The maximum value of n such that 2^n divides 21! is ________

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21! = 51090942171709440000
21! = 51090942171709440000 and 2^65 = 36893488147419103000 . since 2^65< 21! we can divide 21! by 2^65.
I know this please repost the question..
k...
sinju.....u r wrong...the answer is much less than 65.

Answers

2014-05-04T11:25:26+05:30
The answer is the greatest integer function of the successive powers of 2 that divide 21.
therefore n=[21/2]+[21/4]+[21/8]+[21/16]+[21/32]+...till infinity
so n=10+5+2+1=18
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