This is a question from the free Harvard online abstract algebra lectures. I'm posting my solutions here to get some feedback on them. For a fuller explanation, see this post.

This problem is from assignment 6. The notes from this lecture can be found here.

Prove that every integer $a$ is congruent to the sum of its decimal digits modulo 9.

Let $s$ be an $(n+1)$-digit integer given by $s=a_n\cdot10^n+\dots+a_2\cdot10^2+a_1\cdot10^1+a_0\cdot10^0$ with $a_i\in\mathbb{Z}$ such that $0\leq a_i\leq9$. Then
$ \begin{align} s\,\mathrm{mod}\,9&=\bar{s}\\ &=\overline{a_n\cdot10^n+\cdots+a_2\cdot10^2+a_1\cdot10^1+a_0\cdot10^0}\\ &=\overline{a_n\cdot10^n}+\overline{\cdots}+\overline{a_2\cdot10^2}+\overline{a_1\cdot10^1}+\overline{a_0\cdot10^0}\\ &=\overline{\overline{a_n}\cdot\overline{10^n}}+\overline{\cdots}+\overline{\overline{a_2}\cdot\overline{10^2}}+\overline{\overline{a_1}\cdot\overline{10^1}}+\overline{\overline{a_0}\cdot\overline{10^0}}\\ &=\overline{\overline{a_n}\cdot1}+\overline{\cdots}+\overline{\overline{a_2}\cdot1}+\overline{\overline{a_1}\cdot1}+\overline{\overline{a_0}\cdot1}\\ &=\overline{a_n}+\overline{\cdots}+\overline{a_2}+\overline{a_1}+\overline{a_0}\\ &=a_n\,\mathrm{mod}\,9+\cdots+a_2\,\mathrm{mod}\,9+a_1\,\mathrm{mod}\,9+a_0\,\mathrm{mod}\,9 \end{align} $

Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.


  • 3
    $\begingroup$ Reasoning is sound, although you should point out somewhere that $\overline{10^n}=\overline{10}^n=\overline{1}^n=\overline{1}$. Also, it looks weird to have $\overline{\cdots}$, just $\cdots$ suffices. And if you aren't going to distinguish between the regular + and its modular version then you need to carry that really long \overline through until the last line. $\endgroup$ Commented Jan 17, 2012 at 0:45
  • $\begingroup$ I second @AlexBecker's comments. $\endgroup$
    – user21436
    Commented Jan 17, 2012 at 0:52

5 Answers 5


Just another way.

Clearly $9\vert (10-1)$. Suppose that $9\vert (10^n - 1)$, $n\gt 1$. Then $$10^{n+1}-1 =10\cdot 10^{n}-1=9\cdot 10^n + (10^n - 1)$$ thus $9\vert (10^{n+1}-1)$. Therefore $9\vert (10^n-1)$ for any $n\in\mathbb{N}$.

Now take an integer $$a=\sum_{i=0}^n a_i10^{i},$$ where $a_i\in \{0,1,\ldots,9\}$. Then $$a-\sum_{i=0}^na_i=\sum_{i=0}^n a_i10^{i}-\sum_{i=0}^na_i=\sum_{i=1}^n a_i(10^{i}-1),$$ therefore, by the established above $$a\equiv \sum_{i=0}^na_i \pmod 9$$

  • $\begingroup$ This is an elegant proof I must say! $\endgroup$
    – Robur_131
    Commented May 3, 2020 at 12:20

Your suggestion works if you are satisfied $\overline{a_n\cdot10^n} \equiv \overline{\overline{a_n}\cdot\overline{10^n}} \equiv \overline{\overline{a_n}\cdot1} \equiv \overline{a_n} \mod 9$, which is indeed true.

Another way is

$$\begin{align} s&={a_n\cdot10^n+\cdots+a_2\cdot10^2+a_1\cdot10^1+a_0\cdot10^0}\\ &={a_n\cdot 99\ldots99+\cdots+a_2\cdot 99+a_1\cdot 9 +a_0\cdot 0 }+\sum_{i=0}^n a_i\\ &=9({a_n\cdot 11\ldots11+\cdots+a_2\cdot 11+a_1\cdot 1 +a_0\cdot 0 )}+\sum_{i=0}^n a_i \end{align}$$

so $s \equiv \sum_{i=0}^n a_i \mod 9.$


HINT $\rm\ mod\ 9\!:\ 10\:\equiv\: 1\ \Rightarrow\ f(10)\:\equiv\: f(1)\ $ for all polynomials $\rm\:f(x)\:$ with integer coefficients.

The above applies to radix notation since it clearly has polynomial form in the radix.

Alternatively put $\rm\ x\: =\: 10\ $ in $\rm\ x-1\ |\ f(x)-f(1)\ $ by the Factor Theorem.


I write this answer in keeping with my comment:

You need to prove that $\overline {ab}=\overline{\bar a\bar b}$. There are many ways of looking at this, for instance see, Benjamin Lim's answer, here. I shall write a number theoretic proof here:

Suppose $\bar k$ denotes the remainder obtained when $k$ is divided by$q$, we have, $a=b_1q+\bar a; b=b_2q+\bar b$ . So, you have $$ab=(b_1b_2q+b_1\bar b+b_2\bar a)q+\bar a \bar b$$ and this proves that, $\overline{ab}=\overline{\bar a \bar b}$.

Further, on the same line, you'll need to prove for the sum and the fact that $10^n \cong 1 (mod~~9)$

There are very minor gaps which you can still fill as a beginner, as others have pointed out.

Sorry for the late edits!!


$n \equiv \sum\limits_{k=1}^{\lceil log_{10} n\rceil} 10^{k-1} d_{k} \;(mod \; 9) \quad$ (Here $n = d_{\lceil log_{10} n\rceil}\ldots d_2 d_1$, being decimal representation of $n$)

$\equiv \sum\limits_{k=1}^{\lceil log_{10} n\rceil} (9+1)^{k-1} d_{k} \; (mod \; 9)$

$\equiv \sum\limits_{k=1}^{\lceil log_{10} n\rceil} d_k \; (mod \; 9), \quad$ by binomial expansion.


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