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?

CAT Question

By Akanksha Gusain · 3 days ago ·CAT
Bhagwanti answered Oct 23, 2020
Since we need 2 zeroes extra ⇒ (n + 3)! Should contain a multiple of 25.
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.

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