Exam Week - Futility Closet

A problem submitted by the United States and shortlisted for the 16th International Mathematical Olympiad, Erfurt-Berlin, July 1974:

Alice, Betty, and Carol took the same series of examinations. There was one grade of A, one grade of B, and one grade of C for each examination, where A, B, C are different positive integers. The final test scores were

Alice: 20
Betty: 10
Carol: 9

If Betty placed first in the arithmetic examination, who placed second in the spelling examination?

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Call the number of exams n. Then the total number of points distributed is n(A + B + C) = 20 + 10 + 9 = 39. Since A, B, and C are distinct, their sum must be at least 1 + 2 + 3 = 6, and the only combination of factors that can give us 39 is n = 3 and A + B + C = 13. Assume without loss of generality that A > B > C. We’re told that Betty placed first in arithmetic, so she received A points on that exam. But we also know that she received 10 points overall, which is fewer than A + B + C = 13, so she must have received C points in both remaining exams (one of which is spelling). Carol received 9 points overall, which is also less than A + B + C = 13. But she must have received at least B points on two of the exams (since Betty accounts for two Cs), one of which is spelling. If she’d received A points in the spelling exam, though, she’d have a total of at least A + B + C = 13. That’s a contradiction. So Carol must have scored B points in spelling, placing second. (It can be worked out that Alice, Betty, and Carol scored B + A + A, A + C + C, and C + B + B, respectively, and that A = 8, B = 4, and C = 1.) (From Dušan Djukic et al., The IMO Compendium: A Collection of Problems Suggested for the International Mathematical Olympiads: 1959-2004, 2006.)

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