. Valid.
3.  **Cases4 &5 (median=9):**  
    Set \( \frac{28 +n}{5}=9 \). Multiply by5: \(28 +n=45\),  \(n=17\).  
    Check: \(17>10\), which fits Cases4/5. Valid.
---
### Step 5: Rule out invalid candidates
Check edge values of the ranges to confirm no extra solutions:
- For \( n=3,6,9,10 \): None satisfy mean=median (e.g, \( n=6 \): mean=6.8≠6; \(n=9\): mean=7.4≠9).
- For values between the case boundaries (e.g, \(n=4, 8, 11\)): None satisfy mean=median.
---
### Step 6: Count valid solutions
The valid \( n \) values are 2,7,17: 3 total values.
ANSWER: C

To solve this problem, we need to find the number of possible values of \( n \) such that when \( n \) is added to the set \(\{3, 6, 9, 10\}\), the mean of the resulting set of five numbers equals its median.

**Step 1: Restate the problem**
We need to find \( n \) such that the mean of the set \(\{3, 6, 9, 10, n\}\) equals its median.

**Step 2: Calculate the mean**
The mean of the set \(\{3, 6, 9, 10, n\}\) is given by:
\[
\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}
\]

**Step 3: Determine the median**
To find the median, we need to consider the possible positions of \( n \) in the ordered set. The original set \(\{3, 6, 9, 10\}\) has 4 elements, so when \( n \) is added, the median will be the middle number in the ordered set of 5 elements.

**Step 4: Analyze the position of \( n \)**
We need to consider different cases for the value of \( n \) and how it affects the median.

**Case 1: \( n \leq 3 \)**
The ordered set would be \(\{n, 3, 6, 9, 10\}\), and the median is 6.
\[
\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2
\]
This is not possible since \( n \leq 3 \) and \( n = 2 \) is valid.

**Case 2: \( 3 < n \leq 6 \)**
The ordered set would be \(\{3, n, 6, 9, 10\}\), and the median is \( n \).
\[
\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 28 = 4n \implies n = 7
\]
This is not possible since \( n \leq 6 \).

**Case 3: \( 6 < n \leq 9 \)**
The ordered set would be \(\{3, 6, n, 9, 10\}\), and the median is \( n \).
\[
\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 28 = 4n \implies n = 7
\]
This is possible since \( 6 < 7 \leq 9 \).

**Case 4: \( 9 < n \leq 10 \)**
The ordered set would be \(\{3, 6, 9, n, 10\}\), and the median is 9.
\[
\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17
\]
This is not possible since \( n \leq 10 \).

**Case 5: \( n > 10 \)**
The ordered set would be \(\{3, 6, 9, 10, n\}\), and the median is 9.
\[
\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17
\]
This is not possible since \( n > 10 \).

**Step 5: Conclusion**
The only valid value for \( n \) is 7, which fits within the range \( 6 < n \leq 9 \).

Thus, there is only one possible value for \( n \).

ANSWER: A

**Problem Restatement:**
We have the set {3, 6, 9, 10}. We add a fifth number, \( n \), to make the **mean** of the new five-number set equal to its **median**. We need to find how many possible values \( n \) can take.

---

### Step 1: Understand the Definitions
- **Mean**: The average of the numbers. For the new set {3, 6, 9, 10, \( n \)}, the mean is:
  \[
  \text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}
  \]
- **Median**: The middle number when the set is ordered. For five numbers, the median is the third number in the ordered list.

---

### Step 2: Order the Set
The original set is {3, 6, 9, 10}. When we add \( n \), the new set is {3, 6, 9, 10, \( n \)}. To find the median, we must order the set. The position of \( n \) in the ordered set depends on its value. We consider all possible cases for where \( n \) fits in the ordered list.

---

### Step 3: Consider Cases for \( n \)
We analyze where \( n \) can lie in the ordered set. The original ordered set is 3, 6, 9, 10. Adding \( n \) creates five ordered positions. The median is always the third number in this ordered list.

#### Case 1: \( n \leq 3 \)
Ordered set: \( n, 3, 6, 9, 10 \)
Median = 6
Set mean = median:
\[
\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2
\]
Check if \( n = 2 \leq 3 \): Yes. So \( n = 2 \) is valid.

#### Case 2: \( 3 < n \leq 6 \)
Ordered set: \( 3, n, 6, 9, 10 \)
Median = 6
Set mean = median:
\[
\frac{28 + n}{5} = 6 \implies n = 2
\]
But \( n = 2 \) does not satisfy \( 3 < n \leq 6 \). No solution in this case.

#### Case 3: \( 6 < n \leq 9 \)
Ordered set: \( 3, 6, n, 9, 10 \)
Median = \( n \)
Set mean = median:
\[
\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 28 = 4n \implies n = 7
\]
Check if \( 6 < 7 \leq 9 \): Yes. So \( n = 7 \) is valid.

#### Case 4: \( 9 < n \leq 10 \)
Ordered set: \( 3, 6, 9, n, 10 \)
Median = 9
Set mean = median:
\[
\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17
\]
Check if \( 9 < 17 \leq 10 \): No. No solution in this case.

#### Case 5: \( n > 10 \)
Ordered set: \( 3, 6, 9, 10, n \)
Median = 9
Set mean = median:
\[
\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17
\]
Check if \( n = 17 > 10 \): Yes. So \( n = 17 \) is valid.

---

### Step 4: Verify All Cases
From the cases above, the valid values of \( n \) are:
- \( n = 2 \) (from Case 1)
- \( n = 7 \) (from Case 3)
- \( n = 17 \) (from Case 5)

No other cases yield valid solutions.

---

### Step 5: Count Possible Values
There are **3** possible values for \( n \): 2, 7, and 17.

---

### Step 6: Rule Out Wrong Choices
- A (1): Incorrect, as there are 3 values.
- B (2): Incorrect, as there are 3 values.
- D (4): Incorrect, as there are 3 values.
- E (more than 4): Incorrect, as there are exactly 3 values.

---

ANSWER: C

I need to find how many disks Maria must sell to make a profit of $100.

**Step 1: Understand what is being asked**
I need to find the number of disks Maria must sell so that her total profit (revenue minus cost) equals $100.

**Step 2: Determine the cost per disk**
Maria buys at a price of 4 disks for $5.
- Cost per disk = $5 ÷ 4 = $1.25 per disk

**Step 3: Determine the selling price per disk**
Maria sells at a price of 3 disks for $5.
- Selling price per disk = $5 ÷ 3 = $5/3 ≈ $1.667 per disk

**Step 4: Calculate profit per disk**
- Profit per disk = Selling price - Cost price
- Profit per disk = $5/3 - $5/4

To subtract these fractions, I need a common denominator (12):
- Profit per disk = $20/12 - $15/12 = $5/12 per disk

**Step 5: Find how many disks needed for $100 profit**
If each disk yields a profit of $5/12, then:
- Number of disks = Total profit ÷ Profit per disk
- Number of disks = $100 ÷ ($5/12)
- Number of disks = $100 × (12/5)
- Number of disks = $1200/5
- Number of disks = 240

**Step 6: Verify the answer**
- Cost of 240 disks: 240 × $1.25 = $300
- Revenue from 240 disks: 240 × $5/3 = $1200/3 = $400
- Profit: $400 - $300 = $100 ✓

**Step 7: Rule out wrong answers**
- A. 100: Would give profit of 100 × $5/12 = $500/12 ≈ $41.67 (too small)
- B. 120: Would give profit of 120 × $5/12 = $600/12 = $50 (too small)
- C. 200: Would give profit of 200 × $5/12 = $1000/12 ≈ $83.33 (too small)
- D. 240: Would give profit of 240 × $5/12 = $1200/12 = $100 ✓
- E. 1200: Would give profit of 1200 × $5/12 = $6000/12 = $500 (too large)

ANSWER: D

1. Cost per disk: $5/4 = $1.25.
2. Sell price per disk: $5/3 ≈ $1.667.
3. Profit per disk: 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.
4. Disks needed: 100 ÷ (5/12) = 100 × 12/5 = 240.

Ruling out: A (100) too few; B (120) gives only $50 profit; C (200) gives ~$83; E (1200) way too many.

ANSWER: D

## Step-by-Step Solution

**Step 1: Restate the problem.**
Maria buys disks at 4 for $5 and sells them at 3 for $5. I need to find how many disks she must sell to make a $100 profit.

**Step 2: Find the cost per disk.**
She buys 4 disks for $5, so:
$$\text{Cost per disk} = \frac{\$5}{4} = \$1.25$$

**Step 3: Find the selling price per disk.**
She sells 3 disks for $5, so:
$$\text{Selling price per disk} = \frac{\$5}{3} \approx \$1.6\overline{6}$$

**Step 4: Find the profit per disk.**
$$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20}{12} - \frac{15}{12} = \frac{5}{12} \text{ dollars per disk}$$

**Step 5: Find how many disks she must sell for $100 profit.**
$$\text{Number of disks} = \frac{\$100}{\frac{5}{12}} = 100 \times \frac{12}{5} = \frac{1200}{5} = 240$$

**Step 6: Verify the answer.**
- Buy 240 disks: Cost = $240 × \frac{5}{4} = 60 \times 5 = \$300$
- Sell 240 disks: Revenue = $240 × \frac{5}{3} = 80 \times 5 = \$400$
- Profit = $400 - $300 = **$100** ✓

**Step 7: Eliminate wrong choices.**
- A (100): Profit = $100 × \frac{5}{12} = \$41.67$ — not enough
- B (120): Profit = $120 × \frac{5}{12} = \$50$ — not enough
- C (200): Profit = $200 × \frac{5}{12} = \$83.33$ — not enough
- E (1200): Way too many

ANSWER: D

1. **Restate the problem.**  
   Maria **buys** disks at **4 for \$5**, so each disk costs  
   \[
   \frac{5}{4}\text{ dollars}
   \]
   She **sells** disks at **3 for \$5**, so each disk sells for  
   \[
   \frac{5}{3}\text{ dollars}
   \]

2. **Find the profit per disk.**  
   Profit per disk = selling price − buying price:
   \[
   \frac{5}{3}-\frac{5}{4}
   \]
   Use a common denominator of 12:
   \[
   \frac{20}{12}-\frac{15}{12}=\frac{5}{12}
   \]
   So Maria makes a profit of **\(\frac{5}{12}\) dollar per disk**.

