CAT 2022 Slot 2 · Quant · Question 13

CAT 2022 Slot 2 previous year question on Factorial and Trailing Zeroes, with the answer and a step-by-step solution.

Factorial and Trailing ZeroesNumber SystemMultiple choiceMedium
For some natural number n, assume that \left( 15,000 \right)! is divisible by \left( n! \right)!. The largest possible value of n is
  1. A
    5
  2. B
    7
  3. C
    4
  4. D
    6
Pick an option, then check.
Answer & solution

Answer: Option B

For one factorial to divide another, the smaller one's argument must not exceed the larger one's.

That is, a! divides b! exactly when a\le b, since b! contains a! as a factor.

Here \left( n! \right)! must divide \left( 15000 \right)!, so the condition is n!\le 15000.

Now compute the factorials in turn.

5!=120, comfortably under 15000.

6!=720, still well under.

7!=5040, which is under 15000.

8!=40320, which exceeds 15000.

So the largest workable value is n=7 — the second option.

Shortcut. The nested factorial looks intimidating but reduces to the single inequality n!\le 15000. Since 7!=5040 and 8! is eight times that, the boundary is obvious without computing 8! exactly.

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