The number of 24! Is divisible by

Options

A. `6^(24)`
B. `24^(6)`
C. `12^(12)`
D. `48^(5)`

Answer

B

Explanation

We have,
`E_(2)(24!)=[(24)/(2)]+[(24)/(2^(2))]+[(24)/(2^(3))]+[(24)/(2^(4))]`
`=12+6+3+1=22`
and, `E_(3)(24!)=[(24)/(3)]+[(24)/(3^(2))]=8+2=10`
`:.24!""=2^(22)xx3^(10)=(2^(7))xx3^(10)xx2`
`=(2^(3)xx3)^(7)xx3^(3)xx2=(24)^(7)xx3^(3)xx2`
Clearly, 24! is divisible by `24^(6)`.