Combined digit sum

Let's split $S$ into four subsets: $S_4, S_3, S_2$ , and $S_1$ according to the length of the numbers. There are $4!=24$ four-digit numbers, $4!=24$ three-digit numbers, $4*3=12$ two-digit numbers, and $4$ one-digit numbers.

I'll calculate the sum of the elements of $S$ by separately summing $S_k$ and then combining those four sums. For each $k$ , the sum of $S_k$ is the cardinality of $S_k$ multiplied by the average value of a member.

Lemma: the average value of $S_k = \sum_{i=0}^{k-1} 4 \times 10^i$ (I.e., the average value of $S_1$ is $4$ , the average value of $S_2$ is $44$ , the average value of $S_3$ is $444$ , and the average value of $S_4$ is $4444$ .)

This is pretty obvious: for example, for each $abcd \in S_4$ , $(8-a)(8-b)(8-c)(8-d)$ is also in $S_4$ . The average of this pair is $4444$ . The entirety of $S_4$ can be partitioned into such pairs. Since the average value in each pair is $4444$ , the average over the entire set is also $4444$ .

Thus the sum of $S_4$ is $24*4444$ , the sum of $S_3$ is $24*444$ , the sum of $S_2$ is $12*44$ , and the sum of $S_1$ is $4*4$ . The grand total is $117856$ .

The number of single-digit elements $n_1 = 4$ . Therefore $1$ , $3$ , $5$ , and $7$ each appears once. Then the sum of single-digit elements $S_1 = 1+3+5+7 = 16$ .

The number of two-digit elements $n_2 = 4 \times 3 = 12$ . Therefore $1$ , $3$ , $5$ , and $7$ each appears three times as unit digit and three times as tens digit. Then the sum of two-digit elements $S_2 = 33(1+3+5+7) = 528$ .

The number of three-digit elements $n_3 = 4 \times 3 \times 2 = 24$ . Therefore $1$ , $3$ , $5$ , and $7$ each appears six times as unit, tens, and hundreds digits. Then the sum of three-digit elements $S_3 = 666(1+3+5+7) = 10656$ .

The number of four-digit elements $n_4 = 4 \times 3 \times 2 \times 1 = 24$ . Therefore $1$ , $3$ , $5$ , and $7$ each appears six times as unit, tens, hundreds, and thousands digits. Then the sum of four-digit elements $S_4 = 6666(1+3+5+7) = 106656$ .

Therefore the sum of all elements of set $S$ , $S_{\rm all} = S_1 + S_2 + S_3+S_4 = 16 + 528 + 10656+106656 = \boxed{117856}$ .

• 1 digit: $S_1 = 1 + 3 + 5 + 7 = 16$

• 2 digits: each digit appears 3 times at each place, so $\\ S_2 = 3 \cdot (10+1) \cdot S_1 = 33 \cdot S_1$

• 3 digits: each digit appears 6 times at each place, so $\\S_3 = 666 \cdot S_1$

• 4 digits: each digit appears 6 times at each place, so $\\S_4 = 6666 \cdot S_1$

$\therefore \sum S = \sum\limits_{i=1}^4 S_i \\= 16 \times (1 + 33 + 666 + 6666) \\= 117856$