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If `102!=2^(alpha)*3^(beta)*5^(gamma)*7^(delta)`…, thenA. `alpha=98`B. `beta=2gamma+1`C. `alpha=2beta`D. `2gamma=3delta`

Answer» Correct Answer - 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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