Find the exponent of 3 in 100!

Explanation

`100!=1xx2xx3xx..xx98xx99xx100`
`=(1xx2xx4xx5xx..xx98xx100)xx(3xx6xx9xx..xx96xx99)`
`=Kxx3^(33)(1xx2xx3xx.xx32xx33)`
`=Kxx3^(33)(1xx2xx4xx.xx31xx32)(3xx9xx12xx.xx30xx33)`
`=[K(1xx2xx4xx..xx31xx32)]xx3^(33)xx(3xx9xx12xx..xx30xx33)`
`=K_(1)xx3^(33)xx3^(11)(1xx2xx3xx..xx10xx11)`
`=K_(1)xx(1xx2xx4xx..xx10xx11)3^(33)xx3^(11)(3xx6xx9)`
`=K_(2)xx3^(33)xx3^(11)xx3^(3)xx3`
`=K_(3)xx3^(48)`
Hence, exponent of 3 is 48.
Alternate solution :
Exponent of 3 in 100! is
`[(100)/(3)]+[(100)/(3^(2))]+[(100)/(3^(2))]+[(100)/(3^(4))]=33+11+3+1=48`