Competition problem — Number theory

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Number theory Difficulty 4.5 AIME Find the answer

Find $a_{2012}$ if $a_{n} \equiv a_{n-1}+n(\bmod 2012)$ and $a_{1}=1$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Taking both sides modulo 2012, we see that $a_{n} \equiv a_{n-1}+n(\bmod 2012)$ . Therefore, $a_{2012} \equiv a_{2011}+2012 \equiv a_{2010}+2011+2012 \equiv \ldots \equiv 1+2+\ldots+2012 \equiv \frac{(2012)(2013)}{2} \equiv(1006)(2013) \equiv (1006)(1) \equiv 1006(\bmod 2012)$ .

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.