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$a$ - 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.

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    $\begingroup$ $1+1\neq 7$ $\quad$ $\endgroup$ – lulu Dec 3 '18 at 19:44
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    $\begingroup$ 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. $\endgroup$ – lulu Dec 3 '18 at 19:45
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    $\begingroup$ Maybe you mean $1+1+7=9$? $\endgroup$ – Noah Schweber Dec 3 '18 at 19:48
  • $\begingroup$ yea it was a typo i was typing this fast after class and didnt see it until now $\endgroup$ – Lonnie Dec 3 '18 at 19:57
  • $\begingroup$ If you write your name in base 6 (instead of 10), you get the same thing but with 5 (instead of 9).... which suggests... $\endgroup$ – Arturo Magidin Dec 3 '18 at 22:16
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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 $9$

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  • $\begingroup$ You may not initially get nine, but if you repeat the procedure enough you will eventually get nine. $\endgroup$ – Mike Earnest Dec 3 '18 at 20:35
  • $\begingroup$ 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. $\endgroup$ – Mohammad Riazi-Kermani Dec 3 '18 at 21:43
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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 $f(x)<x.$

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 $f$.

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