3. **Set up the equation for \$100 profit.**  
   If she sells \(n\) disks, then total profit is
   \[
   n\cdot \frac{5}{12}
   \]
   We want this to equal 100:
   \[
   n\cdot \frac{5}{12}=100
   \]

4. **Solve for \(n\).**  
   Multiply both sides by \(\frac{12}{5}\):
   \[
   n=100\cdot \frac{12}{5}=20\cdot 12=240
   \]

5. **Check the answer against the choices.**  
   The number 240 is listed as choice **D**.

ANSWER: D

An elegant and clear step-by-step solution to the problem:

**Step 1: Restate what is being asked**
We need to find the number of computer disks Maria must buy and sell to make a total profit of $100, given that her buying price is 4 disks for $5 and her selling price is 3 disks for $5.

**Step 2: Find a convenient common quantity of disks**
To avoid working with fractions, we can find the cost and revenue for a common multiple of the buying and selling quantities. The least common multiple of 4 and 3 is 12. 

**Step 3: Calculate the profit for 12 disks**
* **Cost Price (CP) for 12 disks:** 
  Since she buys 4 disks for $5, the cost for 12 disks is:
  $$\frac{12}{4} \times \$5 = 3 \times \$5 = \$15$$

* **Selling Price (SP) for 12 disks:** 
  Since she sells 3 disks for $5, the revenue from 12 disks is:
  $$\frac{12}{3} \times \$5 = 4 \times \$5 = \$20$$

* **Profit for 12 disks:** 
  $$\text{Profit} = \text{Selling Price} - \text{Cost Price} = \$20 - \$15 = \$5$$

So, Maria makes a profit of $5 for every 12 disks she sells.

**Step 4: Set up the proportion for a $100 profit**
We want to find the total number of disks, $N$, needed to make a profit of $100. We can set up the ratio:
$$\frac{\text{Number of disks}}{\text{Profit}} = \frac{12}{5}$$

$$\frac{N}{100} = \frac{12}{5}$$

**Step 5: Solve for $N$ and verify**
$$N = 100 \times \frac{12}{5}$$
$$N = 20 \times 12$$
$$N = 240$$

Let's verify this:
* Cost of 240 disks: $240 \div 4 \times \$5 = 60 \times \$5 = \$300$
* Revenue from 240 disks: $240 \div 3 \times \$5 = 80 \times \$5 = \$400$
* Profit: $\$400 - \$300 = \$100$
The calculation is correct.

**Step 6: Rule out the wrong choices**
* **A. 100:** If she sells 100 disks, her profit is $100 \times \frac{5}{12} \approx \$41.67$, which is too low.
* **B. 120:** If she sells 120 disks, her profit is $120 \times \frac{5}{12} = \$50$, which is half of what is needed.
* **C. 200:** If she sells 200 disks, her profit is $200 \times \frac{5}{12} \approx \$83.33$, which is too low.
* **E. 1200:** If she sells 1200 disks, her profit is $1200 \times \frac{5}{12} = \$500$, which is too high.

Therefore, the correct choice is D.

ANSWER: D

The problem requires determining the number of disks Maria must sell to earn a profit of exactly $100, given her buying price of 4 disks for $5 and her selling price of 3 disks for $5.

To solve this, first compute her profit per disk. Her cost per disk is \(5/4\) dollars. Her selling price per disk is \(5/3\) dollars. Subtracting these values gives the profit per disk:
\[
\frac{5}{3} - \frac{5}{4} = 5\left(\frac{1}{3} - \frac{1}{4}\right) = 5\left(\frac{4-3}{12}\right) = \frac{5}{12}
\]
dollars per disk.

Let \(n\) be the number of disks she must sell to earn a profit of $100. Then
\[
n \cdot \frac{5}{12} = 100 \implies n = 100 \cdot \frac{12}{5} = 240.
\]
Thus, she must sell 240 disks.

To confirm via batches (without using per-disk fractions), suppose she buys 12 disks. This costs \(3 \times 5 = 15\) dollars. She can then sell all 12 disks for \(4 \times 5 = 20\) dollars, for a profit of 5 dollars. Scaling up to a total profit of 100 dollars requires
\[
\frac{100}{5} = 20
\]
such batches, or \(20 \times 12 = 240\) disks, matching the earlier result.

Choice A (100) would produce a profit of only \(100 \times 5/12 \approx 41.67\) dollars. Choice B (120) would produce a profit of exactly 50 dollars. Choice C (200) would produce a profit of \(200 \times 5/12 \approx 83.33\) dollars. Choice E (1200) would produce a profit of 500 dollars. Only 240 disks yields exactly 100 dollars of profit.

ANSWER: D

1. **Understand the problem**: We need to find out how many computer disks Maria must sell to make a profit of $100, given that she buys them at 4 for $5 and sells them at 3 for $5.

2. **Calculate the cost and selling price per disk**:
   - The cost price per disk is $5 / 4 = $1.25.
   - The selling price per disk is $5 / 3 ≈ $1.67.

3. **Determine the profit per disk**:
   - Profit per disk = Selling price per disk - Cost price per disk = $1.67 - $1.25 = $0.42.

4. **Calculate the number of disks to be sold to make a $100 profit**:
   - Number of disks = Total profit / Profit per disk = $100 / $0.42.

5. **Perform the calculation**:
   - Number of disks = $100 / ($5/3 - $5/4) = $100 / (($20 - $15)/12) = $100 / ($5/12) = $100 * (12/5) = $2400 / 5 = 240.

6. **Compare with the given choices and rule out incorrect options**:
   - A. 100: $0.42 * 100 = $42, which is less than $100.
   - B. 120: $0.42 * 120 = $50.40, still less than $100.
   - C. 200: $0.42 * 200 = $84, less than $100.
   - D. 240: $0.42 * 240 = $100.80, which is just above $100, making it a plausible answer.
   - E. 1200: $0.42 * 1200 = $504, which is more than $100 but not the minimum required.

7. **Select the correct answer**:
   - Since $0.42 * 240 = $100.80, which is the first to exceed $100 among the given choices, D is the correct answer.

ANSWER: D

1. **Understand the problem:** Maria buys disks at a cost of 4 for $5 and sells them at 3 for $5. We need to find how many disks she must sell to make a $100 profit.

2. **Find cost per disk:** Buying 4 disks for $5 means cost per disk = $5 ÷ 4 = $1.25 (or 5/4 dollars).

3. **Find selling price per disk:** Selling 3 disks for $5 means selling price per disk = $5 ÷ 3 = $1.666... (or 5/3 dollars).

4. **Calculate profit per disk:** Profit per disk = selling price − cost price = (5/3) − (5/4).  
   Compute common denominator 12: (20/12) − (15/12) = 5/12 dollars profit per disk.

5. **Determine number of disks for $100 profit:** Let \(n\) be the number of disks.  
   Total profit = \(n \times \frac{5}{12} = 100\).  
   Solve for \(n\): \(n = 100 \times \frac{12}{5} = 100 \times 2.4 = 240\).

6. **Check against choices:** The calculated 240 matches choice D. Eliminate the others:  
   - A (100) would give profit of 100 × (5/12) ≈ $41.67, too low.  
   - B (120) gives 120 × (5/12) = $50.  
   - C (200) gives about $83.33.  
   - E (1200) gives $500 profit, far too high.

ANSWER: D

**Step 1: Restate the problem**
Maria buys disks at a rate of 4 disks for $5 and sells them at a rate of 3 disks for $5. We need to find how many disks she must sell to earn a total profit of $100.

**Step 2: Find the cost per disk (buying price)**
She buys 4 disks for $5.
Cost per disk = $5 ÷ 4 = $5/4 = $1.25

**Step 3: Find the revenue per disk (selling price)**
She sells 3 disks for $5.
Revenue per disk = $5 ÷ 3 = $5/3 ≈ $1.67

**Step 4: Find the profit per disk**
Profit per disk = Revenue per disk − Cost per disk
Profit per disk = $5/3 − $5/4

To subtract, find a common denominator (12):
= $20/12 − $15/12
= $5/12 per disk

**Step 5: Find the number of disks needed for $100 profit**
Number of disks = Total desired profit ÷ Profit per disk
Number of disks = $100 ÷ ($5/12)
= $100 × (12/5)
= $1200/5
= 240 disks

**Step 6: Verify the answer**
- Cost to buy 240 disks: (240 ÷ 4) × $5 = 60 × $5 = $300
- Revenue from selling 240 disks: (240 ÷ 3) × $5 = 80 × $5 = $400
- Profit: $400 − $300 = $100 ✓

**Step 7: Rule out wrong choices**
- A (100): Profit would be 100 × $5/12 ≈ $41.67 — too low.
- B (120): Profit would be 120 × $5/12 = $50 — too low.
- C (200): Profit would be 200 × $5/12 ≈ $83.33 — too low.
- E (1200): Profit would be 1200 × $5/12 = $500 — too high.
Only D (240) gives exactly $100 profit.

ANSWER: D

**Step 1: Restate the problem.**  
Maria purchases disks at a rate of 4 disks for \\(5 and sells them at a rate of 3 disks for \\)5. We need to find the number of disks she must **sell** to earn an overall profit of exactly \\(100.

