Find the number of trailing zeros in 124!

Question

Question

This question was previously asked in

RPF SI (2018) Official Paper (Held On: 24 Dec, 2018 Shift 1)

Answer 1

Given:

Find the number of trailing zeros in 124!

Formula Used:

Number of trailing zeros in n! = \( \lfloor \frac{n}{5} \rfloor + \lfloor \frac{n}{5^2} \rfloor + \lfloor \frac{n}{5^3} \rfloor + \cdots \)

Calculation:

n = 124

Number of trailing zeros = \( \lfloor \frac{124}{5} \rfloor + \lfloor \frac{124}{25} \rfloor + \lfloor \frac{124}{125} \rfloor \)

⇒ \( \lfloor \frac{124}{5} \rfloor = \lfloor 24.8 \rfloor = 24 \)

⇒ \( \lfloor \frac{124}{25} \rfloor = \lfloor 4.96 \rfloor = 4 \)

⇒ \( \lfloor \frac{124}{125} \rfloor = \lfloor 0.992 \rfloor = 0 \)

Number of trailing zeros = 24 + 4 + 0 = 28

The number of trailing zeros in 124! is 28.

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