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

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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.

$\blacksquare$


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