AMC 10 · 2012 · #20

Grade 7 countingprobability

Archetype Branching State Enumeration └ Sub-archetype Rule-Governed Process Simulation

interval-arithmeticdigit-sum work-backwards ↑ Prerequisites: interval-arithmetic

📏 Long solution 💡 4 insights

Problem

Start with an integer N from 0 to 999. Bernardo goes first and doubles whatever he holds; then Silvia adds 50 to whatever she receives, and they keep alternating. A number counts as legal only while it stays below 1000, and the person who produces the last legal number wins. Find the smallest starting N for which Bernardo wins, then add up the digits of that N.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Work Backwards

The winning condition is stated at the end of the game ("be the last under 1000"), while the question asks for the start value N — a classic Tool #11 (Work Backwards) setup. Working from the finish, Bernardo wins exactly when he hands Silvia a number so large that her +50 pushes it to 1000 or beyond; that pins his winning move to the window 950–999. Tool #4 (Introduce a Variable) turns the alternating doubling-and-adding into tidy expressions in N, so Bernardo's outputs are just 2N, 4N+100, 8N+300, 16N+700. Tool #14 (Extreme Principle) then hunts for the smallest N hitting that window, which happens on the last of those outputs. Tool #6 (Guess and Check) confirms the winner by replaying the actual game for N=16 and for N=15.

1STEP 1

Translate the win into a number window

Bernardo wins when Silvia's +50 dies: his number B stays under 1000 but B + 50 reaches it, so B lands in 950 ≤ B ≤ 999.

B ＜ 1000 and B + 50 ≥ 1000 ⟹ 950 ≤ B ≤ 999

2STEP 2

Name the start and track the moves

Let N be the start and write each move as an expression: 2N, 2N+50, 4N+100, 4N+150, 8N+300, 8N+350, 16N+700.

N → 2N → 2N+50 → 4N+100 → 4N+150 → 8N+300 → 8N+350 → 16N+700

3STEP 3

List Bernardo's four numbers

Bernardo's own outputs are 2N, 4N+100, 8N+300, 16N+700 — the last is largest, so it hits the window at the smallest N.

Bernardo's outputs: 2N, 4N+100, 8N+300, 16N+700

4STEP 4

Solve for the smallest winning start

From 950 ≤ 16N + 700 ≤ 999 we get 250 ≤ 16N ≤ 299, so 15.625 ≤ N ≤ 18.6875 and the smallest integer is N = 16.

950 ≤ 16N+700 ≤ 999 → 250 ≤ 16N ≤ 299 → 15.625 ≤ N ≤ 18.6875 → N_min=16

5STEP 5

Replay the game to check

Replay N=16: 32, 82, 164, 214, 428, 478, 956, and Silvia would need 1006 — Bernardo wins. N=15 only reaches 940, so Silvia's 990 ends it.

N=16: 32,82,164,214,428,478,956; 956+50=1006 ≥ 1000 (B wins) | N=15: 990 (S wins)

6STEP 6

Add the digits

The smallest winning start is N = 16, and the question wants its digit sum: 1 + 6 = 7, choice (A).

N = 16 → 1 + 6 = 7 → (A)

Answer

The window logic and the algebra agree: N=16 gives Bernardo's final number 956, which lands in 950 ≤ B ≤ 999 so Silvia's +50 dies at 1006, and the direct replay confirms all earlier numbers stay legal. The neighbor N=15 produces 940, one short of the window, and Silvia wins — proof that 16 is the true minimum, not just a value that happens to work. Its digit sum 1+6=7 is exactly choice (A), and 7 is the smallest offered digit sum, matching the fact that a small smallest-N should have a small digit total.

💡Key takeaway

Work backwards from the finish: Bernardo wins when his number lands in 950–999, and turning the doubling-and-adding into one expression 16N+700 shows the smallest start is 16, whose digits add to 7.

Translate the win into a number window
Name the start and track the moves
List Bernardo's four numbers
Solve for the smallest winning start
Replay the game to check
Add the digits