# 1979 IMO Problems/Problem 1

## Problem

If $p$ and $q$ are natural numbers so that$$\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319},$$prove that $p$ is divisible with $1979$.

## Solution

We first write \begin{align*} \frac{p}{q} &=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{1318}+\frac{1}{1319}\\ &=1+\frac{1}{2}+\cdots+\frac{1}{1319}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{1318}\right)\\ &=1+\frac{1}{2}+\cdots+\frac{1}{1319}-\left(1+\frac{1}{2}+\cdots+\frac{1}{659}\right)\\ &=\frac{1}{660}+\frac{1}{661}+\cdots+\frac{1}{1319} \end{align*}Now, observe that \begin{align*} \frac{1}{660}+\frac{1}{1319}=\frac{660+1319}{660\cdot 1319}=\frac{1979}{660\cdot 1319} \end{align*}and similarly $\frac{1}{661}+\frac{1}{1318}=\frac{1979}{661\cdot 1318}$ and $\frac{1}{662}+\frac{1}{1317}=\frac{1979}{662\cdot 1317}$, and so on. We see that the original equation becomes \begin{align*} \frac{p}{q} =\frac{1979}{660\cdot 1319}+\frac{1979}{661\cdot 1318}+\cdots+\frac{1979}{989\cdot 990}=1979\cdot\frac{r}{s} \end{align*}where $s=660\cdot 661\cdots 1319$ and $r=\frac{s}{660\cdot 1319}+\frac{s}{661\cdot 1318}+\cdots+\frac{s}{989\cdot 990}$ are two integers. Finally consider $p=1979\cdot\frac{qr}{s}$, and observe that $s\nmid 1979$ because $1979$ is a prime, it follows that $\frac{qr}{s}\in\mathbb{Z}$. Hence we deduce that $p$ is divisible with $1979$.

The above solution was posted and copyrighted by Solumilkyu. The original thread for this problem can be found here: [1]