**Step 2: Find the cost per disk.**  
If 4 disks cost \\)5, the cost for one disk is  
\[
\frac{5}{4}=1.25\text{ dollars.}
\]

**Step 3: Find the selling price per disk.**  
If she sells 3 disks for \\(5, the revenue for one disk is  
\[
\frac{5}{3}\text{ dollars.}
\]

**Step 4: Compute the profit per disk.**  
\[
\text{Profit per disk} = \text{Selling price} - \text{Cost} = \frac{5}{3} - \frac{5}{4}.
\]  
Using a common denominator of 12:  
\[
\frac{5}{3} - \frac{5}{4} = \frac{20}{12} - \frac{15}{12} = \frac{5}{12}\text{ dollars per disk.}
\]

**Step 5: Set up an equation for total profit.**  
Let \(n\) be the number of disks sold. Then  
\[
n \cdot \frac{5}{12} = 100.
\]

**Step 6: Solve for \(n\).**  
\[
n = 100 \div \frac{5}{12} = 100 \times \frac{12}{5} = 20 \times 12 = 240.
\]

**Step 7: Verify with a “batch” method (LCM check).**  
The least common multiple of 4 and 3 is 12.  
- Buying 12 disks costs \((12/4)\times 5 = 3\times 5 = \\)15\).  
- Selling 12 disks earns \((12/3)\times 5 = 4\times 5 = \\(20\).  
So every 12 disks yield a profit of \)20 - \\(15 = \\)5\(.  
To reach \)100 profit: \)100 \div 5 = 20\( batches.  
\[
20 \times 12 = 240\text{ disks.}
\]

**Step 8: Rule out the incorrect choices.**  
- **A. 100 disks:** Profit would be \)100 \times \frac{5}{12} \approx \\(41.67.  
- **B. 120 disks:** Profit would be \)120 \times \frac{5}{12} = \\(50.  
- **C. 200 disks:** Profit would be \)200 \times \frac{5}{12} \approx \\(83.33.  
- **D. 240 disks:** Profit is exactly \)100.  
- **E. 1200 disks:** Profit would be \)1200 \times \frac{5}{12} = \\(500.  

Only **240 disks** produce the required \\)100 profit.

ANSWER: D

Here is the step-by-step solution:

**1. Restate the problem**
We need to determine the number of computer disks Maria needs to sell so that her total revenue exceeds her total costs by exactly $\$100$. We are given the buying rate (cost) and the selling rate (revenue).

**2. Set up the math**
Let $n$ be the number of disks Maria buys and sells.
*   **Cost:** She buys 4 disks for $\$5$. Therefore, the cost per disk is $\frac{5}{4}$ dollars. The total cost for $n$ disks is $\frac{5}{4}n$.
*   **Revenue:** She sells 3 disks for $\$5$. Therefore, the revenue per disk is $\frac{5}{3}$ dollars. The total revenue for $n$ disks is $\frac{5}{3}n$.
*   **Profit:** Profit is defined as Total Revenue minus Total Cost. We want this to equal $\$100$.

Equation:
$$ \text{Profit} = \text{Revenue} - \text{Cost} $$
$$ 100 = \frac{5}{3}n - \frac{5}{4}n $$

**3. Do the computation**
To subtract the terms with $n$, find a common denominator for the fractions $\frac{5}{3}$ and $\frac{5}{4}$. The least common multiple of 3 and 4 is 12.

Rewrite the fractions:
$$ \frac{5}{3}n = \frac{20}{12}n $$
$$ \frac{5}{4}n = \frac{15}{12}n $$

Substitute these back into the profit equation:
$$ 100 = \frac{20}{12}n - \frac{15}{12}n $$
$$ 100 = \frac{5}{12}n $$

Now, solve for $n$ by multiplying both sides by 12 and dividing by 5:
$$ 100 \times 12 = 5n $$
$$ 1200 = 5n $$
$$ n = \frac{1200}{5} $$
$$ n = 240 $$

So, she must sell **240** disks.

*(Alternative "Batch" Method)*:
Consider a batch of 12 disks (since 12 is the least common multiple of the buy-group size 4 and sell-group size 3).
*   **Cost for 12 disks:** Since 4 cost $\$5$, 12 disks (3 groups) cost $3 \times \$5 = \$15$.
*   **Revenue for 12 disks:** Since 3 sell for $\$5$, 12 disks (4 groups) sell for $4 \times \$5 = \$20$.
*   **Profit for 12 disks:** $\$20 - \$15 = \$5$.
To make $\$100$ profit, she needs $\frac{100}{5} = 20$ such batches.
Total disks = $20 \text{ batches} \times 12 \text{ disks/batch} = 240$ disks.

**4. Explain how to rule out the wrong choices**
*   **A. 100**: If we estimate profit per disk as roughly $\$1.67 - \$1.25 = \$0.42$, then 100 disks yield only about $\$42$ profit, which is far less than $\$100$.
*   **B. 120**: Using our batch method, 120 disks is exactly 10 batches of 12. Since each batch yields $\$5$ profit, 10 batches yield only $10 \times \$5 = \$50$ profit.
*   **C. 200**: Plugging 200 into the formula $\frac{5}{12}n$ gives $\frac{1000}{12} \approx \$83.33$, which is not $\$100$.
*   **E. 1200**: This number is too large. 1200 disks would result in a profit of $\frac{5}{12}(1200) = 5 \times 100 = \$500$.
*   **D. 240**: This matches our calculated result exactly ($240 \times \frac{5}{12} = 20 \times 5 = 100$).

ANSWER: D

**Step 1: Find the cost per disk.**  
Maria buys 4 disks for \$5, so the cost per disk is  

\[
\frac{5}{4}=1.25\text{ dollars.}
\]

**Step 2: Find the selling price per disk.**  
She sells 3 disks for \$5, so the selling price per disk is  

\[
\frac{5}{3}\approx1.6667\text{ dollars.}
\]

**Step 3: Compute the profit on one disk.**  

\[
\text{Profit per disk}= \frac{5}{3}-\frac{5}{4}=5\Bigl(\frac{1}{3}-\frac{1}{4}\Bigr)=5\left(\frac{1}{12}\right)=\frac{5}{12}\text{ dollars.}
\]

**Step 4: Determine how many disks must be sold to earn \$100 profit.**  

\[
\text{Number of disks}= \frac{100}{\frac{5}{12}} = 100 \times \frac{12}{5}=240.
\]

**Step 5: Check that 240 works with the “group‑pricing.”**  
- 240 disks is a multiple of 4: she buys \(240/4=60\) groups, costing \(60 \times \$5 = \$300\).  
- 240 disks is also a multiple of 3: she sells \(240/3=80\) groups, receiving \(80 \times \$5 = \$400\).  
- Profit \(= \$400-\$300 = \$100\).  

Thus 240 disks yields exactly the desired profit.

**Eliminate other answer choices:**  
- 100 disks give profit \(100 \times \frac{5}{12} \approx \$41.67\).  
- 120 disks give profit \(120 \times \frac{5}{12}= \$50\).  
- 200 disks give profit \(200 \times \frac{5}{12} \approx \$83.33\).  
- 1200 disks give profit \(1200 \times \frac{5}{12}= \$500\).  
Only 240 disks produces the required \$100 profit.

**Answer:**  

ANSWER: D

1. First, find the cost per disk and the selling price per disk:
   - **Cost per disk**:
     - Maria buys 4 disks for $5. To find the cost of one disk, we divide the total cost by the number of disks. So, the cost per disk, $C=\frac{5}{4}=1.25$ dollars.
   - **Selling price per disk**:
     - Maria sells 3 disks for $5. To find the selling - price of one disk, we divide the total selling price by the number of disks. So, the selling price per disk, $S = \frac{5}{3}\approx1.67$ dollars.
2. Then, find the profit per disk:
     - The profit per disk, $P$, is the selling price per disk minus the cost per disk. So, $P=S - C=\frac{5}{3}-\frac{5}{4}$.
     - To subtract these fractions, we need a common denominator. The least common multiple of 3 and 4 is 12.
     - Rewrite the fractions with the common denominator: $\frac{5}{3}=\frac{5\times4}{3\times4}=\frac{20}{12}$ and $\frac{5}{4}=\frac{5\times3}{4\times3}=\frac{15}{12}$.
     - Then $P=\frac{20}{12}-\frac{15}{12}=\frac{20 - 15}{12}=\frac{5}{12}$ dollars per disk.
3. Next, find the number of disks $n$ needed to make a profit of $100:
     - We know that profit is equal to the profit per disk times the number of disks. That is, $100 = P\times n$.
     - Substitute $P=\frac{5}{12}$ into the equation: $100=\frac{5}{12}\times n$.
     - To solve for $n$, we multiply both sides of the equation by $\frac{12}{5}$. So, $n = 100\times\frac{12}{5}$.
     - $100\times\frac{12}{5}=\frac{100\times12}{5}=20\times12 = 240$ disks.
4. Finally, rule out the wrong choices:
     - Option A: If $n = 100$, the profit is $\frac{5}{12}\times100=\frac{500}{12}\approx41.67\neq100$.
     - Option B: If $n = 120$, the profit is $\frac{5}{12}\times120 = 50\neq100$.
     - Option C: If $n = 200$, the profit is $\frac{5}{12}\times200=\frac{1000}{12}\approx83.33\neq100$.
     - Option E: If $n = 1200$, the profit is $\frac{5}{12}\times1200 = 500\neq100$.

