# Why is the sum of the digits of 654321÷3 congruent to 654321÷3 (mod 3)?

• Module 5 Week 1 Day 2 Your Turn

Why is the sum of the digits of 654321÷3 congruent to 654321÷3 (mod 3)?

• @divinedolphin I agree, the way this is written, it looks a little bit confusing! I have a feeling that you understand the basic principle of what Prof. Loh was talking about, but just aren't very used to the notation. For example, if I write

$$25 + 36 + 37 \text{ ( mod 3 ) },$$

it doesn't mean

$$98.$$

Rather, it means the remainder of $$25 + 36 + 37$$ after dividing by $$3.$$ There's a sneaky clue here that gives this away: the not-equals sign!

Do you see the weird equal sign? It's not a $$=,$$ it's a $$\equiv.$$ This tells you that we are only caring about the remainder of a numerical expression after dividing that expression by a certain number, which we call "the modulus," which in this case is $$3.$$

So,

$$25 + 36 + 37 \text{ ( mod 3 ) } \equiv 2.$$

Now, when Prof. Loh says "the sum of the digits of $$654321 \div 3 \equiv 654321 \div 3 \text{ ( mod 3 )"},$$ he means
that you first find the sum of the digits of $$654321,$$ and then divide that sum by $$3,$$ not that you first divide $$654321$$ by $$3$$ and then take the sum of the digits.

The next key detail here is that Prof. Loh decides to find the sum of the digits of $$654321$$ and find its remainder after dividing by $$9$$ first. Why did he do that? Well, if he tried dividing $$654321$$ first by $$3,$$ you would find that

$$654321 = \text{ something } \times 3 + \text{ remainder},$$

but you wouldn't immediately know anything about the "something," like, for example, if it is divisible by $$3.$$ So the cool math insight here is that if you know that

$$654321 = \text{ something } \times 9 + \text{remainder},$$

at least we can just focus all of our attention on that remainder, because the first term, after dividing by $$3,$$ is automatically divisible by $$3.$$ We only need to worry about the remainder and what it will give after dividing by $$3.$$ We only need to look at the remainder of the remainder, instead of the whole huge gigantic troublesome number.

• Thank you so much.

• @divinedolphin You're most welcome!