Question. What does it mean by "Then, the value of a digit in its place is exactly the sum you get when you add up these stray 1's"?

Module 5 Week 1 Day 2 Challenge Part 3 MiniQuestion Solution
What does it mean by "Then, the value of a digit in its place is exactly the sum you get when you add up these stray 1's"?

@neatlobster Thanks for asking. Maybe I'll give an example. How about looking at the number "734"? I'm going to claim that there are \( 7 + 3 + 4\) stray \(1\)'s. That's \(14\) stray \(1\)'s. Why is this?
To start with, since we only care about the remainder when dividing \(734\) by \(9,\) we can separate \(734\) into parts and look at the remainders from the component parts. These parts are the \(700 + 30 + 4.\) We want to find the remainder when dividing each of these parts by \(4,\) and then sum the remainders to get a total remainder.
Why do we split the number \(734\) like this? It's because we're using the base\(10\) system, which means that tens, hundreds, thousands, etc. are all formed from successive groups of \(10.\) This is significant because \(9\) is one less than \(10!\) So any power of \(10,\) like \(100, 1000, 10,000,\) etc. will leave a remainder of \(1\) when divided by \(9.\)
Let's now figure out the remainder for \(734.\) The first \(700\) is \(7 \times 100.\) Every \(100\) gives a remainder of \(1\) after dividing by \(9,\) which we call a stray \(1.\) There will be \(7\) stray \(1\)'s from \(700.\)
Then, \(30 = 3 \times 10.\) Every \(10\) gives a remainder of \(1\) after dividing by \(9,\) so there are \(3\) stray \(1\)'s from the \(30.\)
Lastly, the ones digit contributes a bunch of stray \(1\)s equivalent to the digit itself, which is \(4,\) so there are \(4\) stray \(1\)'s from the units digits.
Adding up all these stray \(1\)'s, we have \(7 + 3 + 4 = 14\) stray \(1\)'s. Now we can package up \(9\) of them and discard them, since we only care about the remainder. We have left \(5,\) which is the remainder of \(734\) after dividing by \(9.\)

Thanks so much!