Latest TCS Aptitude Question SOLUTION: what is the remainder of 100!/(97^2)?

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(#M40013160) TCS question Keep an EYE

what is the remainder of 100!/(97^2)?

Asked In TCS Sanjana (11 years ago)

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swati singh  (11 years ago)

when we are calculating 100! then it is damm sure that in last definately two zero exits..
and it is also multiply bye 10,20,....90 hence at last digit will like this xxxxx......000000000.
now we have 97^2=9409
hence 10000/9409 = 591

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teza  (11 years ago)

100!/(97^2)=8827

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dhyan sourav  (11 years ago)

100*99*98*97*96!/97^2= 100*99*98*96!/97
if we divide 100by 97 den we get remainder of 3,similarly in case of 99 n 98, we get 2 n 1. 96! is divisible of 96. den 96*3*2*1=576 which gives remaineder 91 when it is divided by 97. n in 1st case we have devided 97..so finally we have to multiply it. den the answer will be 91*97. but is applicable only for a prime number. fortunately 97 is a prime number.qwd.

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Prashant  (11 years ago)

91.
100*99*98*97*96!/97*97
=
100*99*98*96!/97 = 3*2*1*96/97
= 91

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