# $k$ 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...

• I am convinced such a $k$ does not exist. The problem seems easy, but I have yet to find a clever proof. Feb 18, 2015 at 20:15
• Beside the trivial case $k=0$, there is no such $k$ below $10^6$. Feb 18, 2015 at 22:02
• @tong_nor See Wikipedia article on Peano axioms. The axioms assume $0$ is a natural number, although my country indeed teaches that $0$ is not a natural number. It always depends and there is no universal agreement on that. Feb 18, 2015 at 23:01
• I can confirm the result found by Peter. Furthermore, there is also no $k$ satisfying the conditions below $10^8$. @tong_nor I see you have tagged with "contest math". Could you tell us where you have found this question? Feb 19, 2015 at 0:00
• Prove that for any $k\in\mathbb{N}$ there exists $n$ such that all numbers $n,\ 2n,\ \dots,\ kn$ have even sum of digits. ---> It was a contest problem (very easy) for young pupils in Ukraine about 10 years ago, I found it in old Ukrainian contest notes. My question is not from the contest, but inspired by the above contest problem. I also think there is no such number. Feb 19, 2015 at 0:31

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.

• I am stunned by this wonderful proof. Marvellous! Feb 19, 2015 at 20:12
• Note that this is easily extended to number systems with a different even base $l$, by considering an appropriate factorisation with respect to the prime divisors $p_i$ ($p_1=2$) of this base: $k=p_1^{a_1}\cdots p_q^{a_q} m$. Now $$q=l^{\max(\{a_j:\:1\leq j\leq q\})}\sum_{i=1}^m \left(l^{\phi(m)}\right)^i$$ will do the trick. Feb 19, 2015 at 20:17
• I fully agree, it's a wonderful solution, and with explicit example. Undeniably, Kate deserved a reward! Feb 22, 2015 at 22:57

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.

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.

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., $99$).