On the table, 100 matchsticks lie. $A$ and $B$ take turns to remove at least one and at most seven matchsticks. The one who takes the last matchstick from the table wins. Is it true or not that the starting player always wins if they play correctly?
I. Solution: Let $A$ be the starting player. If $A$ leaves 8 matches to the opponent last, that is, $A$ takes the 92nd match and stops here, then $A$ has secured the 100th match as well. Similarly, if $A$ stops at the 84th match, $A$ has secured the 92nd and thus the 100th match. It is clear that $A$ will surely win if $A$ always leaves a number of matches on the table that is divisible by 8. Dividing 100 by 8, the remainder is 4. Therefore, $A$ will surely win if $A$ takes 4 matches first, and then always complements the number of matches taken by $B$ to 8. If $A$ fails to do this even once, $A$ has given the advantage to $B$, and $B$ can win in the same way.

Ferenc Tilesch (Esztergom, I. István g. III. o. t.)

II. Solution: We generalize the problem. Let there be $n$ matches on the table, and the number of matches that can be taken at once is at least 1 and at most $m$ ($m \geq 1$). We divide $n$ by $(m+1)$, and the remainder is one of the numbers $0, 1, 2, \ldots, m$, say $r$. If $r \neq 0$, then by taking $r$ matches, the number of matches left on the table is divisible by $(m+1)$. Therefore, the starting player $A$ will surely win in this case if $A$ first takes $r$ matches, and then always complements the number of matches taken by $B$ to $(m+1)$. In this case, the number of matches left on the table will run through the numbers divisible by $(m+1)$ in descending order, so after one of $A$'s turns, it will be exactly 0. - If $r=0$, then $B$ will always complement the number of matches taken by $A$ to $(m+1)$, and thus $B$ will win. - Therefore, in the case of correct play by both players, the starting player will win if and only if $n$ is not divisible by $(m+1)$.

In our problem, $n=100$, $m=7$, and thus $r=4$.