Xavier Aptitude Test 2014 Paper

Question : 67 of 103

Marks: +1, -0

Read the following instruction carefully and answer the question that follows: Expression

↙{n \ 1}↖13 \ \ 1/n

can also be written as

x /13!

What would be the remainder if x is divided by 11?

Solution:

(d)

∑↙{n=1}↖13 1/n

x/{13!}

\⇒ \ {1}/{1}+\ {1}/{2}+\ {1}/{3}+\......+\ {1}/{11}+\ {1}/{12}+\ {1}/{13}= {x}/{13 !}

\∴ x=\ {13 !}/{1}+\ {13 !}/{2}+\ {13 !}/{3}+\......+\ {13 !}/{11}+\ {13 !}/{12}+ {13 !}/{13}

All the terms in x are divisible by 11 except 13!/11

\ {13 !}/{11}=1.2 .3 .4 ...10.12 .13

According to Wilson theorem,When (p – 1)! is divided by p, remainder will be – 1.
i.e., rem

\ (\ {(\ {p}-1) !}/{\ {p}}\ )=-1

Now,

\ {13 !}/{11}=1.2 .3 .4 \... 10.12 .13=10 !(12)(13)

\∴ {\ {rem}} (\ { {13 !}/{11}}/{11}\ )=\ {10 !(12)(13)}/{11}=(-1) \. 1.2=-2=9

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