62nd Ukrainian National Mathematical Olympiad

Question

Petryk solved 33 problems at the exam. For the lesser part of them, including the first problem, he got a points, while for the rest of them he got b points. It is known that the natural numbers a and b satisfy the condition: 1 b < a 10. After the exam, Petrik calculated the average score for all problems and it turned out to be an integer. For solving how many problems did Petrik get a points? **Answer:** 11.

Accepted Answer

Let Petryk solve n problems for a points and (33-n) problems for b points. Then the average result of Petrik is: S = na + (33 - n)b33 = n(a - b) + 33b33 = n(a - b)33 + b. For this number to be an integer, n(a-b) 0 33 is required. Since 1 a-b 9 and n < 33, then n=11 or n=22. Since for the score of a he solved a smaller part of the problems, there remains only one possible option n=11.