# sum of the digits of an integer

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 $$n$$.

• 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. Feb 21, 2022 at 9:17
• 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! Feb 21, 2022 at 14:47

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

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

and

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

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

$$n=n'10^{l+1}$$

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 $$10{\,\not\mid\,}\sigma(nz)$$.

Proof:

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 $$10$$.