If n and k are integers whose product is 400, which of the following statements must be true?

1. Translate the problem requirements: We need to find which statement MUST be true for ALL possible integer pairs (n,k) where \(\mathrm{n} \times \mathrm{k} = 400\). The key word is "must" - meaning it's true in every single case, not just some cases.
2. Analyze the prime factorization constraint: Since \(400 = 2^4 \times 5^2\), examine what this tells us about the nature of any factors n and k.
3. Test each answer choice systematically: For each option, either find a counterexample that proves it's not always true, or demonstrate why it must always be true.
4. Focus on parity (odd/even) relationships: Since several answer choices involve odd/even properties, analyze how the factors of 2 in 400's prime factorization must be distributed between n and k.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this question is really asking. We have two integers n and k, and when we multiply them together, we get 400. The question wants to know which statement MUST be true - meaning it's true for every single possible pair of integers that multiply to give 400.

This is different from asking what COULD be true or what's true SOMETIMES. We need something that's true ALL the time, no matter which specific values of n and k we choose (as long as \(\mathrm{n} \times \mathrm{k} = 400\)).

Think of it this way: if someone gives us any pair of integers that multiply to 400, we should be able to confidently say our chosen statement applies to that pair.

Process Skill: TRANSLATE - Converting "must be true" to "true for ALL cases"

2. Analyze the prime factorization constraint

Now let's break down 400 into its basic building blocks. When we factor 400 completely, we get:

\(400 = 16 \times 25 = 2^4 \times 5^2\)

This means 400 has exactly four factors of 2 and two factors of 5. Here's the key insight: when we multiply n and k to get 400, we're essentially dividing these prime factors between n and k.

For example:

• n could get all four 2's and both 5's: \(\mathrm{n} = 2^4 \times 5^2 = 400, \mathrm{k} = 1\)
• n could get two 2's and one 5: \(\mathrm{n} = 2^2 \times 5 = 20, \mathrm{k} = 2^2 \times 5 = 20\)
• n could get no 2's and no 5's: \(\mathrm{n} = 1, \mathrm{k} = 2^4 \times 5^2 = 400\)

The crucial point is that since 400 contains factors of 2, at least one of n or k must contain factors of 2 (making it even).

Process Skill: INFER - Drawing the non-obvious conclusion about factor distribution

3. Test each answer choice systematically

Let's check each option by either finding a counterexample or proving it's always true:

Choice A: \(\mathrm{n} + \mathrm{k} > 0\)

Counterexample: \(\mathrm{n} = -20, \mathrm{k} = -20\). Check: \((-20) \times (-20) = 400\) ✓. But \(\mathrm{n} + \mathrm{k} = -20 + (-20) = -40 < 0\). So this isn't always true.

Choice B: \(\mathrm{n} \neq \mathrm{k}\)

Counterexample: \(\mathrm{n} = 20, \mathrm{k} = 20\). Check: \(20 \times 20 = 400\) ✓. But \(\mathrm{n} = \mathrm{k}\), so this isn't always true.

Choice C: Either n or k is a multiple of 10

Counterexample: \(\mathrm{n} = 16, \mathrm{k} = 25\). Check: \(16 \times 25 = 400\) ✓. But \(16 = 2^4\) is not a multiple of 10, and \(25 = 5^2\) is not a multiple of 10. So this isn't always true.

Choice D: If n is even, then k is odd

Counterexample: \(\mathrm{n} = 20, \mathrm{k} = 20\). Check: \(20 \times 20 = 400\) ✓. Here n is even (20) but k is also even (20), not odd. So this isn't always true.

Choice E: If n is odd, then k is even

Let's think about this carefully. If n is odd, it means n has no factors of 2. Since \(400 = 2^4 \times 5^2\), and we need \(\mathrm{n} \times \mathrm{k} = 400\), all four factors of 2 must go to k. This means k must have at least one factor of 2, making k even.

Example: \(\mathrm{n} = 25\) (odd), then \(\mathrm{k} = 400 \div 25 = 16\) (even). ✓

Example: \(\mathrm{n} = 5\) (odd), then \(\mathrm{k} = 400 \div 5 = 80\) (even). ✓

Example: \(\mathrm{n} = 1\) (odd), then \(\mathrm{k} = 400 \div 1 = 400\) (even). ✓

Process Skill: CONSIDER ALL CASES - Systematically checking each possibility

4. Focus on parity (odd/even) relationships

The key insight is understanding how the factors of 2 in 400's prime factorization must be distributed:

Since \(400 = 2^4 \times 5^2\), the product \(\mathrm{n} \times \mathrm{k}\) must account for all four factors of 2. Here are the only possibilities:

• Both n and k are even (each gets some factors of 2)
• One is odd and the other is even (the even one gets all factors of 2)
• Both cannot be odd (because then no factors of 2 would be accounted for)

This confirms that if n is odd (has zero factors of 2), then k must be even (has all four factors of 2). Statement E must always be true.

Process Skill: APPLY CONSTRAINTS - Using the prime factorization constraint to determine what's possible

5. Final Answer

The answer is E: If n is odd, then k is even.

This statement must be true because of the prime factorization of 400. Since 400 contains exactly four factors of 2, if one number (n) is odd and contains no factors of 2, then the other number (k) must contain all four factors of 2, making it even.

