$k$ with an even sum of digits for all multiples of $k$? 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...
 A: 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.
A: There is no such $k \in \mathbb{N}$ : given any $k$, we can calculate a value $n$ such that $kn$ has an odd sum of digits.
Factorize $k$ as $k = 2^a 5^b m$. Then consider the following integer
$$q = 10^{\max(a, b)} \sum_{i=1}^{m} {\left(10^{\phi(m)}\right)}^{i}$$
where $\phi(m)$ is Euler's totient function. We claim that $q$ is divisible by $k$. Since $\gcd(2^a 5^b, m) = 1$, if we can show that $2^a 5^b \mid q$ and $m \mid q$ we will be done.
The left term in the product is divisible by $2^a 5^b$ by construction; now we consider the right term.
Since $\gcd(m, 10) = 1$, Euler's theorem tells us that $10^{\phi(m)} \equiv 1 \pmod m$. Additionally, for any power of $i$, $1^i \equiv 1 \pmod m$. So all $m$ summands have remainder one, meaning that the right term is equivalent to $1 \cdot m \equiv 0 \pmod m$, and $m$ divides $q$.
Consequently, $q$ is divisible by $k$, which we can write as $q = kn$ for some integer $n$.
What is the sum of the digits in $q$? The right term is a sum of $m$ distinct powers of ten, whose digits will sum to $m$. Multiplying by the left term, a power of ten, doesn't change the sum, and so the sum remains $m$ – which is odd by definition.
A: What I would try instead is to come up with a method such that for any $k$ I can find an $n > 0$ such that $kn$ has an odd digit sum. It doesn't have to be the smallest possible $n$, it just needs to be an $n$ that you can neatly show $kn$ has an odd digit sum.
My hunch is that, if nothing else, $$n = \frac{10^{\lceil \log_{10} k \rceil} - 1}{3} + 4$$ might make $kn$ have an odd digit sum. For example, $k = 11$ looks very stubborn, especially if you stop before $n = 19$. But with $n = 37$, you get $kn = 407$, which has a digit sum of 11. But like I said, this is just a hunch, it could be wrong.
Although... it might not hurt to tabulate, up to some small value of $k$, say, $k = 99$, what is the smallest $n$ such that $kn$ has odd digit sum. This might or might not suggest a pattern.
A: There is no such number. All the number crunching in the world is no proof, but when someone crunches up to a largish number, it looks very unlikely.
The way I would start to attack this problem is to show that for every positive $k$ there are infinitely many $kn$ with odd digital root. It's of course possible for $kn$ to have an odd digital root but an even digit sum (e.g., $99$).
