Suppose $N$ is a large integer and let $k$ be an integer s.t. $k > N$. Under what conditions can we conclude that $1^N, 2^N, \ldots, k^N$ are all divisors of $k!$?
Determining when $1^N, 2^N, \ldots, k^N$ are divisors of $k!$
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real-analysis
elementary-number-theory
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1Your last statement there is correct, and if one of $k-1, k-2, \dots ,k/2$, is a prime, you are going to struggle to get them all as divisors, for large $N$. In fact, by [Bertrand's postulate](http://en.wikipedia.org/wiki/Bertrand's_postulate), it is probably not difficult to prove that it is never true, at least for $N\geq 2$. – 2012-08-30
1 Answers
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Trivially, the result is true for $N=1$, so consider the case where $N\geq 2$.
Suppose $k$ is even, then by Bertrand's Postulate, there is a prime $p$ satisfying $k/2 < p < k$, so that $2p > k$, and it follows that $p$ is a factor of $k!$, but $p^2$ is not, since there is only one multiple of $p$ amongst the numbers $1, 2, \dots, k$.
A similar argument works if $k$ is odd, say $k=2m+1$, since there is then at least one prime satisfying $m+1 < p < 2m+2$.
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0It just relates to some infinite series. – 2012-08-31