1011.3: Tough Grader Solution

In our last puzzle, you were asked to determine four quiz grades, given four clues. The first two clues are that the product of three grades is 2450, and the sum of those three is twice the fourth. The third clue, which many people missed, is that the first two clues do not determine a solution. The fourth clue is that the product of the two lowest grades is less than the highest grade. Considering the first clue and the prime factorization of 2450 (2 ∙ 52 ∙ 72), there are only twelve possible combinations of the first three grades:

1 25 98 Sum = 124 1 35 70 Sum = 106 1 49 50 Sum = 100 2 25 49 Sum = 76 2 35 35 Sum = 72 5 5 98 Sum = 108 5 7 70 Sum = 82 5 10 49 Sum = 64 5 14 35 Sum = 54 7 7 50 Sum = 64 7 10 35 Sum = 52 7 14 25 Sum = 46

Only two of those triples, {7, 7, 50} and {5, 10, 49} have the same sum, 64, half of which is 32. Since A knew B’s last grade, if B’s first three grades were any other triple, the first two clues would have determined a solution. Thus, the last grade must have been 32. Only the former triple meets the last requirement, that the product of the two lowest grades is less than the highest grade.

So B’s four grades are 7, 7, 32, and 50.

This puzzle was deceptively difficult. I received solutions from twenty-two students, but none of them got the unique correct solution! I suspect that some students are pretty busy with other things these days. So, no prize was awarded. Honorable mention goes to Alex Host '13, Ian Thresher '11, and Brandon Wilson '14, for good (if not fully satisfactory!) reasoning.

Good luck finishing your semesters, and have a great break! I'll have new puzzles for you in the spring.