# Eli5:Rule of nines?

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This is something I have been very interested in for a long time and have never found an answer to. If you multiply 9 at any number and then add the digits it will always come back to nine. I.E. 9 X 2 = 18
1+ 8 = 9. Nine is the only number I have found is the only number that does this. Any ideas why?

In: Mathematics
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Because it is one less than 10.

To add 9 and 9, you borrow one. Now 9 and 10 are being added. You gained 1 in the tens place and lost 1 in the ones place.

Repeat every time. Borrow 1, gain 1.

To be more baffled, look at the rule of 3.

Any number where the digits all add up to a number divisible by 3, is itself divisible by 3.

So, 163947261 is divisible by 3, I typed that randomly, then added a 1 at the end to make sure it met the rule of 3. I haven’t checked, but I know it will work out.

Just to reword the top comment into something more bite sized, the values that change when you add nine change by one, a value goes up by one, a value goes down by one. Start at 09. Add 9 to 09. The zero goes up one, the 9 goes down one, so you get 18. Now add 9 to that. The 1 goes up one, the 8 goes down one. Because you’re adding one and also subtracting one, it stays in an equalibrium of always equaling 9

It’s actually something that transfers across bases as well. Base 8? Multiplying by 7 will do the same thing (all digits add up to a multiple of 7).

Why does this work? Because as stated, 9 is one less than 10.

Specifically, check this out:

x * 9 = x * (10 – 1)
= (10 * x) – (1 * x)
= (10 * (x – 1 + 1)) – (1 * x)
= (10 * (x – 1) + 10) – x
= (10 * (x – 1)) + (10 – x)

This may seem a little silly to see it represented this way, but that this show is that for single digits, you’re putting one number lower in the 10’s place, and putting the missing part to reach 10 in the 1’s place. So for 8, 7 would go in the 10’s place, and 2 goes in the 1’s place. Regardless if what single digit value you pick, you’ll end up totaling to 9.

But what about when you have more digits? Well, let’s pick it apart. What does it mean to have more than one digit?

* Each digit actually represents that digit multiplied by however many 10’s over it is

So, if we’re looking at a number like 357, that’s actually 300 + 50 + 7.

Since 9 * 357 is the same as 9 * 300 + 9 * 50 + 9 * 7, we can already tell that each of those products individually have to have their digits sum up to 9.

… but what about when they’re added together? What happens?

Well, let’s take a look at what happens with addition, and how things look in relation to our decimal system… let’s take this addition:

57
+ 64
—-
121

But how did that happen? The secret is in how we carry numbers over. 7 + 4 is 11, but we carry over the 10 over to the next column. Remember that we’re focused on whether the sum of the digits corresponds to whether it’s divisible by 9, right? Well, that 10 that’s being carried over is turning into a 1 for the next column. **Both have the same remainder when dividing by 9**. That’s the critical point.

Because the remainder for 8, or 80, or even 800 will always be 8 when you try to divide by 9, it doesn’t matter where in the columns that number is actually added if all you’re checking for is divisibility.

Just for a sanity check though… why is this? Well, let’s break that apart:

x * 10 = x * (9 + 1)
= (x * 9) + x

We know (x * 9) will be divisible by 9, so the remainder when we divide (x * 10) by 9 will always result in x.

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*WHEW!!* So we’ve got this far. What does all this mean?

* Multiplying single digits by 9 will always result in the sum of the digits totaling to 9
* Multiplying any number by 10 will always result in the same remainder when dividing by 9
* Adding numbers together will result in the sum maintaining the same divisibility by 9

With all of that, we can now definitively say that any number multiplied by 9 will have its digits sum up to a multiple of 9!

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But wait… what about the other way around? How do we know whether if the sum of the digits is 9, if the number itself is divisible by 9?

Well, it works in reverse. If you break down something like 357 into each digit like we did before (300 + 50 + 7), we know that for dealing with their remainders when dividing by 9, it doesn’t matter how many times we’ve multiplied by 10.

So, 357/9 will have the exact same remainder as (3+5+7)/9. That means if the remainder is 0 (i.e. you get a multiple of 9 from the sum), that must mean that the original number is also a multiple of 9!

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As a bonus, because the sum of the digits gives the same remainder as the number itself, if you end up adding all the digits and you still have a big number, you can add *those* digits together and keep going until you get your result! So for example… 529847382848247 gives you 81, which would then give you 9, so it’s divisible by 9!