TITLE: $k$ with an even sum of digits for all multiples of $k$?
QUESTION [8 upvotes]: Is there a number $k\in\mathbb{N}$ such that $k\cdot n$ has an even sum of digits for all $n\in\mathbb{N}$?
I would be grateful for any ideas of how to attack this problem...

REPLY [5 votes]: First, we may assume that $k$ does not end with $0$, since $n \cdot (k\times 10^a)$ has an even sum of digits if and only if $nk$ has an even sum of digits.
Let $u$ be a multiple of $k$ such that its last digit is not $0$ and its second-to-last digit is not $9$: if $k$ satisfies this condition, take $u = k$; if $k$ ends with one of the combinations $91, 92, \dots, 98$, take $u = 9k$ (it will end with $19, 28,\dots, 82$, correspondingly); if $k$ ends with $99$, take $u = 11k$ (it will end with $89$).
Let $k < 10^d$. Then there must be an integer $v$ such that $9 \cdot 10^d \leq v < 10^{d + 1}$ and $v$ is a multiple of $k$. Note that $v$ starts with $9$.
Now consider the number $w = 10^d u + v$. Sunce $u, v$ are multiples of $k$, $w$ is also a multiple of $k$. The last non-zero digit of $10^d u$ is at the position $d + 1$ from the end, which is also the position of the first digit of $v$ (and this digit is $9$). By doing the long addition, one can see that $S(w) = S(u) + S(v) - 9$, where $S(\cdot)$ is the sum of digits. Thus at least one of the numbers $S(u), S(v), S(w)$ must be odd.