Competition problem — Number theory

Number theory Difficulty 5.3 AIME, harder Prove it Ukraine

How many three-digit numbers are there with non-zero digits which have the following property: after any permutation of its digits one obtains three-digit number which is divisible by $4$ ?

Solution

Obviously, all digits of such number are even because only even digit can be the last one. Also we cannot use $2$ or $6$ because if a number is divisible by $4$ then the last two digits of it are $12$ , $32$ , $...$ , $92$ or $16$ , $36$ , $...$ , $96$ . So such number consists of the digits $4$ and $8$ . So:

Three digits $4$ (or $8$ ) – two numbers.

Two $4$ and $8$ (or vice versa) – six numbers.

Overall – $8$ numbers.

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