The B List - Futility Closet

A problem from the Eighth International Mathematical Olympiad, held in Sofia, Bulgaria, in July 1966 (contributed by the Soviet Union):

In a mathematical contest, three problems, A, B, 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?

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Let N_a, N_b, N_c, N_ab, N_ac, N_bc, and N_abc denote the number of students who have solved exactly the problems identified in the index of each variable. Then: N_a + N_b + N_c + N_ab + N_bc + N_ac + N_abc = 25, N_b + N_bc = 2(N_c + N_bc), N_a – 1 = N_ac + N_ab + N_abc, and N_a = N_b + N_c. Combining the first and third equations tells us that 2N_a + N_b + N_c + N_bc = 26, and from the second and fourth we get 4N_b + N_c = 26, so we know that N_b ≤ 6. At the same time, the second equation tells us that N_b = 2N_c + N_bc, and thence N_c ≤ N_b/2 26 ≤ 9N_b/2, and N_b ≥ 6. Hence N_b must be 6. Via The IMO Compendium.

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