2012 AMC 10B Problem 20

1. Try a small start to see how the game goes. With $N = 1$: Bernardo makes $2$, Silvia $52$, Bernardo $104$, Silvia $154$, Bernardo $308$, Silvia $358$, Bernardo $716$, Silvia $766$, Bernardo $1532$. Bernardo went over $1000$ first, so Silvia made the last number under $1000$ and she wins.
2. Who wins depends on who first goes over $999$. Bernardo wins when Silvia is the one who goes over. Silvia adds only $50$, so this happens when Bernardo hands her a number from $950$ to $999$.
3. Write every number in terms of $N$. Bernardo doubles, Silvia adds $50$:

(Bernardo's fifth number would be $32N + 1500$, which is already over $1000$, so the game is always decided by then.)
4. For any starting $N$, Bernardo's fourth number is his biggest, so it is the one that can reach $950$ with the smallest $N$. Start there. Need $950 \le 16N + 700 \le 999$:
- $16N + 700 \ge 950$ means $16N \ge 250$, so $N \ge 15.6$, that is $N \ge 16$;
- $16N + 700 \le 999$ means $16N \le 299$, so $N \le 18$.
So $N = 16, 17, 18$ all win this way.
5. Could an earlier turn give a smaller $N$? For $8N + 300 \ge 950$ we need $N \ge 82$, and the first two turns need even bigger $N$. So nothing smaller than $16$ works.
6. Check $N = 16$: Bernardo $32$, Silvia $82$, Bernardo $164$, Silvia $214$, Bernardo $428$, Silvia $478$, Bernardo $956$, Silvia $1006$. Silvia went over, so Bernardo's $956$ was the last number under $1000$. Bernardo wins.
7. The digit sum of $16$ is $1 + 6 = 7$.

The answer is $\boxed{\textbf{(A)}\ 7}$.

A process that repeats "double, then add" turns into a simple expression in the starting number after each round, so the whole game fits in one table. The winning condition becomes a window: Bernardo must land just below $1000$, close enough that Silvia's $+50$ pushes her over. Turning "who wins" into an inequality is the key step.