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Question

Which among the following options are correct for N=50! ?

A
Exponent of 2 in N is 47
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B
Exponent of 3 in N is 22
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C
Exponent of 24 in N is 15
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D
Number of zeroes in the end of Nwill be 12
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Solution

The correct option is D Number of zeroes in the end of Nwill be 12
Exponent of 2 in 50! is =[502]+[5022]+......+[5025]
=25+12+6+3+1=47

Exponent of 3 in 50! is =[503]+[5032]+[5033]
=16+5+1=22

Exponent of 2 in 50! is 47
Exponent of 23 in 50! is 15
So, exponent of 24=23×3 in 50! is minimum of exponent of {23,3} =15

Exponent of 5 in 50! is =[505]+[5052]
=10+2=12

Now, number of zeroes in the end of N is equal to the exponent of 10 in N

Exponent of 10=2×5 in 50! is minimum of exponent of {2,5} =12

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