Competition problem — Number theory

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Number theory Difficulty 4.7 AIME Prove it United States

Problem:

If the sum of digits in a decimal representation of a natural number $n$ is equal to 2006, prove that $n$ can't be a perfect square of an integer.

Solution

Solution:

The remainder of $n$ upon division by $3$ is equal to the sum of its digits, i.e. $2006$ . Hence number $n$ has a remainder $2$ upon division by $3$ and no square can have that remainder.

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