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If 102!=2^(alpha)*3^(beta)*5^(gamma)*7^(...

If `102!``=2^(alpha)*3^(beta)*5^(gamma)*7^(delta)`…, then

A

(a) `alpha=98`

B

(b) `beta=2gamma+1`

C

(c) `alpha=2beta`

D

(d) `2gamma=3delta`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`becauseE_(2)(102!)=98,E_(3)(102!)=49`,
`E_(5)(102!)=24 and E_(7)(102!)=16`
`therefore alpha=98,beta=49,gamma=24 and delta=16`
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