# Strange sum that always end up with 9

If we have any number, example 4896, and sum all digits

sum = 4+8+9+6 = 27

and than substract this number from the original number, we always get a number that is divisible by 9:

4896-27=4869 -> 4869/9 = 541.

Moreover now everytime i sum the digits from new number (4869), i will get number divisible by 9:

(4+8+6+9=27[divisible])

71 - (7+1) = 63

485 - (4+8+5) = 468/9=52 , 4+6+8 = 18(divisible)

Similarly interesting is with 2 digit numbers where with the same process resulting number is always 9

45 -> 45-9 = 36 (3+6=9)

87 -> 87-15 = 72 (7+2=9)

Its quite beyond my comprehension. Can anyone explain this phenomena?

• 334-10=324[divisible], 324 -> 3+2+4=9[divisible] – Martin Dec 10 '18 at 22:13
• You might also like to try this. Take a 3 digit nunber and reverse it. Subtract the smaller number from the bigger one. Reverse the result and add. (Example: $532-235 = 297$; $297+792=1089$) Try with a few different numbers that don't read the same backwards as forwards. – timtfj Dec 11 '18 at 0:57

I will give a formal proof for numbers with $$4$$ digits. You can think about how to generalize it (Edit: or refer to @fleablood's answer who gave a formal proof for arbitrarily many digits).

Let $$n=abcd$$ be a number where $$0\leq a,b,c,d\leq 9$$ are it's digits. For example if $$n=4821$$ then $$a=4,b=8,c=2,d=1$$.

Another way to write $$n$$ is as $$n=a\cdot 1000 + b\cdot 100 +c \cdot 10 +d$$.

On the other hand the sum of digits is $$a+b+c+d$$.

so $$n - \text{The sum of digits} = a\cdot 1000 + b\cdot 100 + c\cdot 10 + d -a-b-c-d$$ The right hand side turns out to be $$a\cdot 999 + b\cdot 99 + c\cdot 9$$ which is divisible by $$9$$.

The second thing that interested you is that the sum of digits of the new number is always divisible by $$9$$. For this you should refer to @Ross Milikan answer.

• Voted up because it manages both to be a formal proof and avoid advanced notation. (I think $\Sigma$ may be advanced for the questioner.) – timtfj Dec 10 '18 at 22:47
• @timtfj Thanks. This is basically why I posted this answer despite the existing answers. – Yanko Dec 10 '18 at 22:54
• Thanks to all of you for your answers. As I am not very advanced in formal math notation, this answer is easiest for me to understand. – Martin Dec 12 '18 at 8:35

This is the standard divisibility test for $$9$$. It is also called "casting out 9s". The remainder on dividing by $$9$$ is the same as the remainder on dividing the sum of the digits by $$9$$. It works because $$10^n$$ always has a remainder of $$1$$ when dividing by $$9$$ because, for example, $$100000=99999+1$$

This is very well known:

Consider that the number is $$N= \sum_{k=0}^n a_i 10^k$$ where $$a_i$$ are the digits. Digits. Then the sum of the digits is $$S = \sum_{k=0}^n a_i$$.

Then $$N - S = \sum_{k=0}^n a_i(10^k -1)$$.

Now each of the $$10^k-1$$ is $$99999.......9$$ which is divisible by $$9$$. So the whole number $$N-S$$ is divisible by $$9$$.

(Note the last digit, $$a_0$$ is simply subtracted. You can think of it as $$a_0 - a_0 = a_0*10^0 - a_0 = a_0(10^0 - 1) = a_0(1 - 1) = a_0*0 = 0$$.)

....

As a result, a number is divisible by $$9$$ if and only if the sum of its digits are divisible by $$9$$ so that gives you your second result.

It also helps us realize if we take a number and divide by $$9$$ and take the remainder, then the remainder will be the same as the remainder taking the sum of the digits and dividing by $$9$$ and taking the remainder.

Other answers have explained why the result of subtracting the digits of a number $$n$$ from $$n$$ is always divisible by $$9$$ and that if a number is divisible by $$9$$, so is the sum of its digits.
If $$n$$ only has two digits, it can't be bigger than $$99$$. So when you subtract its digits, you get something that's (i) less than $$99$$, and (ii) divisible by $$9$$. The only possibilities are $$9$$, $$18$$, $$27$$, $$36$$, $$45$$, $$54$$, $$63$$, $$72$$ and $$81$$.
First way: The biggest sum you can get from adding two digits is $$18$$, when both digits are $$9$$. But that needs the result of your first step to be $$99$$, which we've said it can't be. But we know it's a multiple of $$9$$, meaning its digits add up to a multiple of $$9$$. Since they can't add up to $$18$$, the only available multiple of $$9$$ is $$9$$.
Second way: How do you get from, say, $$18$$ to $$27$$? To add $$9$$ you can add $$10$$, which increases the first digit by $$1$$, then subtract $$1$$, which reduces the second digit by $$1$$. When you add the digits, the two changes cancel out and the digits still sum to $$9$$. Thinking of $$9$$ as $$09$$, this works all the way up from $$9$$ to $$90$$ and only goes wrong when you get to $$99$$—which is too big to be one of the options so the sum is always $$9$$.