1979 IMO Problems/Problem 1

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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 pq=112+1314+11318+11319=1+12++113192(12+14++11318)=1+12++11319(1+12++1659)=1660+1661++11319Now, observe that 1660+11319=660+13196601319=19796601319and 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 pq=19796601319+19796611318++1979989990=1979rswhere $s=660\cdot 661\cdots 1319$ 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]

See Also

1979 IMO (Problems) • Resources
Preceded by
First question
1 2 3 4 5 6 Followed by
Problem 2
All IMO Problems and Solutions