ANSWER: D

### Step 1: Restate the Problem
We need to calculate the number of computer disks Maria must sell to earn a $100 profit. Profit equals total revenue (money earned from sales) minus total cost (money spent to buy the disks). Maria buys 4 disks for $5, and sells 3 disks for $5.
---
### Step 2: Calculate Unit Costs and Revenues
First, compute the cost to buy one disk and revenue from selling one disk:
- Cost per disk: She pays $5 for 4 disks, so unit cost = $\frac{\$5}{4} = \$1.25$ per disk.
- Revenue per disk: She earns $5 for 3 disks,  unit revenue = $\frac{\$5}{3} \approx \$1.67$ per disk.
---
### Step 3: Calculate Profit Per Disk
Unit profit = unit revenue - unit cost. Use a common denominator to subtract the fractions:
$$\text{Unit profit} = \frac{5}{3} - \frac{5}{4} = \frac{20}{12} - \frac{15}{12} = \frac{5}{12} \approx \$0.42 \text{ per disk}.$$
To simplify scaling, use the least common multiple of 3 and 4 (12 disks) to verify bulk profit:
- Cost to buy 12 disks: $3 \times \$5 = \$15$ (3 bundles of 4).
- Revenue from selling 12 disks: $4 \times \$5 = \$20$ (4 bundles of 3).
- Total profit for 12 disks: $\$20 - \$15 = \$5$, which matches the unit profit calculation ($12 \times \frac{5}{12} = 5$).
---
### Step 4: Scale to $100 Total Profit
If 12 disks generate $5 in profit, calculate how many 12-disk sets are needed to earn $100:
- Number of 12-disk sets = $\frac{\$100}{\$5} = 20$.
- Total number of disks = $20 \times 12 = 240$.
---
### Step 5: Verify and Rule Out Wrong Choices
Confirm the calculation for 240 disks:
- Total cost: $\frac{240}{4} \times 5 = 60 \times 5 = \$300$.
- Total revenue: $\frac{240}{3} \times 5 = 80 \times 5 = \$400$.
- Profit: $400 - 300 = \$100$, which meets the requirement.
Eliminate incorrect options:
  - A (100): Profit would be ~$41.67, far less than $100.
  - B (120): Profit is exactly $50, half the required amount.
  - C (200): Profit is ~$83.33, still below $100.
  - E (1200): Profit is $500, five times the required amount.
---
ANSWER: D

### Step 1: Restate the problem and define variables
We need to find the number of disks Maria sells to earn a $100 profit. Let \( x \) = number of disks she sells (which equals the number she buys, since she cannot sell disks she does not own).
---
### Step 2: Calculate cost per disk and selling price per disk
- Maria buys disks at 4 for $5, so the cost per disk is \( \frac{5}{4} = \$1.25 \).
- She sells disks at 3 for $5, so the selling price per disk is \( \frac{5}{3} \approx \$1.67 \).
---
### Step 3: Calculate profit per disk
Profit per disk = selling price per disk - cost per disk:
\[
\text{Profit per disk} = \frac{5}{3} - \frac{5}{4}
\]
Use a common denominator of 12 to subtract:
\[
\frac{5}{3} - \frac{5}{4} = \frac{20}{12} - \frac{15}{12} = \frac{5}{12} \approx \$0.4167 \text{ per disk}
\]
---
### Step 4: Set up the profit equation and solve for \( x \)
Total profit = (profit per disk) * number of disks sold. We want total profit = $100:
\[
\frac{5}{12}x = 100
\]
Solve for \( x \):
1.  Multiply both sides by 12: \( 5x = 1200 \)
2.  Divide by 5: \( x = 240 \)
---
### Step 5: Verify with group logic (to confirm)
The least common multiple of 3 (sell group size) and 4 (buy group size) is 12. For 12 disks:
- Total cost: \( \frac{12}{4} \times 5 = 3 \times 5 = \$15 \)
- Total revenue: \( \frac{12}{3} \times 5 = 4 \times 5 = \$20 \)
- Profit for 12 disks: \( 20 - 15 = \$5 \)
To earn $100, we need \( \frac{100}{5} = 20 \) groups of 12 disks: \( 20 \times 12 = 240 \). This matches our earlier result.
---
### Step 6: Rule out wrong choices
- A (100): Profit = \( \frac{5}{12} \times 100 \approx \$41.67 \), too low.
- B (120): Profit = \( \frac{5}{12} \times 120 = \$50 \), too low.
- C (200): Profit = \( \frac{5}{12} \times 200 \approx \$83.33 \), too low.
- E (1200): Profit = \( \frac{5}{12} \times 1200 = \$500 \), too high.
Only D (240) gives exactly $100 profit.
ANSWER: D

1. **Restate the problem**: Maria buys computer disks at a price of 4 for $5 and sells them at a price of 3 for $5. We need to determine how many computer disks she must sell to make a profit of $100.

2. **Calculate the cost price per disk**:
   - Maria buys 4 disks for $5.
   - Therefore, the cost price per disk is \( \frac{5}{4} = 1.25 \) dollars.

3. **Calculate the selling price per disk**:
   - Maria sells 3 disks for $5.
   - Therefore, the selling price per disk is \( \frac{5}{3} \approx 1.6667 \) dollars.

4. **Calculate the profit per disk**:
   - Profit per disk = Selling price per disk - Cost price per disk
   - Profit per disk = \( 1.6667 - 1.25 = 0.4167 \) dollars.

5. **Determine the number of disks needed to make a $100 profit**:
   - Let \( n \) be the number of disks Maria needs to sell.
   - The total profit is given by the profit per disk multiplied by the number of disks sold.
   - We set up the equation: \( 0.4167n = 100 \).

6. **Solve for \( n \)**:
   \[
   n = \frac{100}{0.4167} \approx 240
   \]

7. **Verify the solution**:
   - If Maria sells 240 disks, her profit will be \( 240 \times 0.4167 \approx 100 \) dollars.

8. **Rule out the wrong choices**:
   - A. 100: This would yield a profit of \( 100 \times 0.4167 \approx 41.67 \) dollars, which is less than $100.
   - B. 120: This would yield a profit of \( 120 \times 0.4167 \approx 50 \) dollars, which is less than $100.
   - C. 200: This would yield a profit of \( 200 \times 0.4167 \approx 83.34 \) dollars, which is less than $100.
   - E. 1200: This would yield a profit of \( 1200 \times 0.4167 \approx 500 \) dollars, which is more than $100.

The correct number of disks Maria must sell to make a profit of $100 is 240.

ANSWER: D

**Problem Restatement:**
Maria buys computer disks in bulk (4 disks for $5) and sells them in smaller bundles (3 disks for $5). We need to find out how many disks she must sell to make a profit of exactly $100.

---

**Step 1: Calculate the cost price per disk**
Maria buys 4 disks for $5.
Cost per disk = Total cost / Number of disks
= $5 / 4 disks
= **$1.25 per disk**

---

**Step 2: Calculate the selling price per disk**
Maria sells 3 disks for $5.
Selling price per disk = Total revenue / Number of disks
= $5 / 3 disks
≈ **$1.666... per disk**

---

**Step 3: Calculate the profit per disk**
Profit per disk = Selling price per disk – Cost price per disk
= $1.666... – $1.25
= **$0.4166... per disk**

---

**Step 4: Calculate the total number of disks needed for $100 profit**
Let \( n \) be the number of disks sold.
Total profit = Profit per disk × Number of disks
$100 = ($0.4166...) × \( n \)
\( n \) = $100 / $0.4166...
≈ 240 disks

---

**Step 5: Verify the calculation**
To avoid rounding errors, work with exact fractions:
Profit per disk = (5/3) – (5/4) = (20/12) – (15/12) = 5/12 dollars per disk.
For $100 profit: \( n \) × (5/12) = 100
\( n \) = 100 × (12/5) = 240 disks.

---

**Step 6: Rule out wrong choices**
A. 100 disks → Profit = 100 × (5/12) ≈ $41.67 (too low)
B. 120 disks → Profit = 120 × (5/12) = $50 (too low)
C. 200 disks → Profit = 200 × (5/12) ≈ $83.33 (too low)
D. 240 disks → Profit = 240 × (5/12) = $100 (correct)
E. 1200 disks → Profit = 1200 × (5/12) = $500 (too high)

---

ANSWER: D

I need to find the highest page number Pat can reach before running out of 2's, given that he only has 22 two's.

**Step 1: Understand what we're counting**
Pat is numbering pages: 1, 2, 3, 4, ..., and I need to find the largest page number he can reach before using more than twenty-two 2's total.

**Step 2: Count 2's used in different ranges**

*Pages 1-9:* Only page 2 has one 2.
- Count: 1 two

*Pages 10-19:* Only page 12 has one 2.
- Count: 1 two

*Pages 20-29:* 
- The tens digit is 2 in all: 20, 21, 22, 23, 24, 25, 26, 27, 28, 29 (10 twos)
- Page 22 also has another 2 in the units digit (1 two)
- Total for 20-29: 11 twos

*Pages 30-99:*
- Only the units digit can be 2: pages 32, 42, 52, 62, 72, 82, 92
- Count: 7 twos

**Total through page 99:** 1 + 1 + 11 + 7 = 20 twos

So far Pat has used 20 twos and can still use 2 more.

**Step 3: Continue to page 100+**

*Pages 100-109:* Only page 102 has one 2.
- Count: 1 two (total: 21)

*Pages 110-119:*
- Page 112 has one 2
- Page 120 would have a 2, but that's beyond 119
- Count: 1 two (total: 22)

**Step 4: Verify the count to page 119**

Let me recount to be sure:
- Pages 1-9: page 2 (1 two)
- Pages 10-19: page 12 (1 two)
- Pages 20-29: pages 20,21,22,23,24,25,26,27,28,29 each have a 2 in tens place (10 twos), plus page 22 has a 2 in units (1 two) = 11 twos
- Pages 30-99: pages 32,42,52,62,72,82,92 (7 twos)
- Pages 100-109: page 102 (1 two)
- Pages 110-119: page 112 (1 two)

Total: 1 + 1 + 11 + 7 + 1 + 1 = 22 twos ✓

**Step 5: Check page 120**

Page 120 would require one more 2 (in the tens place), but Pat has already used all 22 twos.

