## Reverse Equivalence

#### Problem

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.

#### Solution

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) = 10`a` + `b` + 10`b` + `a` = 11`a` + 11`b` is divisibly by 11 is trivial.

The problem requires the sum, `S` = (`ab`) + (`cd`) = (`ba`) + (`dc`), such that (ab) (cd) (ba)

`S` = (10`a` + `b`) + (10`c` + `d`) = (10`b` + `a`) + (10`d` + `c`).

Therefore 9`a` + 9`c` = 9`b` + 9`d` 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` = (10`a` + `b`) + (10`c` + `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?