Latest IBM Aptitude Question SOLUTION: If 100 ! = 100 x 99 x 98 x … x 2 x 1, the maximum power of 20 which will divide 100 ! is 1] 21 2] 22 3] 23 4] 24

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If 100 ! = 100 x 99 x 98 x … x 2 x 1, the maximum power of 20 which will divide 100 ! is
1] 21 2] 22 3] 23 4] 24

Asked In IBM (12 years ago)

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lost  (11 years ago)

20 can be written as 2^2 * 5. In 100! we have several even factors. Therefore we need not check for 2. no of factors of 5 will determine the divisibility. there are 20 multiples of 5. plus 4 factors of 25(100, 75, 50, 25). Each of them giving one extra multiple of 5. Therefore total of 24 factors of 5.

therefor 100! is divisible by 20^24.

hence (d).

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