Difference between a non-increasing number and it's mirror (digit reversal)

Question

Consider a subtraction between a non-increasing number $N$ and its mirror $M$ (a non-decreasing number) where N has half or more digits equal to $0$.

Example:$\qquad N-M = 76620000-00002667 = D$

I need to prove that the answer $D$ has the following properties:

• The number is bi-descending (concatenation of 2 non-increasing numbers)

Example:$\qquad 76620000-00002667 = 76617333 = 7661 |7333$ (bi-descending)

• The respective digit sum of $D$ and its mirror $R$ is either $8,9,10$ or $18$

Example 1:$\qquad N= 76620000 \longrightarrow D =76617333, R = 33371667$

$N_1 + M_1 = 7+3=10$

$N_2 + M_2 = 6+3=9$

$N_3 + M_3 = 6+3=9$

$N_4 + M_4 = 1+7=8$

Example 2:$\qquad N= 21000000 \longrightarrow D =20999988, R = 88999902$

$N_1 + M_1 = 2+8=10$

$N_2 + M_2 = 0+8=8$

$N_3 + M_3 = 9+9=18$

$N_4 + M_4 = 9+9=18$

It can be seen that 1 digit pair will sum to 10, 1 pair will sum to 8 and all others will sum to either 9 or 18.

Any help proving these results is greatly appreciated. Thank you in advance.

Answer 1

$N$ can be written as $$N=10^m(a_n\times 10^{n-1}+a_{n-1}\times 10^{n-2}+\cdots +a_{1})$$ where $9\ge a_n\ge a_{n-1}\ge\cdots \ge a_1\ge 1$ and $m\ge n$.

Then, we get $$M=a_1\times 10^{n-1}+a_2\times 10^{n-2}+\cdots +a_n$$

Noting that $m\ge n$, we get $$\begin{align}D&=N-M\\\\&=10^m(a_n\times 10^{n-1}+a_{n-1}\times 10^{n-2}+\cdots +a_{1}) \\&\qquad\quad-(a_1\times 10^{n-1}+a_2\times 10^{n-2}+\cdots +a_n) \\\\&=10^m(a_n\times 10^{n-1}+a_{n-1}\times 10^{n-2}+\cdots +a_{2}\times 10)+(a_1-1)\times 10^m \\&\qquad\quad+10^m-(a_1\times 10^{n-1}+a_2\times 10^{n-2}+\cdots +a_n) \\\\&=10^m(a_n\times 10^{n-1}+a_{n-1}\times 10^{n-2}+\cdots +a_{2}\times 10)+(a_1-1)\times 10^m \\&\qquad\quad+9\times 10^{m-1}+9\times 10^{m-2}+\cdots +9\times 10^n \\&\qquad\quad +(9-a_1)\times 10^{n-1}+(9-a_2)\times 10^{n-2}+\cdots +(9-a_{n-1}) \\&\qquad\quad+(10-a_n) \end{align}$$ which is $$\overline{a_na_{n-1}\cdots a_2(a_1-1)\underbrace{99\cdots 9}_{m-n \ 9\text{'s}}(9-a_1)(9-a_2)\cdots (9-a_{n-1})(10-a_n)}$$

This is not bi-descending in general.

Counterexamples are numbers such that $a_n=a_{n-1}$ such as $N=22000$ where we get $D=22000-22=21978$ which is not bi-descending.

If $9-a_{n-1}\ge 10-a_n$, i.e. $a_n-a_{n-1}\ge 1$ holds, then $D$ is bi-descending as follows : $$\overline{a_na_{n-1}\cdots a_2(a_1-1)}\mid \overline{\underbrace{99\cdots 9}_{m-n \ 9\text{'s}}(9-a_1)(9-a_2)\cdots (9-a_{n-1})(10-a_n)}\qquad\square$$

Also, we get $$\begin{align}R&=(10-a_n)\times 10^{m+n-1}+(9-a_{n-1})\times 10^{m+n-2}+\cdots +(9-a_1)\times 10^m \\&\qquad\quad +9\times 10^{m-1}+9\times 10^{m-2}+\cdots +9\times 10^n \\&\qquad\quad +(a_1-1)\times 10^{n-1}+a_2\times 10^{n-2}+a_3\times 10^{n-3}+\cdots +a_n\end{align}$$ which is $$\overline{(10-a_n)(9-a_{n-1})\cdots (9-a_1)\underbrace{99\cdots 9}_{m-n \ 9\text{'s}}(a_1-1)a_2a_3\cdots a_n}$$

So, the respective digit sum of $D$ and its mirror $R$ is as follows :

• $a_n+(10-a_n)=10$
• $a_{n-1}+(9-a_{n-1})=9$

$\qquad\vdots$

• $a_2+(9-a_2)=9$
• $(a_1-1)+(9-a_1)=8$
• $9+9=18$ only if $m\gt n$

Therefore, if $m\gt n$, then the respective digit sum of $D$ and its mirror $R$ is either $8,9,10$ or $18$. If $m=n$, then the respective digit sum is either $8,9$ or $10$.$\qquad\square$