Therefore, Pat can number pages up to and including page **119**.

**Ruling out wrong answers:**
- A (22): Too low, he can use 2's beyond page 22
- B (99): Too low, he still has 2 twos left after page 99
- C (112): Too low, he hasn't used all 22 twos yet
- E (199): Too high, he would run out of 2's before reaching page 199

ANSWER: D

1. **What's asked:** Pat has unlimited supply of every digit except 2, of which he has only 22. He numbers pages 1, 2, 3, ... and we need to find how far he can go before running out of 2's.

2. **Count the 2's used as we number pages.**

3. **Pages 1–9:** Only the number "2" uses a 2. That's 1 two. Total so far: 1.

4. **Pages 10–99:** Count 2's in two-digit numbers.
   - Tens digit is 2: pages 20–29, that's 10 numbers, each with a 2 in tens place = 10 twos.
   - Units digit is 2: pages 12, 22, 32, 42, 52, 62, 72, 82, 92 = 9 twos.
   - So pages 10–99 use 10 + 9 = 19 twos.

5. **Running total through page 99:** 1 + 19 = 20 twos used. He has 22, so 2 remain.

6. **Continue past 99:** Pages 100, 101, ... none use a 2 until page 102 (uses one 2). After page 102: 21 twos used, 1 remaining.

7. **Next 2 appears:** Pages 103–111 use no 2's. Page 112 uses one 2. After page 112: 22 twos used, 0 remaining.

8. **Page 113 onward:** would need... page 113, 114, ... wait, page 120 needs a 2, but actually let's check: after 112 he has used all 22 twos. The next page needing a 2 is page 120 (and 121, 122...). He cannot write page 120. But he CAN write pages 113–119 since none contain a 2!

9. **Check pages 113–119:** 113, 114, 115, 116, 117, 118, 119 — none contain digit 2. So he can number all of these. He stops at 119, since page 120 requires a 2 he doesn't have.

10. **Verify total through 119:** twos used = 20 (through 99) + page 102 (1) + page 112 (1) = 22. Pages 113–119 use none. Page 120 would need one more — unavailable.

11. **Eliminate wrong choices:**
   - A (22): that's the number of 2's, not pages.
   - B (99): only uses 20 twos, can go further.
   - C (112): can still write 113–119 without 2's.
   - E (199): page 120 already needs a 2 he lacks.

ANSWER: D

## Step-by-Step Solution

**Step 1: Understand what's being asked.**
Pat can use unlimited copies of digits 0, 1, 3, 4, 5, 6, 7, 8, 9, but only twenty-two 2's. I need to find the highest page number he can reach before running out of 2's.

**Step 2: Count how many 2's appear in page numbers 1 through each candidate answer.**

I need to count the total number of 2's used when numbering pages sequentially.

**Step 3: Count 2's from pages 1–99.**

Pages 1–9: Only page 2 uses a "2" → **1 two**

Pages 10–99 (two-digit numbers):
- 2's in the tens place: pages 20–29 → **10 twos**
- 2's in the units place: pages 12, 22, 32, 42, 52, 62, 72, 82, 92 → **9 twos**

Total through page 99: 1 + 10 + 9 = **20 twos**

**Step 4: Continue counting from page 100 onward.**

After page 99, Pat has used 20 twos, with 2 remaining.

**Page 100:** No 2's → still 2 remaining  
**Page 101:** No 2's → still 2 remaining  
**Page 102:** One 2 → now 1 remaining  
**Page 103:** No 2's → still 1 remaining  
...continuing...  
**Page 112:** One 2 → now 0 remaining  

**Step 5: Check page 112.**
Page 112 contains the digit "2" once (in the units place). After numbering page 112, Pat has used exactly 22 twos and has none left.

**Step 6: Can he go further?**
Page 113 has no 2's, page 114 has no 2's, ... page 119 has no 2's. Wait — let me check if pages 113–119 contain any 2's.

Pages 113–119: None of these contain a 2. So Pat *could* number these pages too!

**Step 7: Recheck — when does he truly run out?**

After page 112, he has 0 twos left. Pages 113, 114, 115, 116, 117, 118, 119 contain **no 2's at all**. So he can continue numbering through page 119.

**Page 120** would require a "2" (tens digit), but he has none left. So he **cannot** number page 120.

**Step 8: Verify total 2's through page 119.**
- Pages 1–99: 20 twos
- Pages 100–119:
  - Page 102: one 2
  - Page 112: one 2
  - Pages 120+ not reached
  - Total in 100–119: **2 twos**
- Grand total: 20 + 2 = **22 twos** ✓

He can number through page 119, and page 120 would require a 2 that he doesn't have.

**Step 9: Rule out wrong answers.**
- A (22): Way too low — he'd only use 2 twos by page 22
- B (99): He's only used 20 twos by page 99, can go further
- C (112): He's used all 22 twos here, but pages 113–119 need no 2's, so he CAN continue
- **D (119): Correct** — he can reach page 119; page 120 needs a 2
- E (199): Would require more than 22 twos

ANSWER: D

1. **What is being asked?**  
   We need to find the **largest page number** Pat can write using his digits, given that he has:
   - unlimited 0s, 1s, 3s, 4s, 5s, 6s, 7s, 8s, 9s
   - but only **twenty-two 2s**

2. **Which digit matters?**  
   Since all digits except 2 are unlimited, the only thing that can stop him is running out of **2s**.  
   So we count how many 2s are needed to number pages from 1 upward.

3. **Count the 2s from 1 to 99**  
   The digit 2 appears:
   - once in the numbers **2, 12, 20–29, 32, 42, ..., 92**
   
   A standard way to count:
   - In the **ones place**, 2 appears in: 2, 12, 22, 32, 42, 52, 62, 72, 82, 92 → **10 times**
   - In the **tens place**, 2 appears in: 20–29 → **10 times**

   But note **22** has two 2s, and it is counted once in each place, which is correct.

   So from 1 to 99, there are:
   \[
   10 + 10 = 20
   \]
   digits 2.

4. **How many 2s are used up to 99?**  
   Exactly **20** of his 22 2s are used.

5. **Can he write 100?**  
   The number **100** uses no 2s, so yes.  
   The number **101** also uses no 2s.  
   The number **102** uses **one 2**.

   After 99, he has:
   \[
   22 - 20 = 2
   \]
   2s left, so he can write numbers until he uses those last 2s.

6. **Check 102 and onward**  
   - 100: 0 twos
   - 101: 0 twos
   - 102: 1 two
   - 103, 104, ... 109: still only that one 2 in 102 so far
   - 112 uses **one more 2**, making a total of 2 extra 2s after 99

   So he can certainly get through **112**.

7. **Can he write 113?**  
   To go from 100 up through 113, the twos used are:
   - 102 → 1 two
   - 112 → 1 two  
   Total after 99: **2 more twos**, for a grand total of:
   \[
   20 + 2 = 22
   \]
   exactly all of his 2s.

8. **Can he write 114?**  
   The number 114 uses no 2s, but the problem asks how far he can **number the pages** in order. To write page numbers consecutively, once he reaches 113, the next page 114 would be fine, but the issue is the next time he needs another 2 would be **120**, which needs one 2, and that would exceed his supply.  
   So the last page number he can reach is **119**.

9. **Match with the choices**  
   The largest possible page number is **119**, which is choice **D**.

ANSWER: D

To find how far Pat Peano can number the pages of his scrapbook, we need to count how many times the digit $2$ is used as we number the pages starting from page $1$. He has exactly twenty-two $2$'s.

### **Step 1: Count the number of $2$'s used in pages 1 to 99**
We can count the occurrences of the digit $2$ in the units place and the tens place separately for numbers from $1$ to $99$:
* **Units place:** The digit $2$ appears in the units place of:
  $$2, 12, 22, 32, 42, 52, 62, 72, 82, 92$$
  This is a total of $10$ times.
* **Tens place:** The digit $2$ appears in the tens place of:
  $$20, 21, 22, 23, 24, 25, 26, 27, 28, 29$$
  This is a total of $10$ times.

*(Note: The number $22$ is counted once in the units place and once in the tens place, which correctly accounts for the two $2$'s used to write "$22$").*

Adding these together, the number of $2$'s used for pages $1$ through $99$ is:
$$10 + 10 = 20 \text{ times}$$

At page $99$, Pat has used $20$ of his $22$ available $2$'s. He has $2$ remaining.

---

### **Step 2: Count the $2$'s used for pages 100 and beyond**
Now we continue numbering the pages from $100$ onward:
* Pages $100$ and $101$ do not contain the digit $2$.
* Page $102$ contains one $2$. 
  * **Total $2$'s used so far:** $20 + 1 = 21$.
* Pages $103$ through $111$ do not contain the digit $2$.
* Page $112$ contains one $2$.
  * **Total $2$'s used so far:** $21 + 1 = 22$.

At page $112$, Pat has used all $22$ of his $2$'s.

---

### **Step 3: Determine the maximum page number**
Since Pat has no $2$'s left, he can continue numbering pages only as long as they do not require the digit $2$:
* Pages $113, 114, 115, 116, 117, 118,$ and $119$ do not require any $2$'s.
* Page $120$ requires a $2$ (in the tens place), which Pat does not have.

Thus, the last page he can completely number is **119**.

