1.

Number of zeros in the expansion of `100!` isA. 22B. 23C. 24D. 25

Answer» In terms of prime factors 100! Can be written as `2^(a)3^(b)5^(c)7^(d).......`
Now,
`E_(2)(100!)=[(100)/(2)]+[(100)/(2^(2))]+[(100)/2^(3)]+[(100)/(2^(4))]+[(100)/(2^(5))]+[(100)/(2^(6))]`
`impliesE_(2)(100!)=50+25+12+6+3+1=97`
and, `E_(5)(100!)=[(100)/(5)]+[(100)/(5^(2))]=20+4=24`
`:.100!""=2^(97)xx3^(6)xx5^(24)xx7^(d)xx......`
`implies100!""=2^(73)xx(2xx5)^(24)xx7^(d)xx....`
`implies100!""=10^(24)xx2^(73)xx3^(b)xx7^(d)xx....`
Thus, the number of zeros at the end of 100! is 24.


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