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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