# 132 is Sum of all 2-Digit Numbers formed from its Digits

## Theorem

$132$ is the smallest sum of all the $2$-digit (positive) integers formed from its own digits.

## Proof

It is necessary to postulate that such (positive) integers have $3$ digits or more, as the $2$ digit solution is trivial.

Let $n = \sqbrk {abc}$ be a $3$-digit number.

Let $\map s n$ denote the sum of all the $2$-digit (positive) integers formed from the digits of $n$.

Then:

 $\displaystyle \map s n$ $=$ $\displaystyle \sqbrk {ab} + \sqbrk {ac} + \sqbrk {bc} + \sqbrk {ba} + \sqbrk {ca} + \sqbrk {cb}$ $\displaystyle$ $=$ $\displaystyle \paren {10 a + b} + \paren {10 a + c} + \paren {10 b + c} + \paren {10 b + a} + \paren {10 c + a} + \paren {10 c + b}$ $\displaystyle$ $=$ $\displaystyle 2 \times \paren {10 a + a} + 2 \times \paren {10 b + b} + 2 \times \paren {10 c + c}$ $\displaystyle$ $=$ $\displaystyle 22 \times \paren {a + b + c}$

So for $n = \map s n$, $n = \sqbrk {abc}$ needs to be divisible by both $22$ and $a + b + c$.

 $\displaystyle 22 \times 4$ $=$ $\displaystyle 88$ too small $\displaystyle 22 \times 5$ $=$ $\displaystyle 110$ but $\map s {110} = 11 + 10 + 01 = 22$ $\displaystyle 22 \times 6$ $=$ $\displaystyle 132$ and $\map s {132} = 12 + 13 + 21 + 23 + 31 + 32 = 132$

and the result is apparent.

$\blacksquare$