# 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 Mini-Question 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!