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$\quad$A couple days ago, I heard of a simple numbers game(or trick), and decided to prove it. I succeeded (I think), but don't really like the way I did it, so I was wondering if someone could think of another way.

But first, here is the game:

$\quad$You pick any 2 digit whole number (let's say, for example purposes, you pick 55). Then, add both digits of your number and subtract the result from the original number. $(55 - (5+5) = 45)$ After that, add the digits of your new name and you will always get 9 $(4+5 = 9)$.

and my proof:

Our first number will be N, composed of two digits, x and y. So:

$N = 10x + y\quad,\quad N,x,y \in \Bbb N^0 \;,\; 10 \le N \le 99\;, \; 1 \le x \le 9\;,\; y \le 9 $

Our second number, N', is composed of digits a and b.

$N' = 10a + b\quad,\quad N',a,b \in \Bbb N^0 \;,\; 10 \le N \le 99\;, \; a \le 9\;,\; b \le 9 $

We want to prove that $a+b = 9, \forall x,y \;\text{in the conditions above} $ :

$$N' = N - (x + y) = 10x +y - x - y = 10x - x + y - y <=> N' = 9x \quad <=> 10a + b = 9x <=>\text{(isolating a+b)} 9a + a + b = 9x <=> a + b = 9x - 9a \quad <=> a + b= 9(x-a) $$

$\quad$So, if we can prove that x-a = 1, we prove that a+b = 9 and, in turn, the game. This is where I have a problem. The only way I can find is not one I like. It goes something like this:

a is the first digit of N' = 9x. So, since there are only 9 possible values for x, we do this table

$$\begin{array}{c|c|cc} x & \times9= & a & b \\ \hline 1 & \times9= & 0 & 9 \\ 2 & \times9= & 1 & 8 \\ 3 & \times9= & 2 & 7 \\ 4 & \times9= & 3 & 6 \\ 5 & \times9= & 4 & 5 \\ 6 & \times9= & 5 & 4 \\ 7 & \times9= & 6 & 3 \\ 8 & \times9= & 7 & 2 \\ 9 & \times9= & 8 & 1 \\ \end{array}$$

$\quad$A quick glance at the table is enough to conclude that it is true that $a = x-1 <=> x-a=1$, meaning $a+b = 9$, and so the game is proved.

$\quad$What I'm looking for is a way to prove this without resorting to a table, or anything like that: not writing out all the possible cases. If you read until here, thank you for your time, I sincerely hope you can help me.

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HINT :

Let AB be the chosen 2 digit number then

$ 10A + B - (A+B) = 9 \times A $

So a multiple of 9 , which will sum to 9 for any A between 1 to 9

ILLUSTRATION: For example if A = 9 ( B can be anything because any way in the subtraction it gets cancelled) , 9* 9 = 81 which is 8+1 =9.

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    $\begingroup$ Yes, I got the result of the number being 9 times x, or a multiple of 9. I suppose what I really want, and just figured that out while writing this comment, is proof that the digits of a multiple of 9 add up to 9, which a quick Google search promptly revealed. Thank you for the perspective given by your choice of words. $\endgroup$ – Sampaio Aug 24 '13 at 23:06
  • $\begingroup$ Good Happy learning . Thanks . $\endgroup$ – Harish Kayarohanam Aug 24 '13 at 23:08

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