Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Find the answer

The pairwise products $a b, b c, c d$ , and $d a$ of positive integers $a, b, c$ , and $d$ are $64,88,120$ , and 165 in some order. Find $a+b+c+d$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

The sum $a b+b c+c d+d a=(a+c)(b+d)=437=19 \cdot 23$ , so $\{a+c, b+d\}=\{19,23\}$ as having either pair sum to 1 is impossible. Then the sum of all 4 is $19+23=42$ . (In fact, it is not difficult to see that the only possible solutions are $(a, b, c, d)=(8,8,11,15)$ or its cyclic permutations and reflections.)

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