# Difference between revisions of "1966 IMO Problems/Problem 1"

In a mathematical contest, three problems, $A$, $B$, and $C$ were posed. Among the participants there were $25$ students who solved at least one problem each. Of all the contestants who did not solve problem $A$, the number who solved $B$ was twice the number who solved $C$. The number of students who solved only problem $A$ was one more than the number of students who solved $A$ and at least one other problem. Of all students who solved just one problem, half did not solve problem $A$. How many students solved only problem $B$?

## Solution

Let us draw a Venn Diagram.

Let $a$ be the number of students solving both B and C. Then for some positive integer $x$, $2x - a$ students solved B only, and $x - a$ students solved C only. Let $2y - 1$ be the number of students solving A; then $y$ is the number of students solving A only. We have by given $\[2y - 1 + 3x - a = 25\]$ and $\[y = 3x - 2a.\]$ Substituting for y into the first equation gives $\[9x - 5a = 26.\]$ Thus, because $x$ and $a$ are positive integers with $x-a \ge 0$, we have $x = 4$ and $a = 2$. (Note that $x = 9$ and $a = 11$ does not work.) Hence, the number of students solving B only is $2x - a = 8 - 2 = \boxed{6}.$