Competition problem — Number theory

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Number theory Difficulty 5.0 AIME, harder Prove it Ukraine

Find minimal number $n$ such that $n^3 + n^2 + 330n + 330$ is divisible by $2011$ ?

Solution

$n^3 + n^2 + 330n + 330 = (n+1)(n^2 + 330),$
So this expression is divisible by $2011$ if at least one bracket is divisible by $2011$ . If first bracket is divisible by $2011$ then minimal $n=2010$ , for $n^2 + 330$ , since $n^2$ is increasing, we find that for $n=41$ , $n^2 + 330 = 2011$ .

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