Let's determine the remainder of the sum $1+22+333+4444+\ldots+\underbrace{s s s . \ldots s}_{s \text { times }}$ when divided by $(s-1)$ in the base $s+1$ number system.

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Determine the remainder of the sum $1+22+333+4444+\ldots+\underbrace{s s s . \ldots s}_{s \text { times }}$ when divided by $(s-1)$ in the base $s+1$ number system.
If $s=2$, then $s-1=1$ and thus the remainder is 0. Let's assume from now on that $s>2$. Let $S_{i}$ denote the number $\underbrace{i i i \ldots i}_{i \text{-times}}$ written in the $(s+1)$-base number system.

$$
S_{i}=i \cdot \underbrace{111 \ldots 1}_{i \text{-times}}=i \cdot\left((s+1)^{i-1}+(s+1)^{i-2}+\cdots+1\right)
$$

Dividing $(s+1)$ by $(s-1)$ gives a remainder of 2, so $S_{i}$ and $i\left(2^{i-1}+2^{i-2}+\cdots+1\right)=i\left(2^{i}-1\right)$ give the same remainder when divided by $(s-1)$.

Therefore, $1+22+\cdots+s s \ldots s$ gives the same remainder when divided by $(s-1)$ as

$$
1\left(2^{1}-1\right)+2\left(2^{2}-1\right)+\cdots+s\left(2^{s}-1\right)=1 \cdot 2+2 \cdot 2^{2}+\cdots+s \cdot 2^{s}-\frac{s(s+1)}{2}
$$

Let

$$
B=1 \cdot 2^{1}+2 \cdot 2^{2}+\cdots+s \cdot 2^{s}-2, \text{ then } 2 B=1 \cdot 2^{2}+2 \cdot 2^{2}+\cdots+s \cdot 2^{s+1}-4
$$

The difference between the two equations:

$$
B=-2^{1}-2^{2}-\cdots-2^{s}+s \cdot 2^{s+1}-2=(s-1) 2^{s+1}
$$

which is divisible by $(s-1)$.

Therefore, the quantity to be determined in the problem is the remainder of $\left(2-\frac{s(s+1)}{2}\right)$ when divided by $(s-1)$.

This is 1 if $s$ is even, and $\frac{s+1}{2}$ if $s$ is odd.

Based on the work of Máthé András (Budapest, ELTE Apáczai Csere J. Gimm., 11th grade)