1.

Find the exponent of 3 in 100!A. 47B. 48C. 49D. 46

Answer» Let `E_(p)(n)` denote the exponent of p in n. Then,
`E_(p)(n!)=[(n)/(p)]+[(n)/(p^(2))]+....+[(n)/(p^(s))]`
where s is the largest positive integer such that `p^(s)lenltp^(s+1)`
Here, n=100, p=3
`:.3^(4)lt100lt3^(5)`
`:.s=4`
So, `E_(3)(100!)=[(100)/(3)]+[(100)/(3^(2))]+[(100)/(3^(3))]+[(100)/(3^(4))]`
`=33+11+3+1=48`
Hence, the exponent of 3 in 100! is 48.


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