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?
|
SelectClick for Answer> |
Let Na, Nb, Nc, Nab, Nac, Nbc, and Nabc denote the number of students who have solved exactly the problems identified in the index of each variable. Then:
Na + Nb + Nc + Nab + Nbc + Nac + Nabc = 25,
Nb + Nbc = 2(Nc + Nbc),
Na – 1 = Nac + Nab + Nabc, and
Na = Nb + Nc.
Combining the first and third equations tells us that 2Na + Nb + Nc + Nbc = 26, and from the second and fourth we get 4Nb + Nc = 26, so we know that Nb ≤ 6. At the same time, the second equation tells us that Nb = 2Nc + Nbc, and thence
Nc ≤ Nb/2
26 ≤ 9Nb/2, and
Nb ≥ 6.
Hence Nb must be 6.
Via The IMO Compendium.
|