When the digits of a two-digit number are reversed and the reversed number is added to the original number, by which number is the sum always divisible?

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A 11
B 7
C 9
D None of the above

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Answer: A. 11

Let's say a two digit number is 10a+b. When it is reversed it becomes 10b+a. (10a+b)+(10b+a) = 11a+11b => 11(a+b). Hence, the sum of the original and the reversed number is always divisible by 11.

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When the digits of a two-digit number are reversed and the reversed number is added to the original number, by which number is the sum always divisible?

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