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).
