InterviewSolution
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InterviewSolution
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Prove that 33! is divisible by `" 2^15`. what is the largest integer n such that 33! is divisible by `2^n`. |
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Answer» For prime number `p` and `n in N`, the highest power of `p` that divides `n!` is given by, `sum_(k ge 1) [n/p^k]`. Here, `n = 33` and `p = 2`.So, highest power that divides `33! = [33/2]+[33/2^2]+[33/2^3]+[33/2^4]+[33/2^5]` `=16+8+4+2+1 = 31` So, `33!` can be divided by `2^31`. So, `33!` is divisible by `2^15` and maximun value of `n` is `31` such that `33!` is divisible by `2^31`. |
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