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Check this article for details: Variations in Factorial Manipulation



Solution



To find
We need to determine
    • The number of trailing zeros of 50!

Approach and Working out
Theoretically, the number of trailing zeros of any factorial is determined by the highest power of 10 in that factorial.
    • As 10 = 2 x 5, we need to determine the number of instances of 2 and 5, and check how many pairs of 2 x 5 can be formed
    • However, in a factorial value the instances of 5 are always less than the instances of 2. Hence, calculating the instances of 5 will be sufficient.

Number of 5s in 50! = \(\frac{50}{5} + \frac{50}{5^2}\) = 10 + 2 = 12

Thus, option D is the correct answer.

Correct Answer: Option D
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GMATinsight
How many trailing zeros does 50! have? e.g.301000 has three trailing zeros

A) 5
B) 6
C) 10
D) 12
E) 13
50/5 = 10
10/5 = 2

So , Total No of trailing 0's will be 12 , Answer must be (D)
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To find the number of trailing zeros we need to check power of 5 when divided by 50!

So 50!/5 = 10+2 = 12

Option D

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