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Integers, Integers

This topic has 1 expert reply and 2 member replies

Integers, Integers

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If m > n, then is mn divisible by 3?
(1) The remainder when m + n is divided by 6 is 5.
(2) The remainder when m - n is divided by 6 is 3.


I will post the solution later, but it is lengthy and involved trial and error (or some very enlightened guessing) which seemed very time consuming. I was hoping for a more elegant generalized approach.

Any takers?

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(1) & (2) alone are not sufficient, this can be determined easily by plugging numbers.
Consider when they are combined:

(1) gives us m+n+1 is div by 6
(2) gives us m-n+3 is div by 6

If you add them we get 2m+4 is div by 6. => m+2 is div by 3, hence m is not div by 3
Subtracting gives, 2n - 2 is div by 6 => n - 1 is div by 3. hence n is not div by 3

So m & n both are not div by 3, hence m*n is also not div by 3.

So answer should be C.

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You can tackle this one pretty quickly just by plugging in simple numbers:

If m > n, then is mn divisible by 3?

(1) The remainder when m + n is divided by 6 is 5.

Choose m = 3, n = 2. That works, because 3+2 when divided by 6 does give you a remainder of 5. In that case mn is divisible by 3.

Choose m = 4, n = 1. In that case, mn is not divisible by 3. INSUFFICIENT

(2) The remainder when m - n is divided by 6 is 3.

Again, choose m = 4, n = 1. In that case, mn is not divisible by 3.

Choose m = 3, n = 0. In that case, mn is divisible by 3. INSUFFICIENT

(1) and (2) together:

We have to choose numbers that fit both statements:

m = 4, n = 1 --> mn not divisible by 3

m = 7, n = 4 --> mn not divisible by 3

At this point, you can be pretty sure that you'll continue to get that mn is not divisible by 3, and thus you'd choose C as your answer.

If you want a solid mathematical reason for the two statements being sufficient when combined:

(2) says that m - n divided by 6 yields a remainder of 3. That means m - n is an odd multiple of 3 (i.e. 3, 9, 15, etc).

Because m - n yields a multiple of 3, either one of two things is true: Both m and n are multiples of 3 ... or neither m nor n is a multiple of 3.

But m and n can't both be multiples of 3, because that would contradict Statement (1) (i.e. m+n would have to be a multiple of 3, which it isn't).

Therefore, we can conclude that neither m nor n is a mutiple of 3, and thus mn cannot be a multiple of 3.

Done and done!

Make sense?

_________________
Rich Zwelling
GMAT Instructor, Veritas Prep

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Thanks guys, it's clearer now. Whenever I see problems like this my mind freezes, the whole trying to figure numbers that yield a certain remainder vexes me. You have offered me an alternative strategy to the heavily algebraic one I have been using.

My main problem with plugging in values is figuring out the values in the first place. Anyways thanks!

Below is the proposed solution from my text, although I prefer the two above solutions. Wish I could see this on my own!


From Statement (1) alone, we have that when m + n is divided by 6 the remainder is 5. Suppose m =
10 and n = 1 (Then m + n equals 11 and when divided by 6 yields a remainder of 5). Then mn = 10, and this
number is not divisible by 3. Now, suppose m = 15 and n = 2 (Then m + n equals 17 and when divided by 6
has a remainder of 5). Then mn = 30, and this number is divisible by 3. Hence, we have a double case, and
therefore Statement (1) alone is not sufficient.

From Statement (2) alone, we have that when m - n is divided by 6 the remainder is 3. Suppose m = 10, and
n = 1 (Then m - n equals 9 and when divided by 6 has a remainder of 3). Then mn = 10 and this number is
not divisible by 3. Now, suppose m = 12 and n = 3 (Then m - n equals 9 and when divided by 6 has a
remainder of 3). Then mn = 36 and this number is divisible by 3. Hence, we have a double case, and
therefore Statement (2) alone is not sufficient.

From the statements together, we have that since the remainder when m + n is divided by 6 is 5, m + n can
be expressed as 6p + 5; and since the remainder when m - n is divided by 6 is 3, m - n can be expressed as
6q + 3. Here, p and q are two integers. Adding the two equations yields 2m = 6p + 6q + 8. Solving for m
yields m = 3p + 3q + 4 = 3(p + q + 1) + 1 = 3r + 1, where r is an integer equaling p + q + 1. Now, let’s
subtract the equations m + n = 6p + 5, and m - n = 6q + 3. This yields 2n = (6p + 5) - (6q + 3) =
6(p - q) + 2. Solving for n yields n = 3(p - q) = 3t + 1, where t is a positive integer equaling p - q.

Hence, we have
mn = (3r + 1)(3t + 1)
= 9rt + 3r + 3t + 1
= 3(3rt + r + t) + 1 by factoring out 3
From this equation, it is clear that mn is not divisible by 3, and actually leaves a remainder of 1 when
divided by 3.

Hence, the statements together answer the question. The answer is (C).

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