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In the product of 1 to 100 integers, how many zeroes are there in the ending

Answers (1)

To find the number of trailing zeros in a product of integers we use the below formula

\frac n5+\frac n{5^2} +\frac n{5^3}+…  

The numerator should be greater than denominator in all cases in the above formula.

n is the integer for which the factorial needs to be identified.

n = 100

\frac{100}{5} +\frac{100}{5^2}  = No. of zeros

53= 125 which is greater than 0. Hence, we will add till .52

       20 + 4 = 24  

Hence 24 zeros are available in the product of integers from 1 to 100.

Posted by

Satyajeet Kumar

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