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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

`alpha=97`

B

`beta=(1//2)(alpha-1)`

C

`gamma=(1//2)beta`

D

`delta=(1//3)beta`

Text Solution

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The correct Answer is:
a,b,c,d
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