# Why is modulo 9 the Digital root of all numbers?

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heres what i mean:

132189 –> 1 + 3 + 2 + 1 + 8 + 9 = 24 –> 2 + 4 = 6
132189 % 9 = 6

https://en.wikipedia.org/wiki/Digital_root

In: 7

Because nine is one less than ten, and we work in a base ten system.

132,189 means:

> One hundred and thirty-two thousand, one hundred and eighty-nine

Or in terms of symbols:

> 1 * 10^5 + 3 * 10^4 + 2 * 10^3 + 1 * 10^2 + 8 * 10^1 + 9 * 10^0

So let’s look at our most general number:

> a + b * 10 + c * 100 + d * 1000 + e * 10,000 + …

The sum of the digits of this number will be:

> a + b + c + d + e + …

So if we take that away from the number itself we will get:

> b * 9 + c * 99 + d * 999 + e * 9999 + …. = 9(b + 11c + 111d + 1111e + …)

Meaning the difference between any number and the sum of its digits will always be a multiple of nine!

Now if we have a number that is n modulo 9, and we take away a multiple of 9 from that number, our new number will still be n modulo 9 (that’s kind of what modulo 9 means).

So if our original number is n modulo 9, then the sum of its digits will also be n modulo 9 (as that is a multiple of 9 less than our starting number), as will the sum of the digits of the sum of the digits, and so on.
Now if we have a number that is n modulo 9, and we take away a multiple of 9 from that number, our new number will still be n modulo 9 (that’s kind of what modulo 9 means).

So if our original number is n modulo 9, then the sum of its digits will also be n modulo 9 (as that is a multiple of 9 less than our starting number), as will the sum of the digits of the sum of the digits, and so on.

It is not. read the wiki: In base 10, this is equivalent to taking the remainder upon division by 9 (except when the digital root is 9, where the remainder upon division by 9 will be 0).

The nice thing about modulo arithmetic is you can largely delay applying the modulo to whatever numbers you’re working with and the result is valid.

Now, your number, 132189 could be written as the sum of each digit as you did, but you need to multiply by the tens/hundreds/thousands for each number. This is the actual number as you wrote it, without modulo math.

100000 + 3*10000 + 2*1000 + 100 + 8*10 + 9

Now it should be fairly obvious that 9, 99, 999, etc are divisible by 9, or at equivalent to 0 modulo 9. These are also all 1 shy of the numbers 10, 100, 1000 etc. So all those numbers are equivalent to 1 module 9. Since we can apply the modulo arithmetic at any point, I choose to do it to my breakdown above on each individual number. Remember, 10, 100, 1000 etc are all equal to 1.

1 + 3*1 + 2*1 + 1 + 8*1 + 9 = 1 + 3 + 2 + 1 + 8 + 9 = 24 (modulo 9)

I do that all over again to `24` get 6.

In any number system base N, N-1 is the digital root. Humans count with N=10. In hexadecimal (N=16) the digital root would be 15. Indeed, 255 (1 less than 16*16) is `15 * 17` and so on.

Let’s break it up and look at the individual digits and see what their mod 9’s are.

9 mod 9 = 0
80 mod 9 = 8
100 mod 9 = 1
2000 mod 9 = 2
30000 mod 9 = 3
100000 mod 9 = 1

You can see that any digit `n00000…000 mod 9 = n`.

If we have `n * 10^x`, we can represent it another way. It would also be `m999999…999 + 1`, where `m=n-1` (m could be 0). Rewrite it a little more and you’ll get `m000…000 + 999…999 + 1`. Take the mod9 of that, and that `999…999` disappears.

So now you’re left with the mod of `m000…000 + 1`. Basically we’ve substituted the earlier problem with *m* in place of *n*, but now we’ve added 1. For any starting value n, if we iterate this process *n* times, then we end up adding 1 *n* times, and we reduce *m* to 0. This, our modulus after all of that is just *n*.

So that means that `n*10^x mod 9 = n`. Since modulus can be done both before or after adding, then taking the mod 9 of a number is the same as adding all the digits.

Also, gotta watch out for any numbers that are equal to 0 when mod 9’d. Their digital root is actually 9, but in a mod 9 world that’s 0. The number 0 is also equal to 0 when mod 9’d, so watch out.

It’s because our numbers are in base 10, and we only have 9 digits.

When we have 9 of something and add one, we represent that as 1 ‘ten’ and zero ones. So the sum of the digits between 9 and 10 goes from 9 back to 1. This is the core of modulo 9.

The pattern continues for all numbers – the sums will climb and then drop every time you cross a threshold from a number ending in a 9 to one ending in a zero.

The pattern even works when you cross multiple thresholds at once, as you’re just doing this drop from a nine to a one multiple times.

For example, when you go from 99 to 100, the 9 in the ones place turns into a one in the tens place. That in turn makes that nine into a one in the hundreds place. Once more, you’ve lost two sets of nine and gone back to one – in terms of digit sums, you’re going from 99; 9+9 = 18; 1+8 = 9, back to 100; 1+0+0 = 1 again.

So modulo 9 is baked into our number system because there is no single digit that is bigger than 9 – we just shift positions over and start back at 1.

Any given number is made up of some amount of 9s and then some extra ones. Doing successive digit addition or modulo 9 are equivalent ways of getting rid of sets of nine until you see what’s left.

It is explained in the link you provided.
Look at the section on Congruence Formula.

Our number system is, by convention base 10 and therefore the digital root will generally be modulo (1 less than the base) which is 9. If we used base 12 numbers, then the digital root will be modulo 11.

10 is just the conventional base that we currently use. There is nothing particularly special about 10 – we’re just used to it. Other cultures (in the past) have used base 12 or base 60 number systems. In fact we still use a base 60 system when we tell the time.

First, don be confused by the name, “digital root” is only distantly related to concepts like root of square numbers or the concept of digital in computers.

It just means taking the digit sum over and over again until you have just a single digit left.

Due to the way decimal works, you can tell if a number is divisible by 9 or 3, by looking at its digit sum.

Every time you add 9 to a number ending on 0. you increase its digit sum by 9

Every time you add 9 to a number ending on a last digit larger than 0 you subtract one from the last digit and add one to the 2nd last one. If that 2nd last digit is already 9 it becomes zero and the next one gets one added. And so on.

In any case adding 9 either keeps the digit sum the same or adds or subtracts multiples of 9.

So taking the digit sum of a number that is divisible by 9 without rest results in a smaller number also divisible by 9 without rest.

Any rest from 1 to 8 you add to a number divisible by 9 will be preserved in either the modulo or the digital root method.

This is useful to tell if a large number is divisible by 9 (or 3 if the digital root is 3 or 6 or 0/9).

This is just an aspect of our decimal system. In hex you can take the digit sum of a number to figure out if it is divisible by 15 or (5 or 3). In octal it would be 7.

Because every time one of the digits in our base-10 numbers grows bigger than 9, it “rolls over” to the next digit. Therefore, the digits in the numbers end up corresponding to a count of how many times we had to “roll over” from 9 (or a multiple of 9).