Problem 6

We call a positive integer alternating if every two consecutive digits in its decimal representation are of different parity. Find all positive integers

n

which have an alternating multiple.

Step 1 of 5: Rule out

20 \mid n

Previous step Next step

20\mid n \implies 20\mid kn \implies \text{last digit}=0,\ \text{tens digit even} \implies \text{not alternating}

Detailed analysis

If $20\mid n$ , then every multiple of $n$ is a multiple of $20$ . A multiple of $10$ ends in $0$ (which is even), and being divisible by $20$ means its last two digits form a two-digit multiple of $20$ , namely $00,20,40,60,80$ , in all of which the tens digit is also even. Thus the last two digits have the same parity, so no multiple of $n$ can be alternating.