Assume that a positive integer $n$ can be written in decimal notation as $${n_k}\cdots {n_1}{n_0} = {n_k}10^{k} + \cdots + {n_1}10 +{n_0},$$ and define $${\sigma}(n)={{\sum}^{k}_{j=0}}{n_j}.$$ If ${\sigma}(n)\equiv {0}({\rm mod} 10)$ I am trying to prove that there exists a positive integer $k$ such that ${\sigma}(kn)\neq {0}({\rm mod} 10).$ For example, consider $n = 55.$ Then ${\sigma}(55)=10\equiv {0}({\rm mod}10).$ If we choose $k=2$, then $2\cdot 55=110$, and ${\sigma}(110)=2\neq {0}({\rm mod}10).$ I realized that multiplying by powers of $10$ does not change the value of the function $\sigma$ taken $({\rm mod}10)$. But not sure how this helps me find, or prove the existence of the required positive integer $k$ given an arbitrary positive integer $n$.

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    $\begingroup$ No idea about a general solution yet, but : $9$ seems to work in most cases, the smallest counterexample is $1111111111$ (ten ones) because the digitsum of $9n$ must be of the form $90m$ , if $9$ fails. In this case, $n=19$ does the job, if we have $10101010101010101010$ , $109$ does the job and this continues if the number of zeros between the ones increases. $\endgroup$
    – Peter
    Feb 21, 2022 at 9:17
  • $\begingroup$ Thanks I was struggling with this example trying to work out the number of "carries" etc. to make it work. but your solution solved my problem! $\endgroup$
    – student
    Feb 21, 2022 at 14:47

2 Answers 2


I'll stick with $k$ being the highest power of $10$ in the decimal representation of of $n$ and use $f$ for potential factors. We assume for now that $n_0 \neq 0$, that means $n$ is not divisible by $10$.

Consider $f_1=10^{k+1}-1$ and $f_2=10^{k+2}-1$.

Then the decimal representations of $f_1n$ and $f_2n$ are

$$f_1n=[n_kn_{k-1}\ldots n_1(n_0-1)(9-n_k)(9-n_{k-1})\ldots (9-n_1)(10-n_0)]_{10}$$


$$f_2n=[n_kn_{k-1}\ldots n_1(n_0-1)9(9-n_k)(9-n_{k-1})\ldots (9-n_1)(10-n_0)]_{10}.$$

You can see that by adding $n=[n_k\ldots n_0]_{10}$ to each number and do the summation digit by digit as we all learned in school.

As you can see, those numbers are made of exactly the same digits, except that $fn_2$ has an extra $9$. So the digit sum of both can't be $\equiv 0 \pmod {10}$.

Now, if $n$ is divisible by $10$, then we can divide the highest power of $10$ out of it. So if $n_0=n_1=\ldots =n_l=0$, but $n_{l+1} \neq 0$. We have


with $n'$ not being divisible by $10$. The factor $f$ that works for $n'$ also works for $n$, as $fn$ is just $fn'$ with $l+1$ zeros at the end, which don't change the digit sum.


Claim:$\;$For all positive integers $n$, there exists a positive integer $z$ such that $10{\,\not\mid\,}\sigma(nz)$.


Let $n$ be a positive integer.

If $10{\,\mid\,}n$, then for any positive integer $z$, we have $$ \sigma(nz)=\sigma\Bigl({\small{\frac{n}{10}}}z\Bigr) $$ hence we can assume $10{\,\not\mid\,}n$, else replace $n$ by ${\large{\frac{n}{10}}}$ and start again.

Choose a positive integer $x$ such that $$ \begin{cases} nx\equiv 75\;(\text{mod}\;100)&\qquad\text{if}\;5{\,\mid\,}n\\[4pt] nx\equiv 8\;(\text{mod}\;100)&\qquad\text{otherwise}\\ \end{cases} $$ and choose a positive integer $y$ such that the leading digit of $ny$ is at least $5$.

Let $v=x10^{{\large{s-1}}}$, where $s$ is the number of digits in the base-$10$ representation of $ny$,

Note that the right-most nonzero digit of $nv$ positionally overlaps the left-most digit of $ny$, and the sum of those two digits results in carry of $1$, so $$ \sigma(nv+ny) = \sigma(nv)+\sigma(ny)-9 = \sigma(nx)+\sigma(ny)-9 $$ Thus, letting $w=v+y$, we have $$ \sigma(nw) = \sigma(nx)+\sigma(ny)-9 $$ hence at least one of $\sigma(nw),\sigma(nx),\sigma(ny)$ is not divisible by $10$.


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