Latest TCS Aptitude Question SOLUTION: find the smallest number such that n! is divisible 990

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find the smallest number such that n! is divisible 990

Asked In TCS SHUBHAM SONI (9 years ago)

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SUMANTA SAMANTA  (9 years ago)

The factor of 990 is : 11*5*3*3*2
That why the maximum factor of 990 is 11 .
so, n=11
That means n!=11!= (11*10*9*8*7*6*5*4*3*2*1)
And then n!/990 = (11!/990)= (11*10*9*8*7*6*5*4*3*2*1)/(11*5*3*3*2)
So Ans is 11 .

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debayan khanra  (9 years ago)

The factor of 990 is = 11*5*3*3*2
=11*5*3^2*2
maximum factor of 990 is 11 .
to divisible by 990 n must be multiple of 11..
here we required smallest no that's why n=11..
so, n=11
And then n!/990 = (11!/990)= (11*10*9*8*7*6*5*4*3*2*1)/(11*5*3*3*2)
So Ans is 11 .

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Manasa Gupta  (9 years ago)

990=11*9*10
Smallest number is 11! so that it can be divisible by 11,9&10.

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RITESH KUMAR  (9 years ago)

ans is 11
because 990=2 * 3^2 * 5* 11
so number should contain at least 11
and 11! contains other factors also....

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