1979 IMO Problems/Problem 1
Problem
If and
are natural numbers so that
prove that
is divisible with
.
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 and
, 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
and
are two integers. Finally consider
, and observe that
because
is a prime, it follows that
. Hence we deduce that
is divisible with
.
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 | ||
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