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The Quest for 700: Weekly GMAT Challenge (Answer)

Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:

Given the size of the answer choices, we should look for a way to simplify the task: testing that the product is divisible by 24. We could plug numbers; this method will eliminate all the wrong answers eventually, but it could take a while (and in fact will take a while in this case).

To be divisible by 24, the right answer needs to be divisible by one 3 and three 2’s (the prime factors of 24) under all circumstances. Let’s focus on the 3 first. To guarantee divisibility by 3 no matter what integer k is, we need to have at least one of the factors always be divisible by 3.

We need a “spread” of factors with respect to their possible remainders after division by 3. For example, what would be bad is for all the factors to differ by multiples of 3—say, k – 3, k, k + 3, and k + 6. Then all of these factors would be divisible by 3 when k itself is a multiple of 3, but if k isn’t a multiple of 3, then all the factors would be out of luck. So you want remainders of 0, 1, and 2 to be represented. Take each case, imagining k to be 0 for a moment (even though k is restricted to positive integers):

(A) -4, 0, 3, 7 gives us remainders of 2, 0, 0, and 1. Covered.
(B) -4, -2, 3, 5 gives us remainders of 2, 1, 0, and 2. Covered.
(C) -2, 3, 5, 6 gives us remainders of 1, 0, 2, and 0. Covered.
(D) 1, 3, 5, 7 gives us remainders of 1, 0, 2, and 1. Covered.
(E) -3, 1, 4, 6 gives us remainders of 0, 1, 1, and 0. Not covered. So E can be eliminated, because if k takes on say the value of 1, then there is no 3 in the product.

Now consider the three 2’s. We need them in all circumstances, whether k is even or odd.

Let’s first consider even and odds alone. Again, imagine that k is 0, for argument’s sake.

(A) even*even*odd*odd = two evens, might be enough.
(B) even*even*odd*odd = two evens, might be enough.
(C) even*odd*odd*even = two evens, might be enough.
(D) odd*odd*odd*odd = odd. Not enough. So if k = 2, for instance, then we get all odds.

Now, where do we go from here? The key is to notice that every other even is a multiple of 4 – so if you have a product of consecutive evens, then you’ll get one 2 from one of the evens and two 2’s from the multiple of 4. If the evens differ by a multiple of 4, though, then they’ll either both be multiples of 4 or both not be multiples of 4 – in which case, you’ll only get two 2’s.

(A) k – 4 and k differ by a multiple of 4. So they’ll both be multiples of 4 (or they both won’t be), in which case you’ll only get one 2 out of each of them. Bad.
(B) k – 4 and k – 2 differ by 2, so that’s good: consecutive evens. Likewise, k + 3 and k + 5 differ by 2, so we’d get consecutive evens from that product if k is odd. Looking good.
(C) k – 2 and k + 6 differ by 8, so they’ll both be multiples of 4 (or they both won’t be), in which case you’ll only get one 2 out of each of them. Bad.

The only case that always works to give you a 3 and three 2’s is (B).

The correct answer is B.




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