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The teacher gave us British Mathematical olympiad $1992$ Round $1$ Problem $3$.

Find four distinct positive integers whose product is divisible by the sum of every pair of them. Can you find a set of five or more numbers with the same property?

I can't do the question, and my friends don't know either.

Can someone help me? Any help is appreciated.

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    $\begingroup$ Try: 4, 12, 20, 28. $\endgroup$
    – Michael
    Aug 10, 2019 at 2:54
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    $\begingroup$ Also: 6, 14, 22, 26, 30 $\endgroup$
    – Michael
    Aug 10, 2019 at 3:08
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    $\begingroup$ Find an $N$ for which $N,2N,3N,4N$ works. $\endgroup$
    – Empy2
    Aug 10, 2019 at 3:19
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    $\begingroup$ $(2a,6a,10a,14a)$ works for all $a$ $\endgroup$
    – Nilotpal Sinha
    Aug 10, 2019 at 3:22
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    $\begingroup$ $13!,2\times13!,3\times13!,\dots,8\times13!$. $\endgroup$
    – Gerry Myerson
    Aug 10, 2019 at 4:16

1 Answer 1

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Summarizing the results of the comments (where $a$ is an arbitrary positive integer):

  • 4 numbers: $(2a,6a,10a,14a)$ [from Nilotpal]

  • 5 numbers: $(6a,14a,22a,26a,30a)$ and $(2a,6a,10a,14a,18a)$.

  • 6 numbers: $(6a,14a,22a,26a,30a,34a)$ and $(2a,6a,10a,14a,18a,22a)$

Generalizing the Empy2 method to show it can work for any number of integers $n \geq 2$:

  • $n$ numbers: $(x, 2x, 3x, ...., nx)$ where $x = \prod_{(i,j) \in D_n} (i+j)$, where $D_n$ is the set of all distinct pairs $(i,j)$ for $i,j \in \{1, ..., n\}$ and $i< j$. Indeed, choosing any $ix$ and $jx$ for $(i,j) \in D_n$ we get $$ \frac{(x)(2x)(3x)\cdots(nx)}{(ix + jx)}=\frac{x^n n!}{(ix + jx)} = \frac{x^{n-1}n!}{i+j} = \mbox{integer}$$
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  • $\begingroup$ The choice of $x$ in the general case above gives in an easy proof, but it is often larger than necessary. Of course, if $(y[1],...,y[n])$ works then $(ay[1], ay[2], ..., ay[n])$ also works (for any positive integer $a$) as emphasized in the Nilotpal comment. $\endgroup$
    – Michael
    Aug 10, 2019 at 5:03

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