Trivial riddle. Can you find out the formula?

Others have already solved the riddle, but not provided the function. This is also easy, if a bit ugly. The point is that we need to have a way to find the number of digits of the sum. Let

g(m,n)=number of digits in m+n,

then clearly the desired function is

f(m,n)=(m-n)*10^(g(m,n)) + (m+n).

The number of digits of a positive integer is roughly the base-10 logarithm. Precisely,

g(m,n)=\floor{ \log(m+n) } + 1.

So, if you like, the function is

f(m,n)=(m-n)*10^( floor(log(m+n))+1 ) + m+n

EDIT: Er, I guess I don't know what to do if m<n. But you haven't given any such examples, so I guess this technically "does the job" :P

EDIT 2: I like your extension! (But the ceiling is not the floor+1 when the argument is an integer, see e.g. your first example.)

The first 2 digits (from the right) are the sum of the 2 numbers.

The Other digits are the difference between the 2 numbers.

For Example:

A) 9 + 8 = 117

9+8=17, 9-8=1 ---> 117

B) 15 + 3 = 1218

15+3=18, 15-3=12 ---> 1218

Looks like you have:

m + n = ABC...

m + n = BC

m - n = A

=> (m-n)(m+n) would be the 3+ digit result

Example:

9 + 8 = (9-8)(9+8) = (1)(17) = 117

subtract the second number from the first, and write it down. then append the sum of the two numbers