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2012 AMC 10B Problem 20

2012 AMC 10B Problem 20AMC 10 Step by Step

Name: Date: September 13, 2026

Games & Processes · difficulty 3 of 5 · about 5 min · position P16-20

Bernardo and Silvia play the following game. An integer between 0 and 999, inclusive, is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds 50 to it and passes the result to Bernardo. The winner is the last person who produces a number less than 1000. Let $N$ be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of $N$ ?

A)
$7$
B)
$8$
C)
$9$
D)
$10$
E)
$11$

Answer

Key insight

Bernardo wins exactly when one of his outputs 2N, 4N+100, 8N+300, 16N+700 lies in [950, 999], because Silvia adds only 50; the smallest N is 16.

Solution

Write out the numbers produced, starting from $N$ :

Bernardo	$2N$	$4N+100$	$8N+300$	$16N+700$	$32N+1500$
Silvia	$2N+50$	$4N+150$	$8N+350$	$16N+750$

The outputs increase, and the game ends at the first output of $1000$ or more. Bernardo wins when that first big output is Silvia's, which happens exactly when some Bernardo output $b$ satisfies $b < 1000 \le b + 50$ , i.e. $950 \le b \le 999$ . (Bernardo's own output $32N + 1500$ is always at least $1000$ , so the game never continues past Silvia's fourth turn.)

Test each Bernardo output for the smallest $N$ :
- $16N + 700 \ge 950 \iff N \ge 15.625$ , and $16N + 700 \le 999 \iff N \le 18$ . So $N = 16, 17, 18$ work.
- $8N + 300 \ge 950$ needs $N \ge 82$ ; the earlier outputs need even larger $N$ .

The smallest winning $N$ is $16$ : Bernardo produces $32, 114, 278, 956$ and Silvia then produces $1006$ . The digit sum of $16$ is $1 + 6 = 7$ .

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

Why this works

Iterated "double, then add" processes become linear expressions in the starting value after a few rounds, so the whole game can be tabulated symbolically. The win condition is then a window: Bernardo needs to land just under $1000$ , close enough that Silvia's $+50$ overshoots. Turning "who wins" into an inequality on the last safe output is the key reformulation.

The trap

Finding the winning range 16 ≤ N ≤ 18 and reporting the digit sum of the wrong endpoint, or forgetting that Silvia's next number must reach 1000.

Common mistakes

Finding the winning range 16 ≤ N ≤ 18 and reporting the digit sum of the wrong endpoint, or forgetting that Silvia's next number must reach 1000.
Requiring only that Bernardo's number be under $1000$ (giving $N = 0$ ) without checking that Silvia's response goes over.

Techniques

Bound the quantity above/below or estimate to pin it down · Start from the end state / desired conclusion and reverse