Given an expression 13∑n=11/n which can also be written as a/13!
What would be the remainder if a is divided by 11?
(A)
3
(B)
5
(C)
7
(D)
9
Answer: (D)
Explanation:
We have, a/13! = 1/1 + 1/2 + 1/3 … + 1/13 a = 13!/1 + 13!/2 + 13!/3 … + 13!/13 Clearly, all the terms in the summation are divisible by 11 except 13!/11. We can rewrite it as 13*12*11*…*2*1 / 11 = 13*12*10! We have to find the remainder for 13*12*10! / 11. But, we know, remainder for ((n-1)!/n) = -1. Then, Remainder for 13*12*10! / 11 = 2*1*(-1) = -2. Converting the negative remainder to positive, -2+11 = 9.
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