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How to show that these sets are nonempty (here $\mid $ means "divides")?

Here N is an arbitrary large integer and q is some fixed integer.

$R = \lbrace k \in {\mathbb N}:(kN\mid k!) \wedge ((k - 1)N\mid k!) \wedge \cdots \wedge (N\mid k!) \wedge (k > Nq)\rbrace$

$S = \lbrace k \in {\mathbb N}:({(2k - 1)^2}N\mid k!) \wedge ({(2k - 3)^2}N\mid k!) \wedge \ldots \wedge (N\mid k!) \wedge (k > Nq)\rbrace$

$T = \lbrace k \in {\mathbb N}:({k^5}N\mid k!) \wedge ({(k - 1)^5}N\mid k!) \wedge \ldots \wedge (N\mid k!) \wedge (k > Nq)\rbrace$

They exist by the axiom schema of separation, but how do I determine which $k$ to choose so that it satisfies all the properties? Is there a general approach?

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I don't think it is true for $S$. Let $p$ be a prime between $k$ and $2k$. (Such a $p$ exists by Betrand's Postulate.) Then $p\not\mid k!$, so definitely $p^2N\not\mid k!$ – Thomas Andrews Jul 23 '12 at 19:10
Similarly, I don't think it is true that $T$ is non-empty, since if $p$ is a prime between $k/2$ and $k$ then $p^2\not\mid k!$ so definitely $p^5N\not\mid k!$ – Thomas Andrews Jul 23 '12 at 19:13
That does not mean ${p^2}\nmid k!$, does it? – glebovg Jul 23 '12 at 19:14
What does not mean it? If $p\not\mid k!$ then $p^2\not\mid k!$. @glebovg – Thomas Andrews Jul 23 '12 at 19:16
Choose $k > {p^2}$, ${p^2}$ is an integer so ${p^2}\mid k!$. – glebovg Jul 23 '12 at 19:19
up vote 2 down vote accepted

For example, for $R$, you want $k!/(k-j)$ to be a multiple of $N$ for each $j$ from $0$ to $k-1$. That will certainly be true if $k \ge 2N$.

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Could you explain your reasoning in more detail? How would I apply the same reasoning to $S$ and $T$? Also, I think you mean $k>2N$ because $q$ is an arbitrary integer. – glebovg Jul 23 '12 at 19:00

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