Home > Golden Oldie of the Week Archive Detail

 << Prev 5/22/2011 Next >>

## A One-Question End-of-Year Exam

The following question was the ONLY question on a Russian Graduation Exam in Algebra in 1910. Again, students had to correctly answer this one question...which tried to sum up the entire year!

A certain school has over 100 but fewer than 200 students. If the number of students is divided by the second term of the geometric progression (1) which has as many terms as twice the solution of the equation 2(1 + 9/x) + 3SQRT[(x+9)/x)] = 14, (2) whose first three terms add up to 35, and (3) whose remaining terms add up to 280, then the remainder will be equal to 5. But if the number of students is divided by the solution of the equation 9[1/(8-y)] – 3[1/(3-y)] = 0, then the remainder will be equal to the solution of the equation A(m,4) = 12A(m,2) (Where A(n,k) is the number of permutations of n elements taken k at a time). How many students are there in this school?

Could you pass this exam?

Solution: Many who has tried this problem report it is not solvable. Perhaps there was an error in its translation. Sorry....

Source: A. Karp's "Exams in Algebra in Russia," Int. J. for Hist. of Math. Ed. 2007, Vol 2#1