We verified this by:

1. Finding counterexamples that disproved options A, B, C, and D
2. Proving through prime factorization logic that option E must always hold
3. Testing specific examples to confirm our reasoning

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding "must be true" vs "could be true"

Students often confuse questions asking what "must be true" with what "could be true." They might pick an answer that works for some values of n and k but not all values. For example, seeing that \(\mathrm{n} = 20, \mathrm{k} = 20\) works, they might think "\(\mathrm{n} = \mathrm{k}\)" is always true, missing that this is just one specific case. The key insight is that "must be true" means the statement holds for EVERY possible pair where \(\mathrm{n} \times \mathrm{k} = 400\).

Faltering Point 2: Overlooking the significance of prime factorization

Many students jump straight into testing numbers without first analyzing what \(400 = 2^4 \times 5^2\) tells us about the constraints. They miss that this prime factorization creates fundamental limitations on what values n and k can take. Without this foundational understanding, they end up guessing rather than systematically reasoning about parity (odd/even relationships).

Faltering Point 3: Focusing only on positive integer pairs

Students often assume n and k must both be positive, limiting their consideration to pairs like (1,400), (4,100), (20,20), etc. They forget that negative integers are also valid, missing counterexamples like \(\mathrm{n} = -20, \mathrm{k} = -20\) that disprove statements like "\(\mathrm{n} + \mathrm{k} > 0\)." The problem states "integers," not "positive integers."

Errors while executing the approach

Faltering Point 1: Insufficient counterexample testing

Students may find one example that supports a statement and conclude it's always true, without thoroughly testing edge cases. For instance, they might see that \(\mathrm{n} = 1, \mathrm{k} = 400\) satisfies "either n or k is a multiple of 10" and pick choice C, without testing cases like \(\mathrm{n} = 16, \mathrm{k} = 25\) where neither is a multiple of 10.

Faltering Point 2: Misunderstanding conditional statements ("if-then" logic)

When evaluating choices D and E, students often struggle with conditional logic. They might think "if n is even, then k is odd" means ALL even values of n must pair with odd values of k, missing that this can be disproven by finding just one even n that pairs with an even k (like \(\mathrm{n} = 20, \mathrm{k} = 20\)).

Faltering Point 3: Arithmetic errors in factorization

Students may make basic calculation mistakes when finding factor pairs or checking their arithmetic. For example, incorrectly calculating that \(16 \times 25 \neq 400\), or making errors when dividing 400 by various factors. These computational errors can lead them to dismiss valid counterexamples or accept invalid ones.

Errors while selecting the answer

Faltering Point 1: Picking the first statement that "seems reasonable"

After doing some calculations, students might select an answer that appears to work for the cases they tested, without completing their analysis of all options. They might stop at choice C ("either n or k is a multiple of 10") because it works for several common factor pairs, not realizing they haven't found the statement that works for ALL pairs.

Faltering Point 2: Second-guessing correct logical reasoning

Even when students correctly determine that "if n is odd, then k is even" must always be true based on prime factorization, they might doubt their conclusion because it seems "too mathematical" or because they expect a simpler answer. They may switch to a seemingly more straightforward option that's actually incorrect.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose strategic factor pairs of 400

Since we need to test which statement MUST be true for ALL pairs (n,k) where \(\mathrm{n} \times \mathrm{k} = 400\), let's systematically choose specific factor pairs and test each answer choice.

Key factor pairs of 400:

• \(\mathrm{n} = 1, \mathrm{k} = 400\)
• \(\mathrm{n} = -1, \mathrm{k} = -400\)
• \(\mathrm{n} = 4, \mathrm{k} = 100\)
• \(\mathrm{n} = 5, \mathrm{k} = 80\)
• \(\mathrm{n} = 8, \mathrm{k} = 50\)
• \(\mathrm{n} = 25, \mathrm{k} = 16\)

Step 2: Test each answer choice with our smart numbers

Choice A: \(\mathrm{n} + \mathrm{k} > 0\)

Test with \(\mathrm{n} = -1, \mathrm{k} = -400\): \(\mathrm{n} + \mathrm{k} = -401 < 0\) ✗

Choice B: \(\mathrm{n} \neq \mathrm{k}\)

Test with \(\mathrm{n} = 20, \mathrm{k} = 20\): \(\mathrm{n} \times \mathrm{k} = 400\) and \(\mathrm{n} = \mathrm{k}\) ✗

Choice C: Either n or k is a multiple of 10

Test with \(\mathrm{n} = 25, \mathrm{k} = 16\): Neither 25 nor 16 is a multiple of 10 ✗

Choice D: If n is even, then k is odd

Test with \(\mathrm{n} = 4\) (even), \(\mathrm{k} = 100\) (even): The condition fails ✗

Choice E: If n is odd, then k is even

Test all cases where n is odd:

• \(\mathrm{n} = 1\) (odd) → \(\mathrm{k} = 400\) (even) ✓
• \(\mathrm{n} = 5\) (odd) → \(\mathrm{k} = 80\) (even) ✓
• \(\mathrm{n} = 25\) (odd) → \(\mathrm{k} = 16\) (even) ✓
• \(\mathrm{n} = -1\) (odd) → \(\mathrm{k} = -400\) (even) ✓
• \(\mathrm{n} = -5\) (odd) → \(\mathrm{k} = -80\) (even) ✓

Step 3: Verify the pattern

Since \(400 = 2^4 \times 5^2\), any odd factor n can only contain powers of 5, leaving all four factors of 2 to k, making k necessarily even.

Answer: E