Question:
Trailing zeroes in 100! (product of 1 to 100)?
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: Mar 24, 2026
- 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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