, 2023

Solution:
Note $2 s(n)=s(10 n)+s(n)=s(11 n)$, so there cannot be any carries when adding $n$ and $10 n$. This is equivalent to saying no two consecutive digits of $n$ sum to greater than $9$.

We change the problem to nonnegative integers less than $10^{4}$ (as both $0$ and $10^{4}$ satisfy the condition) so that we simply consider $4$-digit numbers, possibly with leading $0$s. Letting our number be $a b c d$, we need $a+b \leq 9$, $b+c \leq 9$, and $c+d \leq 9$. Letting $b' = 9-b$ and $d' = 9-d$, this means $a \leq b'$, $b' \geq c$, $c \leq d'$.

Summing over all possible values of $b'$ and $d'$, we want
$$\sum_{x, y=1}^{10} x \cdot \min (x, y)$$
The sum over pairs $(x, y)$ with $x>y$ is
$$\frac{(1+2+\cdots+10)^{2}-\left(1^{2}+2^{2}+\cdots+10^{2}\right)}{2}=\frac{55^{2}-55 \cdot 7}{2}=55 \cdot 24 .$$
The sum over pairs $x \leq y$ is
$$\sum_{k=1}^{10} k^{2}(11-k)=11 \cdot 55 \cdot 7-55^{2}=55 \cdot 22$$
The final answer is $55 \cdot(24+22)=55 \cdot 46=2530$.

Solution 2:
Here is another way to calculate the sum. By doing casework on $(b, c)$, the sum is
$$\sum_{b+c \leq 9}(10-b)(10-c)$$
This is the coefficient of $x^{9}$ of
$$\begin{aligned} \left(\sum_{n \geq 0}(10-n) x^{n}\right)^{2}\left(\sum_{n \geq 0} x^{n}\right) & =\frac{(10-11 x)^{2}}{(1-x)^{5}} \\ & =(10-11 x)^{2} \sum_{n \geq 0}\binom{n+4}{4} x^{n} \end{aligned}$$
Thus, the answer is
$$100\binom{13}{4}-220\binom{12}{4}+121\binom{11}{4}=2530$$