Amazing 22 ( a preceding topic for amazing 99)?

Question

Take three diffrent numbers ( eg 672)
make two numbers from the right without repeating
eg 67,72, and 62 and add them together ( 97+72+62= 201)
flip the numbers above and add them together
that is ( 67,72,and 62 becomes 76,27, and 26 respectively)
now adding ( 76+27+26 = 129)
Add the results together ( 201+129= 330)
Divide the result by the sum of the three digits you chose (for this case 6+7+2 which is 15)
The answer is always 22 ( for this case 330/15 = 22)

This follows for all three diffrent numbers.
Look at the other posting for amazing 99, why are all these manipulations a multiple of 11?
is 11 a certain number where all numbers revolve around?)
These was my original work, I was just messing with numbers when the Amazing 99 and amazing 22 happened to be the result)

Thanks for reading and answering too.

Answer 1

That was good.

But lets take any three two-digit numbers (Why trouble a three digit number?). Say 23, 45, 69.

Adding them, we have 23+45+69 = 137

Reversing the numbers and adding them, we have 32+54+96 = 182


Adding these, 182 + 137 = 319, which is again a multiple of 11 (11*29=319).

And, adding the digits of the numbers we took : 3+2+5+4+9+6 = 29.

Coincidence??? Guess ...

And I would just call it amazing 11 !!!

Keep posting.

---------------------------------------------------------------------------------

My friends it seems; still think that its 22 that has all the magic.

Well, as I said earlier, it really does not matter if you take a three digit number and make three 2 digit numbers out of it. You may pick three 2 digit numbers randomly.

The reason you get 22 as the final answer is because while summing the digits, you add only three digits (instead of six). This is because in your case these digits repeat.

As say from 123, you take 12, 23, 31, but while summing up the digits, you add 1+2+3(=6), instead of 1+2+2+3+3+1(=12), therefore you divide (the summation of numbers) by half the sum and then get 22 instead of 11 as the final answer.

Answer 2

Do you want to know why?

Choose three different digts a, b, and c. Make three numbers with the digits (using each pair only once); so ab, ac, and bc. Now reverse these numbers: ba, ca, cb.

Add all these numbers together:
ab+ac+bc+ba+ca+cb = (a+a+b+b+c+c)•10 + (b+c+c+a+a+b)= 20(a+b+c)+2(a+b+c) = 22(a+b+c). Divide this by the sum of the three number (a+b+c) and you get 22.

The numbers don't have to be different, but you have to choose appropriate combinations.

For example, use 112. You need to make groups 11 (this is ab), 12 (this is ac), and 12 (this is bc). You have two 12s because there are two 1s.
The sum of these numbers is 11+12+12=35.
Reversing the numbers and adding: 11+21+21=53
Adding together 35+53=88
Dividing by 1+1+2=4: 88÷4=22.

If you did this with 4 digit numbers, you should get "Amazing 33." I'm not really positive on it, but try it. (for the "reversing" stage, you have to shift all one direction, and then do it again, cause you will be dealing with 3 digit numbers).