N! is ending with x zeroes. (N + 3)! is ending with (x + 2) zeroes. How many such N's will exist, given N is a two-digit number?

- a)5
- b)12
- c)10
- d)8

Correct answer is option 'B'. Can you explain this answer?

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Bhagwanti
answered
Oct 23, 2020

Since we need 2 zeroes extra ⇒ (n + 3)! Should contain a multiple of 25.

So,** **

So,

So, There will exist 3 values of N corresponding to every multiple of 25.

So, N = 22, 23, 24, 47, 48, 49, 72, 73, 74, 97, 98, 99

Hence the answer is 12.

So, N = 22, 23, 24, 47, 48, 49, 72, 73, 74, 97, 98, 99

Hence the answer is 12.

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