Competition problem — Number theory

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Number theory Difficulty 5.7 AIME, harder Prove it Philippines

Problem:

Seven points on a circle are numbered $1$ to $7$ in the clockwise direction. A grasshopper jumps in the counterclockwise direction, from one point to another on the circle. If the grasshopper is on an odd-numbered point, it moves one point, and moves two points if it is on an even-numbered point. If the grasshopper begins at the point $7$ , where will it be after $2011$ jumps?

Solution

Solution:

It will only land at the points $6, 4, 2, 7$ , landing at $7$ after every fourth jump. $2011 \equiv 3 \pmod{4}$ , so it is at $2$ by the $2011$ th jump.

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