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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?