---

### **Step 4: Evaluate the choices**
* **A. 22:** Incorrect. Pat can go much further because most numbers do not contain the digit $2$.
* **B. 99:** Incorrect. At page $99$, Pat has only used $20$ of his $2$'s and still has $2$ left.
* **C. 112:** Incorrect. Although he runs out of $2$'s at page $112$, he can still write pages $113$ through $119$ as they do not require any $2$'s.
* **D. 119:** Correct. This is the last page he can write before needing another $2$ for page $120$.
* **E. 199:** Incorrect. He would need many more $2$'s to reach $199$ (e.g., for $120, 121, 122$, etc.).

ANSWER: D

The problem requires determining the largest page number \(N\) that can be reached when numbering pages consecutively from 1 onward, subject to the constraint of using at most 22 digits equal to 2 (with all other digits available without limit).

To solve this, compute the cumulative number of 2 digits used up to various values of \(N\) in the given choices, separating the count into ranges based on the number of digits per page number.

- From pages 1 to 9 there is exactly one occurrence of the digit 2 (on page 2).  
- From pages 10 to 99 the tens digit contributes 10 occurrences of 2 (all of 20–29) and the units digit contributes 9 occurrences of 2 (12, 22, 32, …, 92). This gives a total of 19 occurrences, so the cumulative count through page 99 is \(1 + 19 = 20\).

Exactly two 2 digits therefore remain available.  

- From pages 100 to 119 every page number begins with the digit 1, so 2 digits can appear only in the tens or units place. The only such occurrences are the units digit of 102 and the units digit of 112. These two pages therefore consume exactly the two remaining 2 digits, bringing the cumulative total to 22. Page 119 itself contains no digit 2.

The next page, 120, contains a tens digit equal to 2 and would therefore require a 23rd occurrence of the digit 2, which exceeds the supply. Consequently the largest reachable page is 119.

(The same cumulative count of 22 is reached at 112, but the calculation above shows that the two extra pages 113–119 can be numbered without using any additional 2 digits, so 119 is strictly larger and still feasible.)

This rules out every other choice:  
- 22 and 99 both lie strictly before the point at which all 22 allowed 2 digits have been used.  
- 112 uses only 21 twos and is therefore smaller than the feasible value 119.  
- 199 lies well beyond 120 and would require many more than 22 occurrences of the digit 2.

Thus the maximum reachable page is 119.

ANSWER: D

1. **Understanding the Problem**: We are given that Pat Peano has an ample supply of digits 0, 1, 3, 4, 5, 6, 7, 8, 9, but is limited to twenty-two 2's. We need to determine the maximum number of pages he can number in his scrapbook using these digits.

2. **Setting Up the Math**: To number pages, Pat will start from 1 and continue sequentially until he runs out of the digit '2'. We need to track how many '2's are used in numbering the pages.

3. **Analyzing the Use of '2's**: The digit '2' appears in the numbers 2, 12, 20, 21, 22, ..., 29, 32, ..., 92, 120, 121, ..., 122, etc. We need to count how many times '2' is used.

4. **Counting '2's in the First 100 Numbers**: 
   - '2' is used once in the numbers 2, 12, 20-29 (total of 11 times for 20-29 and 1 time each for 2 and 12, so 1 + 1 + 10 = 12 times for 2, 12, and 20-29), and 32, 42, 52, 62, 72, 82, 92 (7 more times), totaling 2 (for 2 and 12) + 10 (for 20-29) + 7 (for the tens place in other numbers) = 19 times in the first 100 numbers.

5. **Counting '2's Beyond 100**: 
   - For numbers 100-119, '2' is used in 102, 112, 120-122 (3 times here because there are three '2's in 122, but we are limited by the count of '2's available). So, for 102 and 112, that's 2 '2's. For 120, 121, and 122, that's 4 '2's (1 for 120, 1 for 121, and 2 for 122), totaling 6 '2's.

6. **Doing the Computation**: 
   - Pat has 22 '2's. In the first 100 numbers, he uses 19 '2's. This leaves him with 22 - 19 = 3 '2's for numbers beyond 100.
   - He can number 101, 102 (uses 1 '2'), 103, ..., 112 (uses 1 '2'), so he has used 2 '2's by 112. He has 1 '2' left, which he can use for 120 (using the last '2'), but he cannot number 121 or beyond because he will run out of '2's.

7. **Ruling Out Wrong Choices**:
   - A. 22: Too low because he can number beyond 22.
   - B. 99: Possible because he hasn't used up all his '2's by 99, but let's check further.
   - C. 112: This matches our count because he can number up to 112 with the '2's he has (using them for 2, 12, 20-29, 32, 42, 52, 62, 72, 82, 92, 102, 112, and 120).
   - D. 119: Too high because to reach 119, he needs to number 120 (which he can do with his remaining '2's), but he won't have enough '2's to number beyond 120 (like 121 or 122).
   - E. 199: Too high because it requires many more '2's than he has.

8. **Conclusion**: Pat can number up to 112 because he has enough '2's to cover the numbers up to that point.

ANSWER: C

1. **Understand the problem:** Pat has an unlimited supply of digits except for '2', of which he has exactly 22 copies. He numbers scrapbook pages sequentially starting from 1. He can continue as long as he has the necessary digits. The only limitation is the number of '2's. He must stop when the next page number requires a '2' but none are left.

2. **Count the total number of '2's used from page 1 upward:**  
   - Pages 1–9: Only the number 2 uses a '2'. **Total = 1**.  
   - Pages 10–19: Only 12 uses a '2'. **Total = 1** (cumulative = 2).  
   - Pages 20–29: All ten numbers have a '2' in the tens place (10 '2's). Unit digits: only 22 has an extra '2' (1 more). **Total = 11** (cumulative = 13).  
   - Pages 30–39: Only 32 has one '2'. **Cumulative = 14**.  
   - Pages 40–49: 42 → **15**.  
   - Pages 50–59: 52 → **16**.  
   - Pages 60–69: 62 → **17**.  
   - Pages 70–79: 72 → **18**.  
   - Pages 80–89: 82 → **19**.  
   - Pages 90–99: 92 → **20**.  
   So up to page 99, Pat has used 20 of his 22 '2's.

3. **Continue counting:**  
   - Pages 100–109: Only 102 uses a '2'. **Cumulative = 21**.  
   - Pages 110–119: Only 112 uses a '2'. **Cumulative = 22**.  
   The 22nd '2' is used on page 112.

4. **Check remaining pages after 112:**  
   - Pages 113, 114, 115, 116, 117, 118, 119 contain no '2'. Because Pat has plenty of all other digits, he can number these pages without using any more '2's.  
   - Page 120 contains a '2' (tens digit) and would require a 23rd '2', which Pat does not have.

5. **Conclusion:** The farthest he can number sequentially is page 119. Checking choices: A (22), B (99), C (112), D (119), E (199).  
   The correct choice is D.

ANSWER: D

# Counting 2's in Page Numbers

## Step 1: Understand the Problem
Pat can number pages 1, 2, 3, ... using digit cutouts. He has unlimited 0, 1, 3, 4, 5, 6, 7, 8, 9 but only **twenty-two 2's**. I need to find the highest page number he can reach before running out of 2's (and being unable to make the next page number).

## Step 2: Count 2's Used in Pages 1–99

| Page Range | Pages containing a 2 | Number of 2's used |
|---|---|---|
| 1–9 | 2 | 1 |
| 10–19 | 12 | 1 |
| 20–29 | 20, 21, **22**, 23, 24, 25, 26, 27, 28, 29 | 1+1+**2**+1+1+1+1+1+1+1 = **11** |
| 30–39 | 32 | 1 |
| 40–49 | 42 | 1 |
| 50–59 | 52 | 1 |
| 60–69 | 62 | 1 |
| 70–79 | 72 | 1 |
| 80–89 | 82 | 1 |
| 90–99 | 92 | 1 |

**Total 2's through page 99:** 1 + 1 + 11 + 7×1 = **20 twos used**

He has 22 − 20 = **2 twos remaining**.

## Step 3: Count 2's Used in Pages 100+

- Pages 100–109: Only page **102** contains a 2 → uses **1** two. (Running total: 21)
- Pages 110–119: Only page **112** contains a 2 → uses **1** two. (Running total: **22**)

All 22 twos are now used up after page 112.

## Step 4: Can He Continue Past Page 112?

Even though he has no 2's left, he can still number pages that **don't require a 2**:
- 113 → digits 1, 1, 3 ✓
- 114 → digits 1, 1, 4 ✓
- 115 → digits 1, 1, 5 ✓
- 116 → digits 1, 1, 6 ✓
- 117 → digits 1, 1, 7 ✓
- 118 → digits 1, 1, 8 ✓
- 119 → digits 1, 1, 9 ✓

**Page 120** requires digits 1, **2**, 0 — he has no 2's left! He **cannot** number page 120.

## Step 5: Rule Out Wrong Choices

- **A (22):** Far too low; he barely uses 2 twos by page 9.
- **B (99):** Uses 20 twos, but he still has 2 left to continue.
- **C (112):** This is the last page *using* a 2, but pages 113–119 need no 2's — a classic trap!
- **E (199):** Pages 120–129 alone would need many more 2's (120, 121, ..., 129 all have a 2 in the tens place).

ANSWER: D

 **Step 1: Restate the problem.**  
Pat is numbering pages consecutively starting from 1. He has an unlimited supply of every digit except 2, of which he has only 22. We need the largest page number \(N\) such that every page from 1 through \(N\) can be written without exceeding 22 copies of the digit 2. Once the 22nd “2” is used, any later page that contains a 2 cannot be written, and because the pages are consecutive he cannot skip such a page. Thus we seek the greatest \(N\) for which the total count of digit 2’s in the list \(1,2,\dots,N\) is at most 22, and the very next page would push that total over 22.

