By adding the different 2-digit numbers 12 and 32 we get 44. If the digits in each number are reversed we get two different 2-digit numbers, and 21 + 23 also equals 44.
The same is true of 42 + 35 = 24 + 53 = 77.
Prove that the sum of two 2-digit numbers with this property will always be divisible by 11.
Let the 2-digit numbers be (ab) and (cd).
We shall ignore cases like 11 + 11 = 11 + 11 and 12 + 21 = 21 + 12, as proving that (ab) + (ba) = 10a + b + 10b + a = 11a + 11b is divisibly by 11 is trivial.
The problem requires the sum, S = (ab) + (cd) = (ba) + (dc), such that (ab) (cd) (ba)
S = (10a + b) + (10c + d) = (10b + a) + (10d + c).
Therefore 9a + 9c = 9b + 9d and so a + c = b + d.
Hence we obtain a pair of numbers with the required property if the sum of the first digits of each number is equal to the sum of the second digits. For example, 41 + 36 = 14 + 63 = 77, because 4 + 3 = 1 + 6.
Now S = (10a + b) + (10c + d) = 10(a + c) + b + d
As a + c = b + d, S = 10(a + c) + a + c = 11(a + c).
Hence the sum, S, is always divisible by 11.
What adding three 2-digit numbers?
Can you find two 3-digit numbers with this property?