k is a positive integer. When k^4 is divided by 32, the remainder is 0. Which of the following could...

1. Translate the problem requirements: We need to find which remainders (2, 4, or 6) are possible when \(\mathrm{k}\) is divided by 32, given that \(\mathrm{k}^4\) is divisible by 32 (remainder 0).
2. Establish the divisibility constraint: Determine what the condition "\(\mathrm{k}^4\) divisible by 32" tells us about \(\mathrm{k}\)'s prime factorization, specifically its factors of 2.
3. Test each remainder possibility: For each proposed remainder (2, 4, 6), check if there exists a valid \(\mathrm{k}\) that satisfies both the remainder condition and the \(\mathrm{k}^4\) divisibility requirement.
4. Verify through concrete examples: Use specific values of \(\mathrm{k}\) to confirm which remainders are actually achievable under the given constraints.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking us in plain English:

• We have a positive integer \(\mathrm{k}\)
• When we calculate \(\mathrm{k}^4\) (\(\mathrm{k}\) to the fourth power) and divide it by 32, we get no remainder (remainder = 0)
• We want to know: if we divide \(\mathrm{k}\) itself by 32, which of these remainders are possible: 2, 4, or 6?

Think of it this way: imagine \(\mathrm{k} = 36\). When we divide 36 by 32, we get 1 with remainder 4. So the remainder when \(\mathrm{k}\) is divided by 32 would be 4. But we need to check if this \(\mathrm{k}\) would also satisfy our main condition that \(\mathrm{k}^4\) divided by 32 gives remainder 0.

Process Skill: TRANSLATE - Converting the problem statement into clear mathematical understanding

2. Establish the divisibility constraint

Now let's figure out what the condition "\(\mathrm{k}^4\) is divisible by 32" really means for \(\mathrm{k}\).

Since \(32 = 2^5\), saying that \(\mathrm{k}^4\) is divisible by 32 means that \(\mathrm{k}^4\) must contain at least five factors of 2.

Let's think about this step by step:

• If \(\mathrm{k}\) has one factor of 2, then \(\mathrm{k}^4\) has four factors of 2
• If \(\mathrm{k}\) has two factors of 2, then \(\mathrm{k}^4\) has eight factors of 2
• If \(\mathrm{k}\) has three factors of 2, then \(\mathrm{k}^4\) has twelve factors of 2

Since we need \(\mathrm{k}^4\) to have at least five factors of 2, and \(\mathrm{k}^4 = (\mathrm{k})^4\), we need \(\mathrm{k}\) to have at least \(5/4 = 1.25\) factors of 2. Since we can't have fractional factors, \(\mathrm{k}\) must have at least 2 factors of 2.

This means \(\mathrm{k}\) must be divisible by 4 (since \(4 = 2^2\)).

Process Skill: INFER - Drawing the non-obvious conclusion about \(\mathrm{k}\)'s divisibility requirements

3. Test each remainder possibility

Now let's check each proposed remainder when \(\mathrm{k}\) is divided by 32:

Testing remainder 2:

If \(\mathrm{k}\) gives remainder 2 when divided by 32, then \(\mathrm{k} = 32\mathrm{q} + 2\) for some integer \(\mathrm{q}\).

The smallest such value is \(\mathrm{k} = 2\). But we established that \(\mathrm{k}\) must be divisible by 4. Since 2 is not divisible by 4, this doesn't work.

Let's try \(\mathrm{k} = 34\) (next possibility). \(34 = 2 \times 17\), so it has only one factor of 2, not the required two factors. This doesn't work either.

In fact, any number of the form \(32\mathrm{q} + 2\) will always have exactly one factor of 2, never the required two or more.

Testing remainder 4:

If \(\mathrm{k}\) gives remainder 4 when divided by 32, then \(\mathrm{k} = 32\mathrm{q} + 4\).

Let's try \(\mathrm{k} = 4\). Since \(4 = 2^2\), it has exactly two factors of 2, so \(\mathrm{k}^4\) has eight factors of 2. Since \(8 > 5\), this satisfies our constraint!

