Latest TCS Aptitude Question SOLUTION: Find the remainder when 97! is divided by 101

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(#M40026186) TCS Numerical Ability question Keep an EYE

Find the remainder when 97! is divided
by 101

Asked In TCS A G (10 years ago)

Unsolved Read Solution (4)

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Read Solution (4) :

manoj chauhan  (10 years ago)

Use this theorem...
Wilson's Theorem is about primes, namely, for integer n>1,
(n-1)! = -1 mod n iff n is prime.

Since 101 is prime, we can use it, along with the fact that
100*99*98*97! = 100!

Then taking mod 101 and using Wilson's Theorem,
100*99*98*97! = -1 mod 101
(-1)(-2)(-3)97! = -1 mod 101

so that, multiplying both sides by -1,
6*97! = 1 mod 101

Now, we'd like to find the inverse mod 101 of 6. To do this, use the fact that
6*17 = 102 = 1 mod 101, to multiply both sides by 17, yielding

17*6*97! = 97!mod 101 = 17 mod 101

So it's 17.

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Am  (10 years ago)

97! can be found using calculator given in open seesame.
97!=9.619 e+151
97!/101=9.52376 e+149
to find remainder:
subtract the whole number part i.e 9
=(9-9.52376)=0.52376
multiply this by 101
0.52376*101=52.89976
this can be round of to 53
Hence remainder is 53.

Is this solution Helpfull? Yes (5) | No (1)  

Arjun  (10 years ago)

we know that 97! = 97*96*95.........1
it can be write as 97+96.....+2+1 because when you find the remainder 91/101 =97,96/101=96...as it is and so on..
add all by using n/2[2a+(n-1)d] . ..you will find the total than easily you can divide by 101 ans==whatever according to approach
as

Is this solution Helpfull? Yes (3) | No (32)  

pooja gautam  (10 years ago)

total sum of series n/2[2a+(n-1)d]
then sum is 9312*97
then divide by 101
first be divide 9312 by 101 then reminder is 19 and multiply 19 and 97
then divide by 101
and ans is 25

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