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 $\begin{aligned} \frac{p}{q} & = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots - \frac{1}{1318} + \frac{1}{1319} \\ = 1 + \frac{1}{2} + \dots + \frac{1}{1319} - 2 \cdot (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{1318}) \\ = 1 + \frac{1}{2} + \dots + \frac{1}{1319} - (1 + \frac{1}{2} + \dots + \frac{1}{659}) \\ = \frac{1}{660} + \frac{1}{661} + \dots + \frac{1}{1319} \end{aligned}$ Now, observe that $\begin{array}{r} \frac{1}{660} + \frac{1}{1319} = \frac{660 + 1319}{660 \cdot 1319} = \frac{1979}{660 \cdot 1319} \end{array}$ 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 $\begin{array}{r} \frac{p}{q} = \frac{1979}{660 \cdot 1319} + \frac{1979}{661 \cdot 1318} + \dots + \frac{1979}{989 \cdot 990} = 1979 \cdot \frac{r}{s} \end{array}$ 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]

1979 IMO (Problems) • Resources
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1979 IMO Problems/Problem 1

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