If both x and y are positive integers less than 100 and greater than 10, is the sum x + y a multiple of 11?(1) x - y is a multiple of 22. If x=33 and y=11, then the answer is YES but if x=34 and y=12, then the answer is NO. Not sufficient.

(2) The tens digit and the units digit of x are the same; the tens digit and the units digit of y are the same. Note that any two-digit integer can be represented as 10a+b (wher a and b are single digit integers), for example 37=3*10+7, 88=8*10+8, etc. Thus, we are given that x=10a+a=11a and y=10b+b=11b --> x+y=11a+11b=11(a+b). Therefore x+y is a multiple of 11. Sufficient.

OR: from (2) we have that both x and y must be multiples of 11 (11, 22, 33, 44, ..., 99). The sum of two multiples of 11 will give a multiple of 11. Sufficient.,

Answer: B.

GENERALLY:

If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;

OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;

OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

Hope helps.

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