Competition problem — Number theory

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Number theory Difficulty 5.4 AIME, harder Prove it United States

Problem:
A combination lock has a 3 number combination, with each number an integer between $0$ and $39$ inclusive. Call the numbers $n_{1}$ , $n_{2}$ , and $n_{3}$ . If you know that $n_{1}$ and $n_{3}$ leave the same remainder when divided by $4$ , and $n_{2}$ and $n_{1}+2$ leave the same remainder when divided by $4$ , how many possible combinations are there?

Solution

Solution:
There are $40$ choices for the last number, and for each of these we have $10$ choices for each of the first two numbers, thus giving us a total of $4000$ possible combinations. It is interesting to note that these restrictions are actually true for Master locks.

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