The number of zeros at the end of 100!, is

Options

A. 24
B. 51
C. 10
D. 70

Answer

A

Explanation

We have,
`70!""=2^(a)xx3^(b)xx5^(c)xx7^(d)xx.....`
Now,
`c=E_(2)(70!)=[(70)/(2)]+[(70)/(2^(2))]+[(70)/(2^(4))]+[(70)/(2^(5))]+[(70)/(2^(6))]`
`impliesa=35+17+8+4+2+1=67`
and,
`c=E_(5)(70!)=[(70)/(2)]+[(70)/(5^(2))]=14+2=16`
`:.70!""=2^(67)xx5^(16)xx3^(b)xx7^(d)xx......`
`implies70!""=(2xx5)^(16)xx2^(51)xx3^(b)xx7^(d)xx......`
`implies70!""=10^(16)xx2^(51)xx3^(b)xx7^(d)xx......`
Thus, the number of zeros at the end of 70! is 16.