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Prove that 33! Is divisible by 2^(19) an...

Prove that 33! Is divisible by `2^(19)` and what is the largest integer n such that 33! Is divisible by `2^(n)`?

Text Solution

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In terms of prime factors, 33! Can be written as
`2^(a)*3^(b)*7^(d)` . . .
Now, `E_(2)(33!)=[(33)/(2)]+[(33)/(2^(2))]+[(33)/(2^(3))]=[(33)/(2^(4))]+[(33)/(2^(5))]` + . . .
`=16+8+4+2+1+0` . .
=31
Hence, the exponent of 2 in 33!. now , 33! is divisible by `2^(31)` which is also divisible by `2^(19)`
`therefore`Largest value of n in 31.
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