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 5 of 5: Splice head and tail to finish all three families

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In plain words

Splice head and tail to finish all three families

10^k f\!\left(-10^{-k}g(k)\bmod m\right)+g(k),\quad 10^k f\!\left(-10^{-k}h(k)\bmod m\right)+h(k)

Detailed analysis

Use the even alternating head $f(b)$ from Step 3. For the first family concatenate it with $g(k)$ ; for the other two concatenate it with $h(k)$ . The chosen residue makes each concatenation divisible by $m$ , while the tail supplies the required factor of $2$ or $5$ . Because the head ends in an even digit and each tail has even length and odd leading digit, the parity alternation is preserved across the join. Since $n$ divides one of these constructed numbers, every $20 mid n$ has an alternating multiple.