Prove that every positive integer is congruent to the sum of its digits modulo 3.

Question

Prove that every positive integer is congruent to the sum of its digits modulo 3.So I start with rewrite as $n = c_r 10^r + c_{r-1} 10^{r-1}...$?

Generalized version : http://math.stackexchange.com/questions/328562/divisibility-criteria-for-7-11-13-17-19 — lab bhattacharjee, Nov 03 '15 at 04:47

score 2 · Answer 1 · answered Nov 03 '15 at 00:20

2

Yes, start by writing $n = c_r 10^r + c_{r-1} 10^{r-1} + \dots + c_0$ and note that $10^k \equiv 1 \mod 3$ for every $k \geq 0$.

answered Nov 03 '15 at 00:20

Michael Biro

I don't get how is that for every $k\ge0$ could yo provide further detail please? – DISPLAY888 Nov 03 '15 at 00:37
@DISPLAY888 $10 = 3(3) + 1 \equiv 1 \mod 3$, so $10^k \equiv 1^k \equiv 1 \mod 3$. – Michael Biro Nov 03 '15 at 00:39

score 1 · Accepted Answer · answered Nov 03 '15 at 02:04

First note that for any power of $10$ we have

$10^r=99...9+ 1$ (with $r$ digits of $9$) (e.g. $100=99+1, 1000=999+1$)

$=3(33...3)+1$

$=1$ $(mod 3)$

Therefore given any $n$ we have

$n = c_r 10^r + c_{r-1} 10^{r-1}+ ... + c_1 10+c_0=c_r + c_{r-1} + ... + c_1 +c_0$$(mod 3)$

2 Answers2