Question:

Trailing zeroes in 100! (product of 1 to 100)?

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Count powers of 5 only (as 2s are plenty in factorials).
Updated On: Aug 6, 2025
  • 20
  • 24
  • 19
  • 22
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The Correct Option is B

Solution and Explanation

Formula for trailing zeroes in \( n! \): \[ \left\lfloor \frac{100}{5} \right\rfloor + \left\lfloor \frac{100}{25} \right\rfloor + \left\lfloor \frac{100}{125} \right\rfloor = 20 + 4 + 0 = \boxed{24} \]
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