Find the number of zeros at the end of the result `3xx6xx\ 9xx\ 12xx\ 15xxdotxx99xx102.` 4 (b) 6 (c) 7 (d) 10

Explanation

In terms of prime factors, 100! Can be written as `2^(a)*3^(b)*5^(c)*7^(d)` . . .
Now, `E_(1)(100!)=[(100)/(2)]+[100/2^(2)]+[(100)/(2^(3))]+[(100)/(2^(4))]+[(100)/(2^(5))]+[(100)/(2^(6))]`
`=50+25+12+6+3+1=97`
and `E_(5)(100!)=[(100)/(5)]+[(100)/(5^(2))]`
`=20+4=24`
`therefore100!=2^(97)*3^(b)*5^(24)*7^(d)=2^(73)*3^(b)*(2xx5)^(24)*7^(d)` . . .
`=2^(73)*3^(b)*(10)^(24)*7^(d)`
Hence, number of zeros at the end of 100! is 24.
or exponent of 10 in 100!=min(97,24)=24.