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I'm trying to construct a direct proof to show that a five-digit number is divisible by 3 if the sum of its five digits is divisible by 3. What I was thinking of doing was expanding the five-digit number, but then I get stuck in figuring out how to bring in congruence modulo n and the definition of divides into my proof. Any help is appreciated in making a clear and concise proof. Here's what I have so far:

Suppose that you have a five-digit number $n$ that is written $abcde$. Then,

\begin{align*} n&=10^4a+10^3b+10^2c+10d+e\\ &=(9999+1)a+(999+1)b+(99+1)c+(9+1)d+e\\ &=(9999a+999b+99c+9d)+(a+b+c+d+e)\\ &=3(3333a+333b+33c+3d)+(a+b+c+d+e)\\ \end{align*}

Arata Kaizaki
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    you pretty much have everything. So $n=3(\text{integer})+(a+b+c+d+e)$, this means $3$ divides the difference $n-(a+b+c+d+e)$, same as saying that $n \equiv (a+b+c+d+e) \pmod{3}$. So if $3$ divides $n$ then $3$ divides $(a+b+c+d+e)$ and vice-versa. – Anurag A Jan 22 '20 at 06:28
  • Notice that the first part of the sum is of the form $3k$. Therefore, $n$ is divisible by $3$ iff the remaining part $a+b+c+d+e$ is divisible by $3$, and this is the sum of digits of $n$. – Saswat Padhi Jan 22 '20 at 06:29
  • See https://math.stackexchange.com/questions/328562/divisibility-criteria-for-7-11-13-17-19 – lab bhattacharjee Jan 22 '20 at 07:03

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Using congruence:

Since $3 \equiv 0 \pmod 3$, from your working, we have $n \equiv a+b+c+d+e \pmod{3}$

Using direct division.

$$n-(a+b+c+d+e) = 3(3333a+333b+33c+3d)$$

Hence $n-(a+b+c+d+e)=3\alpha$ is a multiple of $3$. If $n$ is a multiple of $3$, then we can write $n = 3\beta$, and we have $a+b+c+d+e=3(\beta-\alpha)$. Similarly if $a+b+c+d+e$ is a multiple of $3$.

In general, if $n = \sum_{i=0}^d n_i \cdot 10^{i}$, then since $10 \equiv 1 \pmod{3}$, we have $n \equiv \sum_{i=0}^d n_i$.

Siong Thye Goh
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