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\(\mathrm{k}\) is a positive integer. When \(\mathrm{k}^4\) is divided by \(\mathrm{32}\), the remainder is \(\mathrm{0}\). Which of the following could be the remainder when \(\mathrm{k}\) is divided by \(\mathrm{32}\)?
Let's break down what this problem is asking us in plain English:
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
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:
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
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.
Let's double-check our conclusion with specific examples:
For remainder 4:
For remainder 2:
For remainder 6:
Our analysis confirms that only remainder 4 is possible.
Process Skill: APPLY CONSTRAINTS - Systematically checking each case against our established requirements
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.
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.
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.
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.