Olympiad Maths Prep

Track / Stage 6 / 82 of 400 #1082 of 2000

Problem 1082

National olympiad, first round
Number theory Difficulty 6.1 Prove it

## Problem 1

Let (a,b,,k)(\mathrm{a}, \mathrm{b}, \ldots, \mathrm{k}) denote the greatest common divisor of the integers a,b,k\mathrm{a}, \mathrm{b}, \ldots \mathrm{k} and [a,b,,k][\mathrm{a}, \mathrm{b}, \ldots, \mathrm{k}] denote their least common multiple. Show that for any positive integers a,b,c\mathrm{a}, \mathrm{b}, \mathrm{c} we have (a,b,c)2[a,b][b,c][c,a]=[a,b(\mathrm{a}, \mathrm{b}, \mathrm{c})^{2}[\mathrm{a}, \mathrm{b}][\mathrm{b}, \mathrm{c}][\mathrm{c}, \mathrm{a}]=[\mathrm{a}, \mathrm{b}, c]2(a,b)(b,c)(c,a)c]^{2}(a, b)(b, c)(c, a).

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

## Solution

If we express a,b,c\mathrm{a}, \mathrm{b}, \mathrm{c} as a product of primes then the gcd\mathrm{gcd} has each prime to the smallest power and the 1 cm1 \mathrm{~cm} has each prime to the largest power. So the equation given is equivalent to showing that 2min(r,s,t)+max(r,s)2 \min (\mathrm{r}, \mathrm{s}, \mathrm{t})+\max (\mathrm{r}, \mathrm{s}) +max(s,t)+max(t,r)=2max(r,s,t)+min(r,s)+min(s,t)+min(t,r)+\max (\mathrm{s}, \mathrm{t})+\max (\mathrm{t}, \mathrm{r})=2 \max (\mathrm{r}, \mathrm{s}, \mathrm{t})+\min (\mathrm{r}, \mathrm{s})+\min (\mathrm{s}, \mathrm{t})+\min (\mathrm{t}, \mathrm{r}) for non-negative integers r,s,t\mathrm{r}, \mathrm{s}, \mathrm{t}. Assume rst\mathrm{r} \leq \mathrm{s} \leq \mathrm{t}. Then each side is 2r+s+2t2 \mathrm{r}+\mathrm{s}+2 \mathrm{t}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.