Manhattan GMAT Challenge Problem of the Week – 16 October 2012

by on October 16th, 2012

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Question

If a, b, and c are positive integers, what is the remainder when a – b is divided by 6?

(1) a = c^3
(2) b = (c-2)^3

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.

Answer

We know that the three variables (a, b, and c) are all positive integers. The question asks for the remainder when ab is divided by 6. In other words, if a – b = 6×(integer) + R, what is R?

Statement 1: NOT SUFFICIENT. We are given an equation that only involves a and c, not b. So we cannot answer a question about a – b, since there’s still too much unknown about a – b (you could make a – b anything you want by changing the value of b).

Statement 2: NOT SUFFICIENT. We are given an equation that only involves b and c this time. Again, we cannot answer a question about a – b, using only information about b.

Statements 1 and 2 TOGETHER: SUFFICIENT. We have two equations and three unknowns, but that shouldn’t stop us from trying to combine the equations, because now we can get a – b all in terms of one variable, c, that we know is an integer. In fact, we should suspect that there’s enough information even before we begin.

First, expand the right side of Statement 2:

b = (c-2)^3
b = (c-2)*(c-2)^3 = (c-2)*(c-2)^2- 4c+ 4 = c^32c^24c^2 + 8c + 4c - 8
b = c^36c^2 +12c-8

Now ab = c^3 – (c^36c^2 +12c-8 = 6c^2 -12c +8

Look at each term in turn. Since c is an integer, so is c^2, and so 6c^2 is a multiple of 6.

12c is also a multiple of 6, so 6c^2 - 12c is a multiple of 6.

A multiple of 6, plus 8, leaves a remainder of 2 after you divide by 6. So we know the remainder, no matter what c or the other variables are.

You can test numbers here to look for a pattern, but it will take some time to compute some cubes and their differences.

Moreover, you can’t be sure that the remainder will always be 2 (although you might suspect so). The algebraic approach here is worth mastering.

The correct answer is C.

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2 comments

  • I got the same answer by plugging in but I don't know if it's the right way to go about doing this question. I'm thinking it could be a fluke but this is how I did it.

    a,b,c

    a = c^3, hence a could be 8 and c could be 2. Not enough info as we do not have a value for b.

    b=(c-2)^3 hence b could be zero or any other number but one can't tell because there is no mention of a.

    Combining both equations, you get a=8, b=0, c=2 and the remainder for 8-0/6 is 2 which is the answer in your explanation.

    Would it be wrong to go by this method?

  • 若3n-2, 当3n-2 devided by 3 的时候, 余数为1, since 3n-2 = 3n-3+1

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