The simplest way to explain arithmetic mean is in terms of "equal sharing":
Abe has 12 cookies, Brianna has 8 cookies, and Chuck has 7 cookies. If they were to redistribute them so that they all have the same amount, how many would each get?
Obviously the way you answer this question is to find the total amount of cookies ($12 + 8 + 7 = 27$) and then divide the cookies among the three people ($27/3 = 9$). That's precisely what the computation of arithmetic mean does.
Is that what you're looking for?
Edited to add:
Here's another viewpoint that might help. We would like to find some number $N$ that is in the "middle" of the set $ \{12, 8, 7 \}$ (using the same numbers from the example above). What does "in the middle" mean? Well, one way to interpret this vague phrase is to imagine that we already had such an $N$ in hand, and we compute the three quantities $12-N, 8-N,7-N$. These three quantities tell us how far $N$ is from each of three pieces of information -- call these the "deviations".
What if we made a bad choice of $N$? For example, if each of the three deviations were positive, then that would mean that $N$ is smaller than each of the three original numbers, which we don't want. If each of the three deviations were negative, then that would mean that $N$ is larger than each of the three original numbers -- again bad. For $N$ to be in the middle, we would want some of the deviations to be positive and some of them to be negative. In fact, if we could choose $N$ so that the positive deviations exactly cancel out the negative deviations, then we will feel like we've really found the "middle".
Let's translate that now into a computation. We want to find $N$ such that
$$(12-N) + (8-N) + (7-N) = 0$$
If you now think about what it would take to solve this equation, you will quickly realize that you end up adding the three numbers in your dataset together and then dividing by 3.
Setting $100A+10B+C$, we have
$$(100A+10B+C)-(100C+10B+A)=100(A-C)+C-A$$$$=100(A-C-1)+9\cdot 10+(10+C-A).$$
So, we have
$$\left(100(A-C-1)+90+(10+C-A)\right)+\left(100(10+C-A)+90+(A-C-1)\right)$$$$=9\cdot 100+90\cdot 2+9=1089.$$
Best Answer
Let "$ab$" be a two digit number.
Then $ab = 10a + b$ and $ab\times 11 = (10a+b)(10 + 1) =$
$ 10a(10 + 1) + b(10+1) =$
$ (100a + 10a)+ (10b + b) =$
$100a + (10a + 10b)+b =$
$100a + 10(a+b) + b$.
And if $a+b < 10$ we get $100a + 10(a+b) + b = a(a+b)b$.
Not it doesn't work if $a+b \ge 10$. Example $84\times 11= 924$ and $9+4 =13$ and not $2$. But notice that $9+4 -11 = 2$.
If $a+b \ge 10$ you get:
$100a + 10(a+b) + b = 100a + 10([a+b-10] + 10)+b$
$=100a + 100 + 10[a+b-10] + b = (a+1)(a+b-10)b$.
If we write this as $cde$ we have $c+e = a+b + 1 = (a+b-10) + 11 = d+11$.
So you can modify to rule to if $cde = K\times 11$ then either $c +e =d$ or $c+e = d+11$
We can extend this further: