If a number is a multiple of 11, that means you can write the number as '11q' where q is an integer. So if m - n = 11x, and x is an integer, that just means that m-n is a multiple of 11. So that's the question we're trying to answer: is m - n divisible by 11?
Statement 1 tells us that the two-digit number m looks something like 11, 22, 33, etc. Those numbers are all multiples of 11, so Statement 1 is just telling us that m alone is divisible by 11. We know nothing about n, however, so we can't say if m - n is divisible by 11.
Statement 2 tells us that m + n is divisible by 11. This is not sufficient, as you can see by generating almost any two examples, one using multiples of 11, and one not. So maybe m = 22 and n = 11, and then m-n is also divisible by 11. But maybe m = 23 and n = 10, and then m - n is not divisible by 11.
Using both statements, we know that m+n and m are both multiples of 11. That guarantees that n is also a multiple of 11. And if m and n are both multiples of 11, then m-n will always be a multiple of 11 as well, so the answer is C.
If it's unclear why n must be a multiple of 11 here, you can see that in one of the following ways:
• perhaps the fastest is to use the fact that, if we subtract one multiple of 11 from another, we always get a multiple of 11. So if (m+n) and (m) are both multiples of 11, then (m+n) - m will be too, but that's just equal to n.
• the longer way is conceptually more useful to understand, because it illustrates why it's true that you always get a multiple of 11 when you add or subtract two multiples of 11:
- if m+n is a multiple of 11, then m+n = 11q for some integer q
- if m is a multiple of 11, then m = 11k for some integer k
So we know:
m + n = 11q
but m = 11k, so we can substitute for m and rearrange:
11k + n = 11q
n = 11q - 11k
n = 11 (q - k)
and we can now see that n is equal to 11 times some integer, so n is a multiple of 11 also.
Of course there's nothing special about '11' here. When you add or subtract two multiples of any integer p, you always get a multiple of p.
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