Number theory proof about splitting a number I have to prove that, given a number, if I split it into two numbers, sum them up and repeat the same thing with the resulting number, I will eventually get a one-digit number that doesn't depend on how I decided to split all the numbers. For instance $$2020\mapsto20+20=40\mapsto4+0=4$$ but also $$2020\mapsto202+0=202\mapsto20+2=22\mapsto2+2=4$$ and so on. I thought that mod $9$ I always get the sum of the digits, but how to go further?
 A: Assume $N_0 \equiv k \pmod 9$ where $k= 1,2,....,9$.
We define a split $N_{m+1}$ noting that if $N_m$ is a $j$ digit number and we take $l$ of then the digits to make a number $B$ and the remaining $j-l$ of the digits to make another number $A$, then $N_m = 10{l} A + B$ and $N_{m+1}= A+B$
If $N_m \equiv k \pmod 9$ then $10^l A + B\equiv k \pmod 9$ but $10^l= 9*\frac {10^l-1}9 + 1 \equiv 1 \pmod 9$ and so $k \equiv N_{m} = 10^l A + B \equiv A + B =N_{m+1} \pmod 9$.
So by induction as $N_0 \equiv k \pmod 9$ then all splits $N_m \equiv k \pmod 9$.
Which means that the final split down to a single digit is $N_{\omega} \equiv k$ and $1 \le N_{\omega} < 10$ and the only single digit $\equiv k$ is $k$ so no matter how you do the splits you get $k$.
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The only real assumption made is that every choice of splits terminates to a single digit and that single digit is not $0$.
Well, every split results in fewer digits and by well ordering of naturals then there is some lowest number of digits through splits.  And that number of digits can't be be more than $1$ as we can always do a split on any digit more than $1$.
That final digit, $N_{\omega}$ cant be $0$, because for any split $N_{m+1} =A+B \ge \max (A,B)$ so if $A+B= 0$ then then $A,B=0$ and $N_{m}=10^lA + B=0$.  So we can only split to $0$ from $0$ and as $N_0 \ne 0$ then no $N_m = 0$ and $N_{\omega} \ne 0$.
