Competition problem — Number theory

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Number theory Difficulty 5.2 AIME, harder Prove it Romania

A positive integer $N$ has the digits $1, 2, 3, 4, 5, 6$ and $7$ , so that each digit $i$ , $i \in \{1, 2, 3, 4, 5, 6, 7\}$ occurs $4i$ times in the decimal representation of $N$ . Prove that $N$ is not a perfect square.

Solution

$N$ has $1 \cdot 4 = 4$ digits equal to $1$ , $2 \cdot 4 = 8$ digits equal to $2$ , $\dots$ , $7 \cdot 4 = 28$ digits equal to $7$ , so the sum of its digits equals $S = 4(1^2 + 2^2 + \dots + 7^2) = 560$ . Since $560 = 3 \cdot 186 + 2$ , the number $N$ is not a square, and the remainder left by a perfect square upon division by $3$ cannot be equal to $2$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.