Why is can number be divided by 3(ofc I mean we get an integer result) if its sum of digits can be divided by 3

Question

Why is can number be divided by 3(ofc I mean we get an integer result) if its sum of digits can be divided by 3? Please post an easy proof since Im in grade 9:) I know there are already proofs but can someone show me a proof for the level of a grade 9 student?

Answer 1

A number such as $2437$ is understood as the sum $2\cdot1000+4\cdot100+3\cdot10+7$. But this can be written

$$2\cdot(9\cdot111+1)+4\cdot(9\cdot11+1)+3\cdot(9+1)+7 \\=9\cdot(2\cdot111+4\cdot11+3)+2+4+3+7.$$

So the remainder of the division of $2437$ by $9$ is the same as the remainder of the division of $2+4+3+7$.

You can generalize.

(Also note that $2+4+3+7=16$, so that the remainder is $1+6$: you may iterate.)

Answer 2

If you have a multiple of $3$, and you add or subtract another multiple of $3$ then you will get a multiple of $3$.

So notice that $9$, and $99$ and $999$, etc are all multiples of $3.$ So if you take a number like $456$ you have:

$$456 = 4\times 100 + 5\times 10 + 6 $$

$$ = 4 \times (99+1) + 5\times (9+1) +6$$

$$=4+5+6+\mbox{ a multiple of } 3.$$

So you have that, if $4+5+6$ is a multiple of 3, then so is $456.$

Answer 3

Any number can be written in powers of $10$, i.e. $2472=2 \cdot 1000 + 4 \cdot 100 + 7 \cdot 10 + 2$. As you noticed, if we divide power of ten by three, the remainder is one. If we divide each member of the sum by $3$ and add remainders, we will get $2+4+7+2=15$ (notice that this is the sum of digits) as the remainder. Thus we can write $2472=k \cdot 3 +15=3(k+5)$ which is divisible by $3$.