Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME Prove it Philippines

Problem:
Find the largest three-digit number such that the number minus the sum of its digits is a perfect square.

Solution

Solution:
919
Let $abc$ be a three-digit number such that the difference between the number and the sum of its digits is a perfect square; that is,
$(100a + 10b + c) - (a + b + c) = 99a + 9b = 9(11a + b)$
is a perfect square. To maximize the number $100a + 10b + c$ , we set $a = 9$ , $b = 1$ , and $c = 9$ .

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