**Step 2: Count the 2’s used from page 1 to page 99.**  
In any block of 100 consecutive numbers (say 0–99), each digit 0–9 appears exactly 10 times in the units place and 10 times in the tens place.  
- Units digit 2: pages 2, 12, 22, 32, 42, 52, 62, 72, 82, 92 → 10 twos.  
- Tens digit 2: pages 20, 21, 22, 23, 24, 25, 26, 27, 28, 29 → 10 twos.  

Total 2’s from 1 to 99: \(10+10 = 20\).  
(Notice that page 22 is counted twice, correctly accounting for its two 2’s.)

**Step 3: Determine how many 2’s remain.**  
\(22 - 20 = 2\) twos are still available after page 99.

**Step 4: Continue counting into the 100s until the 2’s run out.**  
The hundreds digit is 1 for pages 100–199, so it contributes no 2’s. We scan the next pages:

| Page | 2’s on this page | Cumulative 2’s |
|------|------------------|----------------|
| 100–101 | 0 | 20 |
| 102 | 1 | 21 |
| 103–111 | 0 | 21 |
| 112 | 1 | 22 |
| 113–119 | 0 | 22 |

After page 112, Pat has used exactly 22 twos. Pages 113 through 119 contain no digit 2, so he can still write them even though he has zero 2’s left.

**Step 5: Identify the first page that cannot be written.**  
The next page is 120, which has a 2 in the tens place. Writing it would require a 23rd copy of the digit 2, which Pat does not have. Therefore page 120 is impossible.

**Step 6: State the farthest reachable page.**  
The last page Pat can successfully number is **119**.

**Step 7: Rule out the wrong choices.**  
- **A. 22:** Only 6 twos are needed up to page 22; he could go much farther.  
- **B. 99:** Uses 20 twos, leaving 2 unused.  
- **C. 112:** Uses exactly 22 twos, but pages 113–119 need no 2’s, so he can continue past 112.  
- **E. 199:** Pages 1–199 contain 40 twos (20 in 1–99 and 20 more in 100–199), far exceeding 22.  

Only **D. 119** fits the calculation.

ANSWER: D

Here is the step-by-step solution:

**1. Understand the Goal**
We need to find the largest integer $N$ (the last page number) such that the total number of times the digit '2' appears in the page numbers from 1 to $N$ is less than or equal to 22. Pat has an unlimited supply of all other digits.

**2. Count the digit '2' in pages 1 through 99**
We break this down by the number of digits in the page number.

*   **One-digit pages (1–9):**
    The only page with a '2' is page 2.
    *Count:* **1**

*   **Two-digit pages (10–99):**
    *   **Tens place:** The digit '2' appears in the tens place for pages 20, 21, 22, ..., 29. That is 10 pages.
        *Count:* **10**
    *   **Units place:** The digit '2' appears in the units place for pages 12, 22, 32, ..., 92. That is 9 pages. (Note: Page 22 was already counted for its tens digit, but here we are counting the units digit occurrence).
        *Count:* **9**

*   **Total for 1–99:**
    $$1 \text{ (from 1-9)} + 10 \text{ (tens)} + 9 \text{ (units)} = 20$$
    So, numbering up to page 99 uses exactly **twenty** 2's.

**3. Determine the remaining supply of 2's**
Pat started with 22 twos.
After numbering up to page 99, he has used 20.
$$\text{Remaining} = 22 - 20 = 2 \text{ twos}$$

**4. Continue numbering from page 100**
We now proceed page by page (or by checking for the digit 2) to see how far the remaining 2 twos will take us.

*   **Pages 100–101:** Contain no 2's. (Remaining: 2)
*   **Page 102:** Contains one '2' (in the units place).
    *   Use 1 two. **Remaining: 1**.
*   **Pages 103–111:** Contain no 2's. (Remaining: 1)
*   **Page 112:** Contains one '2' (in the units place).
    *   Use 1 two. **Remaining: 0**.
*   **Pages 113–119:** Contain no 2's. Since we have 0 twos remaining but these pages don't require any, we can number all of them successfully.
*   **Page 120:** Contains one '2' (in the tens place).
    *   We need 1 two, but we have 0 remaining. **We cannot number page 120.**

Therefore, the last page Pat can number is **119**.

**5. Evaluate the Choices**
*   **A. 22:** Too low. We have many 2's left after page 22.
*   **B. 99:** We use 20 twos here. We have 2 left, so we can go further than 99.
*   **C. 112:** We use our last 2 on page 112. However, pages 113 through 119 do not contain the digit 2, so we can continue numbering until we hit a page that requires a 2 (which is 120). Thus, 112 is not the maximum.
*   **D. 119:** This matches our calculation. We use the 22nd '2' on page 112, and can freely number pages 113-119.
*   **E. 199:** To reach 199, we would have to number the 120s (which requires eleven 2's just for the tens digit: 120-129). We only had 2 twos left after page 99, which is insufficient.

ANSWER: D

**Step 1 – Understand the problem**

Pat has an unlimited supply of the digits 0, 1, 3, 4, 5, 6, 7, 8, 9, but only **22 copies of the digit 2**.  
He numbers the pages of his scrapbook consecutively starting with page 1.  
We must find the greatest page number \(N\) such that the total number of the digit 2 used to write all page numbers from 1 up to \(N\) does not exceed 22.

---

**Step 2 – Count the 2’s from 1 to 99**

Treat each two‑digit block \(00\) to \(99\).  

* In the units position each digit (0–9) appears exactly 10 times.  
* In the tens position each digit also appears exactly 10 times.

Hence the digit 2 appears  

\[
10\text{ (units)} + 10\text{ (tens)} = 20\text{ times}
\]

in the numbers \(0\) to \(99\).  
Since page 0 does not contain a 2, the count for pages **1 to 99** is also 20.  

Thus after page 99 Pat has used only 20 of his 22 twos.

---

**Step 3 – Add the pages from 100 to 112**

Pages 100–112 are three‑digit numbers (hundreds digit = 1).  
No hundreds digit is a 2, and none of the tens digits is a 2 in this range.

Only the units digit can be a 2. The numbers that contain a 2 are:

* 102 – one 2 (units)
* 112 – one 2 (units)

So from 100 to 112 there are **2 more** twos.

Total twos used up to page 112:

\[
20\ (\text{1–99}) + 2\ (\text{100–112}) = 22.
\]

Pat has now exhausted all his 2’s.

---

**Step 4 – Can we go beyond page 112?**

Pages 113, 114, 115, 116, 117, 118, 119 contain **no** digit 2, so they can be written using only the unlimited digits.  

The next page, 120, needs a 2 (its tens digit). Since Pat has no 2 left, he cannot label page 120 (or any later page that contains a 2).

Therefore the farthest page he can number is **119**.

---

**Step 5 – Eliminate the other answer choices**

* A. 22 – far from the maximum (he can go beyond 22).  
* B. 99 – uses only 20 twos; we can go further.  
* C. 112 – uses exactly the 22 twos, but pages 113–119 are still possible.  
* D. 119 – the highest page that can be labeled without needing another 2.  
* E. 199 – would require many more 2’s (far exceeding 22).

Only **119** satisfies the constraint.

---

**Final answer**

\[
\boxed{\text{ANSWER: D}}
\]

1. **Restate what is being asked**:
   - We need to find out the highest - numbered page that Pat can number his scrapbook with, given that he has only twenty - two 2's and plenty of other digits (0, 1, 3, 4, 5, 6, 7, 8, 9).
2. **Count the number of 2's in single - digit numbers**:
   - The single - digit numbers are from 0 to 9. The only single - digit number with a 2 is 2 itself. So, there is 1 two in single - digit numbers.
3. **Count the number of 2's in two - digit numbers**:
   - The two - digit numbers range from 10 to 99.
   - For the numbers in the form of $x2$ (where $x$ can be 1, 3, 4, 5, 6, 7, 8, 9), there are 8 such numbers (12, 32, 42, 52, 62, 72, 82, 92).
   - For the numbers in the form of $2x$ (where $x$ can be 0, 1, 3, 4, 5, 6, 7, 8, 9), there are 9 such numbers (20, 21, 23, 24, 25, 26, 27, 28, 29).
   - The total number of 2's in two - digit numbers is $8 + 9=17$.
4. **Calculate the total number of 2's used so far**:
   - The total number of 2's used in single - digit and two - digit numbers is $1+17 = 18$.
5. **Determine how many more 2's we can use for three - digit numbers**:
   - We know that Pat has 22 2's in total. After using 18 for single - digit and two - digit numbers, the number of remaining 2's is $22−18 = 4$.
   - In three - digit numbers, if we start with 100, the first number with a 2 in the units place is 102 (uses 1 two), then 112 (uses 1 two), 120 (uses 1 two), 121 (uses 2 twos). But we want to use the remaining 4 2's in the most efficient way.
   - We can use the 2's in the numbers 102, 112, 120, 121. After using these 4 2's, the next number we can start numbering from is 123. But if we consider the last number we can fully number with the available 2's, we note that we can reach 119. In the numbers from 100 - 119, the 2's are in 102 and 112, which uses 2 more 2's in addition to the 18 we used before, and we still have enough 2's to account for all the numbers up to 119.
6. **Rule out the wrong choices**:
   - **Choice A**: 22 is much too small. We know we can number more than just 22 pages since we have used 2's in single - digit and two - digit numbers and still have some left for three - digit numbers.
   - **Choice B**: 99 is incorrect because we have not used up all our 2's by the time we reach 99. We have only used 18 2's by 99, and we have 22 2's in total.
   - **Choice D**: As we calculated above, we can reach 119 with the 22 2's we have.
   - **Choice E**: If we try to reach 199, we will use more than 22 2's. For example, just considering the numbers from 120 - 129, there are 10 numbers with at least one 2, and there are many more 2's in other three - digit numbers starting with 1 and having 2 in other positions.

