Prove that 33! is divisible by `" 2^15`. what is the largest integer n such that 33! is divisible by `2^n`.

Explanation

we know that,
`33! =33×32×31×30×........×8×7×6×5×4×3×2×1`

`33! =33×(2)^5×31×30×.....(2)^3×7×6×5×(2)^2×3(2)^1×1`

let us now consider all the 2's form the

above configuration, we get

`33! =2^5×2^4×2^3×2^2×2^1(33×31×30×......×6×5×3×1)`

`33! =2^(5+4+3+2+1)(33×31×30×......×6×5×3×1)`


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