Let's verify: \(4^4 = 256\), and \(256 ÷ 32 = 8\) with remainder 0. ✓

Testing remainder 6:

If \(\mathrm{k}\) gives remainder 6 when divided by 32, then \(\mathrm{k} = 32\mathrm{q} + 6\).

The smallest such value is \(\mathrm{k} = 6\). Since \(6 = 2 \times 3\), it has only one factor of 2, not the required two factors.

Similarly, any number of the form \(32\mathrm{q} + 6\) will have the same issue.

4. Verify through concrete examples

Let's double-check our conclusion with specific examples:

For remainder 4:

• \(\mathrm{k} = 4\): \(4^4 = 256\), \(256 ÷ 32 = 8\) remainder 0 ✓
• \(\mathrm{k} = 36\): \(36^4 = 1,679,616\), and this is also divisible by 32 ✓

For remainder 2:

• \(\mathrm{k} = 2\): \(2^4 = 16\), \(16 ÷ 32 = 0\) remainder 16 ✗
• \(\mathrm{k} = 34\): This would give an even larger remainder ✗

For remainder 6:

• \(\mathrm{k} = 6\): \(6^4 = 1,296\), \(1,296 ÷ 32 = 40\) remainder 16 ✗

Our analysis confirms that only remainder 4 is possible.

Process Skill: APPLY CONSTRAINTS - Systematically checking each case against our established requirements

Final Answer

Only option II (remainder 4) is possible when \(\mathrm{k}\) is divided by 32, given that \(\mathrm{k}^4\) is divisible by 32.

The answer is B. II only.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the divisibility constraint: Students often fail to recognize that "\(\mathrm{k}^4\) is divisible by 32" imposes specific constraints on \(\mathrm{k}\) itself. They might think they can directly test remainders 2, 4, and 6 without first establishing what types of numbers \(\mathrm{k}\) can be. The key insight that \(\mathrm{k}^4\) being divisible by \(2^5\) means \(\mathrm{k}\) must be divisible by at least \(2^2\) is frequently missed.

2. Confusion about remainder possibilities: Students may assume that since we're looking for remainders when \(\mathrm{k}\) is divided by 32, any remainder from 0 to 31 is equally likely to work. They don't realize that the constraint \(\mathrm{k}^4 ≡ 0 \pmod{32}\) severely limits which remainders are actually possible for \(\mathrm{k}\).

3. Not connecting prime factorization to divisibility: Many students struggle to see the connection between \(32 = 2^5\) and what this means for the prime factorization of \(\mathrm{k}\). They might attempt to work with 32 as a whole number rather than breaking it down into its prime factors to understand the underlying divisibility requirements.

Errors while executing the approach

1. Arithmetic errors in power calculations: When testing specific values like \(\mathrm{k} = 4\), students often make computational mistakes calculating \(4^4 = 256\) or checking if \(256 ÷ 32 = 8\) with remainder 0. These arithmetic errors can lead to incorrect conclusions about which remainders work.

2. Incomplete testing of remainder cases: Students might test only the smallest possible value for each remainder (\(\mathrm{k} = 2\), \(\mathrm{k} = 4\), \(\mathrm{k} = 6\)) and fail to verify their reasoning with additional examples. They may also incorrectly conclude that if one number with a particular remainder doesn't work, then no number with that remainder can work, without proper justification.

3. Misapplying the factor counting rule: When determining how many factors of 2 are in \(\mathrm{k}^4\), students often incorrectly apply the exponent rule. They might think that if \(\mathrm{k}\) has \(\mathrm{n}\) factors of 2, then \(\mathrm{k}^4\) has \(\mathrm{n} + 4\) factors instead of \(4\mathrm{n}\) factors, leading to wrong conclusions about the minimum number of factors \(\mathrm{k}\) needs.

Errors while selecting the answer

1. Misreading the answer choices: Students might correctly identify that only remainder 4 works but then select answer choice D (I and II only) thinking it includes their answer, rather than carefully checking that B (II only) is the precise match for their conclusion.

2. Second-guessing correct analysis: After correctly determining that only remainder 4 works, students might doubt their analysis because they expected multiple remainders to be possible. This self-doubt can lead them to change their answer to include additional options that they didn't properly verify.