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, 2010

Number theory Difficulty 5.5 AIME, harder Prove it Ukraine

Do there exist pairwise distinct natural numbers $a_1, a_2, \dots, a_k$ , greater than $1$ , for which
$a_1 + a_2 + \dots + a_k = 2010 \cdot \left( \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_k} \right)$

a) if $k=2$ ;
b) if $k=12$ .

Solution

Answer: a) any pair of divisors of the number $2010$ with product $2010$ , for example $a_1 = 2$ , $a_2 = 1005$ ;
b) $2$ , $3$ , $5$ , $6$ , $10$ , $15$ , $134$ , $201$ , $339$ , $402$ , $670$ , $1005$ .

The main idea is: if $n$ is not a square of an integer and $d$ is its divisor, $1 < d < n$ , then $d' = \frac{n}{d}$ is also a divisor of $n$ , for which $1 < d' < n$ and $d \neq d'$ .

a) In such a way, for $a_1$ we can take any divisor of $2010$ , that differs from $1$ and $2010$ , and then put $a_2 = \frac{2010}{a_1}$ .

b) Analogously, we can write out pairwise distinct pairs of divisors of $2010$ , that satisfy the condition above. The answer was given before.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.