# Why can you bring any number back to $9$ by doing this formula?

$$a$$ - Digits of $$a$$ = $$9$$ If you take any number add the digits then subtract the sum of the digits from the original number and repeat until you come to a one digit number then the one digit number is always 9, for example $$125$$ and you add the digits $$1 + 2 + 5 =$$8 then do $$125 - 8 = 117$$ then $$1 + 1 + 7 = 9$$, you can always bring it back to $$9$$.

From what I have found, you can do this with any number that has more than one digit. But why does this happen to every number? I can't seem to find a reason.

• $1+1\neq 7$ $\quad$ – lulu Dec 3 '18 at 19:44
• I'm just guessing here, but I think you are after the fact that , for any $N$, $9$ divides the difference between $N$ and the sum of the digits of $N$. That's just because $10^k$ is always one more than a multiple of $9$. It's a very useful fact and the basis for lots of numerical puzzles. – lulu Dec 3 '18 at 19:45
• Maybe you mean $1+1+7=9$? – Noah Schweber Dec 3 '18 at 19:48
• yea it was a typo i was typing this fast after class and didnt see it until now – Lonnie Dec 3 '18 at 19:57
• If you write your name in base 6 (instead of 10), you get the same thing but with 5 (instead of 9).... which suggests... – Arturo Magidin Dec 3 '18 at 22:16

Note that $$100a+10b+c -(a+b+c)= 99a+9b$$ which is a multiple of $$9$$ and as a result it's sum of digits is a multiple of $$9$$.

You do not necessarily get $$9$$ but you get a multiple of $$9$$

When repeated we will end up with nine eventually.

Same argument works for four or more digit numbers as well.

For example $$54321-15=54306$$ which is a multiple of $$9$$

• You may not initially get nine, but if you repeat the procedure enough you will eventually get nine. – Mike Earnest Dec 3 '18 at 20:35
• Very good comment. Yes, that is true because you are subtracting a positive integer every time so eventually you will end up with a nine. – Mohammad Riazi-Kermani Dec 3 '18 at 21:43

If $$j$$ is a non-negative integer then $$10^j$$ is $$1$$ more than a multiple of $$9$$ so let $$10^j-1=9V(j)$$ where $$0\leq V(j)\in \Bbb Z.$$ E.g. $$10^6=1+999999=1+(9)(111111)=1+9V(6).$$

For $$10\leq x\in \Bbb Z^+$$ let $$S(x)$$ be the sum of the digits of $$x$$ and let $$f(x)=x-S(x).$$ Then $$f(x)

And $$f(x)$$ is divisible by $$9.$$ Because if the digit-sequence for $$x$$ is $$(x_n,...,x_0)$$ then $$f(x)=x-S(x)=(x_n\cdot 10^n+...+x_0\cdot 10^0)-(x_n+...+x_0)=$$ $$=x_n(10^n-1)+...+x_0(10^0-1)=$$ $$=x_n\cdot 9V(n)+...+x_0\cdot 9V(0)=$$ $$=(9)(x_n V(n)+...x_0 V(0))\quad \bullet$$ Now if $$x\geq 10$$ then (i) $$n \geq 1$$ and $$x_n\geq 1$$ and $$V(n)\geq V(1)=1,$$ and (ii) no $$x_j V(j)$$ is negative whenever $$0\leq j< n,$$ so by $$\bullet$$ we have $$f(x)\geq (9)(x_nV(n))\geq 9\quad \star$$

So the sequence $$x,f(x), f((f(x)),...$$ (in a manner of speaking, as this sequence might have only the 2 terms $$x$$ and $$f(x)$$)..... is a strictly decreasing sequence in which every term, except possibly the first term, is a multiple of $$9$$. This sequence must reach a value less than $$10$$ where it must stop. But the last value is $$f(y)$$ where $$y$$ is the second-last value, with $$y\geq 10$$. So by $$\star$$ we have $$f(y)\geq 9.$$ That is, $$10>f(y)\geq 9$$.

Remark. The sequence must stop when it reaches a value less than $$10$$ because we deliberately keep values less than $$10$$ out of the domain of the function $$f$$.