Henry Ernest Dudeney/Puzzles and Curious Problems/90 - Summing the Digits/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $90$

Summing the Digits
What is the sum of all the numbers that can be formed with all $9$ digits ($0$ excluded),
using each digit once and once only, in every number?


Solution

$201 \, 599 \, 999 \, 798 \, 400$


Proof

There are $9!$ permutations of the $9$ digits, that is: $362 \, 880$.

Of these permutations, each digit appears in each of the $9$ positions a total of $8!$ times each.

We note that $\ds \sum_{n \mathop = 1}^9 n = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.

Let $S$ be the required sum.

We have:

\(\ds S\) \(=\) \(\ds 8! \times \sum_{n \mathop = 1}^9 n\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 8! \times \sum_{n \mathop = 1}^9 10 n\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \cdots\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 8! \times \sum_{n \mathop = 1}^9 100 \, 000 \, 000 n\)
\(\ds \) \(=\) \(\ds 8! \times \paren {45 + 450 + 4500 + 45 \, 000 + 450 \, 000 + 4 \, 500 \, 000 + 45 \, 000 \, 000 + 450 \, 000 \, 000 + 4 \, 500 \, 000 \, 000}\)
\(\ds \) \(=\) \(\ds 40 \, 320 \times 4 \, 999 \, 999 \, 995\)
\(\ds \) \(=\) \(\ds 201 \, 599 \, 999 \, 798 \, 400\)

$\blacksquare$



This theorem requires a proof.
In particular: Given that we have this problem and the previous one, it's worth setting up a page proving the sum of all numbers made from the digits $1, 2, \ldots, n$ once and once only for $1 \le n \le 9$, that is, for the general $n$, and using that. Include Dudeney's technique (not given here), tightened up and rationalised.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
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Sources

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