ANSWER: D

### Step 1: Restate the problem
Pat only has 22 copies of the digit 2, with unlimited copies of all other digits. We need to find the highest page number he can write sequentially before running out of 2s, meaning we count all occurrences of the digit 2 across page numbers until we use all 22, then find the last valid page number.

### Step 2: Count occurrences of the digit 2 up to 99
For numbers 1–99, 2s appear in two places:
- **Units place**: 2,12,22,32,42,52,62,72,82,92 → 10 total 2s.
- **Tens place**: All numbers 20–29 → 10 total 2s.
Total 2s used by page 99: 10 + 10 = 20. Pat has 22-20=2 2s remaining, so he can write pages past 99, ruling out options A (22) and B (99) immediately.

### Step 3: Count 2s for pages after 99
All numbers 100–199 have a 1 in the hundreds place, so 2s only appear in the units/tens places:
- The next 2 after page 99 is at page 102 (units digit 2) → uses 1 2, leaving 1 remaining.
- The next 2 is at page 112 (units digit 2) → uses the last 2, total used = 22.

### Step 4: Check pages after 112 to find the maximum valid page
Pages 113–119 only use digits 1 and 3–9, which Pat has unlimited copies of. The next page requiring a 2 is 120 (tens digit 2), which would need a 23rd 2 Pat does not have. He can write all pages up to 119.

### Step 5: Rule out remaining wrong options
- Option C (112): Wrong, because Pat can write all pages after 112 up to 119 with no additional 2s.
- Option E (199): Wrong, as reaching 199 would require dozens more 2s (for 120–129, 132, etc.) that Pat does not have.

ANSWER: D

### Step 1: Restate the Problem
Pat is numbering scrapbook pages starting from 1, using physical digit cut-outs. He has unlimited supply of all digits except 2, of which he only has 22. We need to find the highest page number he can write without running out of 2s.
---
### Step 2: Count 2s Used for 1-Digit and 2-Digit Pages (1–99)
We count 2s by digit place (units, tens) to avoid double-counting numbers like 22 (which has two 2s):
1.  **1-digit pages (1–9):** Only the page number 2 uses a 2. Total 2s used: 1.
2.  **2-digit pages (10–99):**
    -   *Tens place:* 2 appears in 20–29, which is 10 pages, contributing 10 twos.
    -   *Units place:* 2 appears in 12, 22, 32, 42, 52, 62, 72, 82, 92: 9 pages, contributing 9 twos (the extra 2 in 22 is already counted in the tens place, so this is correct).
    -   Total 2s for 2-digit pages: 10 + 9 = 19.
3.  **Total 2s used up to page 99:** 1 + 19 = 20. Pat has 22 total 2s, so 2 remain after page 99.
---
### Step 3: Count 2s for 3-Digit Pages (100 onwards)
We now use the remaining 2 twos on 3-digit pages, starting from 100:
-   **100–109:** Only page 102 has a 2 (units place), using 1 two. Remaining 2s: 1.
-   **110–119:** Only page 112 has a 2 (units place), using the last remaining two. Remaining 2s: 0.
-   **Pages 113–119:** None of these numbers contain the digit 2, so no additional 2s are needed to write them.
---
### Step 4: Verify the Maximum Page
-   Up to page 119, exactly 22 twos are used (matching Pat's supply).
-   Page 120 has a 2 in the tens place, which would require a 23rd two Pat does not have, so he cannot write page 120 or any higher page with a 2.
---
### Step 5: Rule Out Wrong Choices
-   **A. 22:** Only uses 6 twos (far fewer than 22), so Pat can go much higher.
-   **B. 99:** Uses 20 twos, but Pat has 2 more, so he can go higher.
-   **C. 112:** Uses all 22 twos, but pages 113–119 have no 2s, so Pat can write these and go higher.
-   **E. 199:** Would require 41 total twos (far more than 22), which is impossible.
---
ANSWER: D

1. **Identify the problem**:  We need to determine the maximum page number Pat can reach given his limited supply of the digit 2.

2. **Count the occurrences of the digit 2**:  We need to count how many times the digit 2 appears in each range of numbers:
   - From 1 to 9: 2 appears 1 time (2).
   - From 10 to 19: 2 appears 2 times (12, 20).

3. **Generalize the count for tens and hundreds**:  We need to extend this counting to higher ranges:
   - From 1 to 99: 
     - Units place: 10 times (2, 12, 20, 21, 22, 23, 24, 25, 26, 27).
     - Tens place: 10 times (20, 21, 22, 23, 24, 25, 26, 27, 28, 29).
     - Total for 1 to 99: \(10 + 10 = 20\).

4. **Check numbers up to 199**:  We now count the occurrences of the digit 2 from 1 to 199:
   - From 1 to 99: 20 times (as calculated above).
   - From 100 to 109: 2 appears 1 time (102).
   - From 110 to 119: 2 appears 1 time (112).
   - From 120 to 129: 2 appears 10 times (120, 121, 122, 123, 124, 125, 126, 127, 128, 129).
   - From 130 to 139: 2 appears 1 time (132).
   - From 140 to 149: 2 appears 1 time (142).
   - From 150 to 159: 2 appears 1 time (152).
   - From 160 to 169: 2 appears 1 time (162).
   - From 170 to 179: 2 appears 1 time (172).
   - From 180 to 189: 2 appears 1 time (182).
   - From 190 to 199: 2 appears 1 time (192).

5. **Sum the occurrences**:  Add up all the occurrences of the digit 2 from 1 to 199:
   - 20 (from 1 to 99) + 10 (from 120 to 129) + 8 (from 102, 112, 132, 142, 152, 162, 172, 182, 192) = 38.

6. **Determine the maximum page number**:  Since Pat has only 22 twos, we need to find the highest number he can reach without exceeding this limit. From our count:
   - Up to 119, the total number of 2's used is 22.

Thus, the maximum page number Pat can reach is 119.

ANSWER: D

rm **ABC**, where **A** is the hundreds digit, **B** is the tens digit, and **C** is the units digit.
- **Hundreds place:**
  - The digit **2** appears in the hundreds place for numbers **200 to 299** → 100 numbers.
  - Each of these contributes **1** to the count of **2's** in the hundreds place.
  - Since we only have **2** remaining **2's**, we cannot use any numbers in the 200s.
- **Tens place:**
  - The digit **2** appears in the tens place for numbers **120-129, 220-229, ..., 920-929** → 10 numbers per hundreds digit.
  - For hundreds digit **1**, the numbers are **120-129** → 10 numbers.
  - Each of these contributes **1** to the count of **2's** in the tens place.
  - We have **2** remaining **2's**, so we can use **2** of these numbers.
  - For example, **120** and **121** would use **2** in the tens place for both numbers, but we only have **2** **2's** left, so we can only use **120** and **122** (but **122** uses two **2's** in the tens and units place, which would exceed our limit).
  - Instead, we can use **120** and **121** (each uses **1** **2** in the tens place).
  - Total **2's** used in tens place: **2** (for **120** and **121**).
- **Units place:**
  - The digit **2** appears in the units place for numbers **102, 112, 122, ..., 192** → 10 numbers.
  - We have already used our remaining **2** **2's** in the tens place, so we cannot use any additional **2's** in the units place.

#### **Cumulative total after 3-digit numbers:**
- 1-digit: 1
- 2-digit: 19
- 3-digit: 2
- **Total:** 1 + 19 + 2 = **22**

#### **Highest page number:**
- We have used all **22** **2's** by numbering up to **121**.
- However, let's verify:
  - **120** uses **1** **2** (tens place).
  - **121** uses **1** **2** (tens place).
  - Total **2's** used: 20 (from 1-digit and 2-digit) + 2 = **22**.
  - The next number, **122**, would use **2** **2's** (tens and units place), which would exceed our limit of **22**.

Thus, the highest page number Pat can label is **119** (since **120** and **121** use up the remaining **2's**, but **119** does not use any **2's** and is the highest number before we start using the remaining **2's**).

Wait, let's re-examine this carefully:

- Up to **99**, we have used **20** **2's**.
- We have **2** **2's** left.
- The next numbers are **100, 101, 102, ..., 119, 120, 121, 122, ...**
- **102** uses **1** **2** (units place).
- **112** uses **1** **2** (units place).
- **120** uses **1** **2** (tens place).
- **121** uses **1** **2** (tens place).
- **122** uses **2** **2's** (tens and units place).

If we use **102** and **112**, that uses **2** **2's**, bringing the total to **22**.
- The highest page number in this case would be **112**.

But wait, the options include **119**, which is higher than **112**. Let's see if we can do better:

Alternative approach:
- Use **120** and **121**, which use **2** **2's** (one in the tens place for each).
- This also brings the total to **22**.
- The highest page number in this case would be **121**.

But **121** is not among the options. The closest option below **121** is **119**.

Wait, let's check the count for **119**:
- Up to **99**: 20 **2's**.
- From **100** to **119**:
  - **102**: 1 **2** (units place).
  - **112**: 1 **2** (units place).
  - Total **2's** used in this range: 2.
  - Cumulative total: 20 + 2 = **22**.
- The highest page number here is **119**, which does not use any **2's** itself.

Thus, the highest page number Pat can label without exceeding **22** **2's** is **119**.

---

### **Step 3: Rule Out Wrong Choices**
- **A. 22**: Too low, as we can go much higher.
- **B. 99**: Uses only **20** **2's**, so we can go higher.
- **C. 112**: Uses **22** **2's** (20 up to 99 + 2 for 102 and 112), but we can do better.
- **D. 119**: Uses **22** **2's** (20 up to 99 + 2 for 102 and 112), and is the highest option that fits.
- **E. 199**: Would require many more **2's** (e.g., 120-129, 200-299, etc.), far exceeding 22.

---

### **Final Answer:**
ANSWER: D