If we arrange the positive distinct divisors of a number A by increasing order, then we get something like: $$1<a_1<a_2<a_3<...<a_{n-2}<a_{n-1}<a_n<A$$How can we prove that $$a_1\cdot a_n=A,$$$$a_2\cdot a_{n-1}=A,$$$$a_3\cdot a_{n-2}=A, etc.$$

Clarification: If number of divisors is odd, just ignore the middle term.

  • $\begingroup$ Can an integer have an odd number of divisors? What would happen with your statement? $\endgroup$ – FormerMath Oct 23 '14 at 15:09
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    $\begingroup$ Look at $$\frac{A}{1} > \frac{A}{a_1} > \dotsc > \frac{A}{a_n} > \frac{A}{A}.$$ $\endgroup$ – Daniel Fischer Oct 23 '14 at 15:13
  • $\begingroup$ @Daniel You should write that as an answer, this is the most clever and straightforward solution IMO. $\endgroup$ – Traklon Oct 23 '14 at 15:27
  • $\begingroup$ @Luis See the Clarification. $\endgroup$ – Jason Oct 23 '14 at 17:38

Let's also write $a_0 = 1$ and $a_{n+1} = A$. If you divide $A$ by each of the divisors, you get the arrangement

$$\underbrace{\frac{A}{a_0}}_{b_0} > \underbrace{\frac{A}{a_1}}_{b_1} > \underbrace{\frac{A}{a_2}}_{b_2} > \dotsc > \underbrace{\frac{A}{a_{n-1}}}_{b_{n-1}} > \underbrace{\frac{A}{a_n}}_{b_n} > \underbrace{\frac{A}{a_{n+1}}}_{b_{n+1}}.$$

All of the $b_i$ are distinct, and all of them are divisors of $A$. Hence the $b_i$ are just the $a_i$ in a (possibly) different order. Since the $a_i$ are indexed in increasing order, and the $b_i$ are indexed in decreasing order, it follows that we have

$$b_i = a_{n+1-i}\tag{$\ast$}$$

for $0 \leqslant i \leqslant n+1$. But since by definition $b_i = \frac{A}{a_i}$, $(\ast)$ is nothing else but

$$\frac{A}{a_i} = a_{n+1-i},\quad 0 \leqslant i \leqslant n+1,$$

which in turn is just another way to write

$$a_i\cdot a_{n+1-i} = A$$

for $0 \leqslant i \leqslant n+1$.

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