From the 2004 Turkey Junior Maths Olympiad:
One evening, more than of the students at a school go to the cinema. On the same evening, more than go to the theatre and more than go to a concert. What is the smallest possible number of students at the school?
When I first saw this problem, I thought it would be fairly easy to solve with a little bit of algebra. It ended up taking much longer than I expected, and I admire the problem as a result.
Let be the number of students at the school, and let be the number of students who go to the cinema, the theatre and the concert respectively. The conditions given amount to:
These turn into
Since all these quantities are positive integers, we have
Adding 110 lots of the first inequality, 33 lots of the second and 30 lots of the third yields
But , so in fact we have
so that . To show that 173 is possible, we replace all the inequalities with equations, to obtain
Solving these gives us . Since 58 is more than of 173, and 52 is more than of 173, and 63 is more than of 173, this is indeed a solution, and so the minimum possible number of students at the school is 173.
This is a great problem. What really makes it tricky is the fact that is almost 1, and since the problem is discrete, a lot of objects are required to provide such sensitive fractions. The discreteness is crucial, since it allows you to convert strict inequalities into the more useful ‘less than or equal to’ inequalities above.