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Difference between revisions of "1979 IMO Problems/Problem 1"
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\frac{p}{q}
 
\frac{p}{q}
 
=\frac{1979}{660\cdot 1319}+\frac{1979}{661\cdot 1318}+\cdots+\frac{1979}{989\cdot 990}=1979\cdot\frac{r}{s}
 
=\frac{1979}{660\cdot 1319}+\frac{1979}{661\cdot 1318}+\cdots+\frac{1979}{989\cdot 990}=1979\cdot\frac{r}{s}
−
\end{align*}</cmath>where <math>s=660\cdot 661\cdots 1319</math> and <math>r=\frac{s}{660\cdot 1319}+\frac{s}{661\cdot 1318}+\cdots+\frac{s}{989\cdot 990}</math> are two integers. Finally consider <math>p=1979\cdot\frac{qr}{s}</math>, and observe that <math>s\nmid 1979</math> because <math>1979</math> is a prime, it follows that <math>\frac{qr}{s}\in\mathbb{Z}</math>. Hence we deduce that <math>p</math> is divisible with <math>1979</math>.
+
\end{align*}</cmath>where <math>s=660\cdot 661\cdots 1319</math> and <math>r=\frac{s}{660\cdot 1319}+\frac{s}{661\cdot 1318}+\cdots+\frac{s}{989\cdot 990}</math> are two integers. Finally consider <math>p=1979\cdot\frac{qr}{s}</math>, and observe that <math>\gcd(s,1979)=1</math> since s consists of products of positive integers less than <math>1979</math>. It follows that <math>\frac{qr}{s}\in\mathbb{Z}</math>. Hence we deduce that <math>p</math> is divisible with <math>1979</math>.
  
 
The above solution was posted and copyrighted by Solumilkyu. The original thread for this problem can be found here: [https://aops.com/community/p6171228]
 
The above solution was posted and copyrighted by Solumilkyu. The original thread for this problem can be found here: [https://aops.com/community/p6171228]
  
 
== See Also == {{IMO box|year=1979|before=First question|num-a=2}}
 
== See Also == {{IMO box|year=1979|before=First question|num-a=2}}

Latest revision as of 09:41, 11 August 2025

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 $\gcd(s,1979)=1$ since s consists of products of positive integers less than $1979$. 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
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