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Romanian Mathematical Olympiad

Algebra Difficulty 5.4 AIME, harder Prove it Romania

Each of the numbers $1, 2, 3, \ldots, 31$ is written on a separate card. Alex and Bogdan pick $15$ cards each and notice that the sum of Alex's cards is three times larger than the sum of Bogdan's cards. Find the number of the remaining card.

Solution

Let $A$ and $B$ be the respective sums. Then $B$ is at least $1+2+3+\ldots+15 = 120$ , so $A$ is at least $3 \cdot 120 = 360$ . But $A$ is at most $17+18+\ldots+31 = 360$ , so $A$ must be $360$ and $B$ must be $120$ . Therefore Alex must pick $17, 18, \ldots, 31$ and Bogdan has to pick $1, 2, \ldots, 15$ , so the remaining card must have $16$ written on it.

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