AMC 10 · 2012 · #20
Grade 7 arithmeticBernardo 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?
Pick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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; they keep alternating. A move is legal only while the number stays below $1000$. The person who makes the last legal number (the last one under $1000$) wins. Find the smallest starting $N$ for which Bernardo wins, then add up the digits of that $N$.
Givens: Start value $N$ is an integer with $0 \le N \le 999$; Bernardo's move: double the number he receives; Silvia's move: add $50$ to the number she receives; Bernardo moves first; they alternate; The winner is the last person to produce a number less than $1000$; Answer choices: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $11$
Unknowns: The smallest starting value $N$ that makes Bernardo the winner; The sum of the digits of that $N$
Understand
Restated: 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; they keep alternating. A move is legal only while the number stays below $1000$. The person who makes the last legal number (the last one under $1000$) wins. Find the smallest starting $N$ for which Bernardo wins, then add up the digits of that $N$.
Givens: Start value $N$ is an integer with $0 \le N \le 999$; Bernardo's move: double the number he receives; Silvia's move: add $50$ to the number she receives; Bernardo moves first; they alternate; The winner is the last person to produce a number less than $1000$; Answer choices: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $11$
Plan
Primary tool: #11 Work Backwards
Secondary: #4 Introduce a Variable, #14 Extreme Principle, #6 Guess and Check
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 $N=15$.
Execute — Answer: A
5.OA.A.2 Step 1 Translate the win into a number window
- Read the finish line backwards.
- Bernardo wins if he is the last to stay under $1000$.
- Right after Bernardo speaks a legal number $B < 1000$, it is Silvia's turn and she must add $50$.
- If $B + 50 \ge 1000$, Silvia cannot make a legal move, so Bernardo's number was the last one — Bernardo wins.
- That happens exactly when $B \ge 950$.
- Combined with $B < 1000$, Bernardo's winning move must land in the window $950 \le B \le 999$.
💡 Bernardo wins by handing Silvia a number too big for her $+50$ to survive, so his last number must sit in the top $50$-slot below $1000$.
6.EE.B.6 Step 2 Name the start and track the moves
- Let $N$ be the starting integer.
- Follow one full cycle: Bernardo doubles to $2N$; Silvia adds $50$ to get $2N+50$; Bernardo doubles that to $4N+100$; Silvia makes $4N+150$; Bernardo doubles to $8N+300$; Silvia makes $8N+350$; Bernardo doubles to $16N+700$.
- Writing the moves as expressions in $N$ keeps the whole alternating game in view at once.
💡 Using a letter for the unknown start lets one line of algebra stand in for the entire back-and-forth.
6.EE.A.2 Step 3 List Bernardo's four numbers
- Pull out just the numbers Bernardo produces, in order: $2N$, $4N+100$, $8N+300$, and $16N+700$.
- Each time the coefficient of $N$ doubles, so the *last* output $16N+700$ is the largest and reaches the $950$ window for the smallest $N$.
- To make $N$ as small as possible, we want that final output $16N+700$ to be Bernardo's winning move.
💡 The output with the biggest multiplier on $N$ climbs into the winning window first, so it gives the smallest possible start.
7.EE.B.4 Step 4 Solve for the smallest winning start
- Set Bernardo's last output inside the winning window: $950 \le 16N + 700 \le 999$.
- Subtract $700$: $250 \le 16N \le 299$.
- Divide by $16$: $15.625 \le N \le 18.6875$.
- The whole numbers in that range are $16, 17, 18$, and by the Extreme Principle the smallest is $N = 16$.
💡 Solving the compound inequality shows the winning starts are a short block of integers, and the smallest end of that block is the answer.
4.OA.A.3 Step 5 Replay the game to check
- Play $N=16$ for real: $16 \to 32 \to 82 \to 164 \to 214 \to 428 \to 478 \to 956$.
- All eight numbers are under $1000$, and Bernardo's $956$ forces Silvia to $956+50 = 1006 \ge 1000$, so Bernardo wins.
- Check $N=15$ fails: it leads to $16(15)+700 = 940 < 950$, then Silvia makes $990$, and Bernardo's double $1980 \ge 1000$ — Silvia's $990$ was last, so Silvia wins.
- So $16$ really is the smallest winning start.
💡 Actually walking the game for $16$ and for $15$ confirms $16$ wins and nothing smaller does.
4.NBT.B.4 Step 6 Add the digits
- The smallest winning start is $N = 16$.
- The question asks for the sum of its digits: $1 + 6 = 7$.
- So the answer is $(\text{A})\ 7$.
💡 The problem's final ask is just the digit sum of the number we found, not the number itself.
5.OA.A.2 Read the finish line backwards. Bernardo wins if he is the last to stay under $1 6.EE.B.6 Let $N$ be the starting integer. Follow one full cycle: Bernardo doubles to $2N$ 6.EE.A.2 Pull out just the numbers Bernardo produces, in order: $2N$, $4N+100$, $8N+300$, 7.EE.B.4 Set Bernardo's last output inside the winning window: $950 \le 16N + 700 \le 999 4.OA.A.3 Play $N=16$ for real: $16 \to 32 \to 82 \to 164 \to 214 \to 428 \to 478 \to 956$ 4.NBT.B.4 The smallest winning start is $N = 16$. The question asks for the sum of its dig Review
Reasonableness: The window logic and the algebra agree: $N=16$ gives Bernardo's final number $956$, which lands in $950\le B\le 999$ 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.
Alternative: Pure Tool #11 (Work Backwards) with no forward algebra: start from Bernardo's winning number in $[950,999]$ and peel off moves. Undo Bernardo's doubling (halve): the number Silvia handed him was $[475,499.5]$. Undo Silvia's $+50$: Bernardo's prior output was $[425,449.5]$. Halve again: $[212.5,224.75]$; undo $+50$: $[162.5,174.75]$; halve: $[81.25,87.375]$; undo $+50$: $[31.25,37.375]$; halve one last time to reach the start: $N \in [15.625,18.6875]$. The smallest integer start is $N=16$ — the same answer, reached entirely by unwinding the game from its end.
CCSS standards used (min grade 7)
5.OA.A.2Write simple expressions that record calculations with numbers (Recording the win condition as a number window: Bernardo's last move must satisfy $950 \le B \le 999$ so Silvia's $+50$ reaches $1000$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Letting $N$ stand for the unknown start and writing each alternating move as an expression in $N$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Listing Bernardo's four outputs $2N,\,4N+100,\,8N+300,\,16N+700$ and comparing their sizes.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Solving the compound inequality $950 \le 16N+700 \le 999$ to find the integer starts $16,17,18$ and the smallest, $N=16$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Replaying the game step by step for $N=16$ (win) and $N=15$ (loss) to confirm the minimum.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the digits of $16$ to get the final answer $1+6=7$.)
⭐ 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 $7$.
⭐ 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 $7$.
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