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?
Integers, Integers
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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.
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?
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
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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).
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).