1979 IMO Problems/Problem 1
Problem
If 
and 
are natural numbers so that![\[\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319},\]](http://latex.artofproblemsolving.com/f/e/7/fe7a47dd9fd28d623ff2b2b7186626d70073ea29.png)
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 | ||
| Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
| All IMO Problems and Solutions | ||
