Competition problem — Number theory

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Number theory Difficulty 5.0 AIME, harder Find the answer Japan

For positive integers $n$ , $m$ , $k$ , let us write $n \equiv m \pmod k$ if $n - m$ is divisible by $k$ . Let $A$ be the sum of all positive integers $a$ less than or equal to $2011$ for which $a \equiv 1 \pmod 3$ , and $B$ be the sum of all positive integers $b$ less than or equal to $2011$ for which $b \equiv 2 \pmod 3$ . Find the value of $A - B$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Note that $2011 \equiv 1 \pmod 3$ . So, we see that $A$ is the sum of the numbers
$3 \cdot 0 + 1,\ 3 \cdot 1 + 1,\ \dots,\ 3 \cdot 669 + 1,\ 3 \cdot 670 + 1,$
while $B$ is the sum of the numbers
$3 \cdot 0 + 2,\ 3 \cdot 1 + 2,\ \dots,\ 3 \cdot 668 + 2,\ 3 \cdot 669 + 2.$
For $0 \le i \le 669$ we have $(3i+1)-(3i+2)=-1$ so we obtain
$A - B = (-1) \times 670 + (3 \cdot 670 + 1) = 1341.$

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