Question

Give the proof for "why a three digit number is divisible by 3 when sum of all the digits in that number is divisible by 3"? 5

Answer 1

For the proof, first we have to consider any number as an example. The condition is not applied on three digit numbers only.

Consider the number 756.

Since  7 + 5 + 6 = 18  is divisible by 3, so is 756.

Here we are going to find why it happens so.

Expand the number 756, so that,

$\begin{aligned}&\ \ 756 \\ \\ =&\ \ 7 \cdot 100+5\cdot 10+6 \\ \\ =&\ \ 7(99+1)+5(9+1)+6 \\ \\ =&\ \ 7\cdot 99+7+5\cdot 9+5+6 \\ \\ =&\ \ 7\cdot 99+5\cdot9+7+5+6 \\ \\ =&\ \ 7\cdot 3\cdot 33+5\cdot 3\cdot 3+7+5+6 \\ \\ =&\ \ 3(7\cdot 33+5\cdot 3)+(7+5+6)\end{aligned}$

From this we can understand that,  7 + 5 + 6  should be a multiple of 3 to get 756 as a multiple of 3, because  3(7 · 33  +  5 · 3)  is a multiple of 3.

Hence proved by an example!

Now it's going to be proved algebraically.

Consider a three-digit number  100a + 10b + c.

$\begin{aligned}&\ \ 100a+10b+c\\ \\ \leadsto&\ \ (99+1)a+(9+1)b+c\\ \\ \leadsto&\ \ 99a+a+9b+b+c \\ \\ \leadsto&\ \ 99a+9b+a+b+c \\ \\ \leadsto&\ \ 3(33a+3b)+(a+b+c)\end{aligned}$

Since  3(33a + 3b)  is a multiple of 3, so should be  a + b + c  to get  100a + 10b + c  as a multiple of 3.

Hence Proved!

The given modulo congruence always holds true that,

$10^n\equiv 1\pmod{3}$

Thus,

$\bullet\ 100a = a \cdot 10^2\ \ \ \ \ \longrightarrow\ \ \ \ \ a\cdot 10^2\equiv a\pmod{3}\ \ \ \ \ [\because\ 10^n\equiv 1\pmod{3}]\\ \\ \bullet\ 10b\equiv b\pmod{3}$

Hence,

$\large 100a+10b+c\ \equiv\ a+b+c\pmod{3}$

We can also prove it in this way.