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 notequals 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!

Man this is too hard to understand