Race to Fifty

Call a total safe if the player who has just named it can force a win.

Claim. The safe totals are exactly those with $t \equiv 6 \pmod{11}$, namely $6, 17, 28, 39, 50$.

Safe totals stay safe. Suppose you have just named such a total $t<50$. Your opponent adds some $k\in\{1,\dots,10\}$. You then add $11-k$, which also lies in $\{1,\dots,10\}$. This brings the total to $t+11$, which is again $\equiv 6 \pmod{11}$ and does not exceed $50$. Repeating this, you reach $50$ yourself, since $50 = 6 + 4\cdot 11$. The opponent can never reach $50$ first: from a safe total $t\le 39$ they can reach at most $t+10<t+11\le 50$.

Other totals are unsafe. Suppose the total $t<50$ is not $\equiv 6 \pmod{11}$. Let $s$ be the next value above $t$ with $s \equiv 6 \pmod{11}$. Then $1\le s-t\le 10$, so the player to move can jump to $s$. The previous strategy then wins for that player.

Your first move. You can name any number from $1$ to $10$, and the only safe one is $6$. Any other choice loses:

• from $1,\dots,5$ your opponent jumps to $6$;
• from $7,\dots,10$ your opponent jumps to $17$.

So you should say $6$. After that, whenever your opponent adds $k$, you add $11-k$, naming $17, 28, 39$ and finally $50$.

Key idea. You can always complete each round to $m+1=11$. Hence the winning totals are those congruent to the target modulo $11$.