Problems A, B and C were posed in a mathematical contest. 25 competitors solved at least one of the three. Amongst those who did not solve A, twice as many solved B as C. The number solving only A was one more than the number solving A and at least one other. The number solving just A equalled the number solving just B plus the number solving just C. How many solved just B?
Answer
6.
Solution
Let a solve just A, b solve just B, c solve just C, and d solve B and C but not A. Then 25 - a - b - c - d solve A and at least one of B or C. The conditions give:
b + d = 2(c + d); a = 1 + 25 - a - b - c - d; a = b + c.
Eliminating a and d, we get: 4b + c = 26. But d = b - 2c ≥ 0, so b = 6, c = 2.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
29 Sep 1998
Last corrected/updated 26 Sep 2003