The sum of two positive integers, m and n, is a multiple of 3. Is n divisible by 3? (1) When (m + 2n) is divided by 3 the remainder is 2. (2) When (2m + n) is divided by 3 the remainder is 1.

A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not. B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not. C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient. D. Either statement BY ITSELF is sufficient to answer the question. E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.

(D) (m + n) is a multiple of 3. In other words m + n = 3k, where k is a positive integer.

The original statement also yields that if n is divisible by 3, then m must be divisible as well and vice versa, otherwise the sum would NOT be divisible by 3. So if at least one of the two integers is NOT divisible by 3 then another one is NOT divisible by 3 as well.

Let’s use statement (1) by itself. m + 2n = (m + n) + n = 3k + n. Therefore the remainder when m + 2n is divided by 3 is the same as the remainder when n is divided by 3. Statement (1) tells us that this remainder is 2. Therefore n is definitely NOT divisible by 3. Statement (1) by itself is sufficient.

Let’s use statement (2) by itself. 2m + n = (m + n) + m = 3k + m. Therefore the remainder when 2m + n is divided by 3 is the same as the remainder when m is divided by 3. Statement (2) tells us that this remainder is 1. Therefore m is definitely NOT divisible by 3. As we have already analyzed above, this means that n is NOT divisible by 3 as well. Statement (2) by itself is sufficient.

Each statement by itself is sufficient, so the correct answer is D. ---------- How do you jump from: m + 2n = (m + n) + n = 3k + n to saying that there is a remainder of 2?

How do you jump from: m + 2n = (m + n) + n = 3k + n to saying that there is a remainder of 2?

If we consider divisibility of x by t, then adding or subtracting a multiple of t does not affect the remainder.

What we have: (m + 2n) = 3k + n

So when we add 3k to n, the remainder when dividing by 3 stays the same. We know that the remainder for (m + 2n) is 2, so it must be 2 for n as well.

Here is another way of explanation: "(1) When (m + 2n) is divided by 3 the remainder is 2" results in (m + 2n) = 3b + 2, where b is an integer.

Combine it with the result we have, m + 2n = 3k + n, to get 3b + 2 = 3k + n|rearrange n = 3(b – k) + 2 (b – k) is an integer, so n has 2 as a remainder, when divided by 3.

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