2021 AMC 10 B Problem 24

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To help you master 2021 AMC 10 B Problem 24, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

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2021 AMC10 B Problem 24

Solution

Solution 1

Sprague-Grundy and XOR Method

Replace each wall by a number that records the strategic behavior of that wall. Because gaps separate the bricks into independent walls, combine these numbers with XOR and look for the unique configuration having total XOR $0$.

For a finite game in which both players have the same moves and the last player to move wins, each position has a Sprague-Grundy value. The value is the smallest nonnegative integer missing from the values of all positions reachable in one move. Independent components combine by XOR, denoted $\oplus$. A total XOR of $0$ means the next player is in a losing position under perfect play.

Step 1: Define the value of one wall

Let the value assigned to a wall depend only on its length. An empty wall has value zero. For every positive length, examine every position reachable in one move and take the smallest nonnegative integer that does not occur among those resulting values.

$$g(0)=0$$

$$g(n)=\mathrm{mex}(R_n)$$

Step 2: Translate a removal into smaller walls

If one brick is removed, the remaining bricks form a left wall and a right wall whose lengths add to one less than the original length. If two adjacent bricks are removed, those lengths add to two less than the original length. The value of the resulting pair is the XOR of the two wall values.

$$R_n=\{g(i)\oplus g(n-1-i):0\le i\le n-1\}\cup \{g(i)\oplus g(n-2-i):0\le i\le n-2\}$$

Step 3: Compute the wall values through length six

Starting with the empty wall, use the recurrence one length at a time. The middle column lists all distinct values reachable by one legal move. The missing smallest nonnegative integer gives the value in the last column.

$$\begin{array}{ccc}n& R_n& g(n)\\ 0& -& 0\\ 1& \{0\}& 1\\ 2& \{0,1\}& 2\\ 3& \{0,1,2\}& 3\\ 4& \{0,2,3\}& 1\\ 5& \{0,1,2,3\}& 4\\ 6& \{0,1,2,4\}& 3\end{array}$$

$$g(1)=1,g(2)=2,g(3)=3,g(6)=3$$

Step 4: Evaluate the five starting configurations

The walls in each choice are independent, so XOR their individual values. Beth plays second, so she needs Arjun to begin with total value zero.

$$\text{A:}g(6)\oplus g(1)\oplus g(1)=3\oplus 1\oplus 1=3$$

$$\text{B:}g(6)\oplus g(2)\oplus g(1)=3\oplus 2\oplus 1=0$$

$$\text{C:}g(6)\oplus g(2)\oplus g(2)=3\oplus 2\oplus 2=3$$

$$\text{D:}g(6)\oplus g(3)\oplus g(1)=3\oplus 3\oplus 1=1$$

$$\text{E:}g(6)\oplus g(3)\oplus g(2)=3\oplus 3\oplus 2=2$$

Step 5: Explain why zero XOR guarantees Beth’s win

Choice B starts with zero XOR. Any move by Arjun changes exactly one component and makes the total XOR nonzero. The Sprague-Grundy theorem then guarantees that Beth can make a move restoring the XOR to zero. She repeats this response after every Arjun move. Since the game must eventually end, Beth makes the final move. As an additional check, every other choice lets Arjun move immediately to zero XOR.

$$\text{A:}(6,1,1)⟶(1,4,1,1),1\oplus 1\oplus 1\oplus 1=0$$

$$\text{C:}(6,2,2)⟶(1,4,2,2),1\oplus 1\oplus 2\oplus 2=0$$

$$\text{D:}(6,3,1)⟶(6,3),3\oplus 3=0$$

$$\text{E:}(6,3,2)⟶(6,3),3\oplus 3=0$$

Correct answer: B. $(6,2,1)$
Only choice B has total Sprague-Grundy XOR equal to $0$, so Arjun begins in a losing position and Beth can always return the game to XOR $0$. The displayed first moves for A, C, D, and E confirm that Arjun can win from each of those configurations. Therefore the answer is $\boxed{\text{B.}(6,2,1)}$.

Similar Problems in Real Exams

The following five AMC 10 B problems are closely related to AMC 10 B 2021 Question 24. These real AMC 10 B problems test the same key concept—Combinatorics. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2020 AMC 10 B Problem 15

2020 AMC 10 B Problem 16

2025 AMC 10 B Problem 16

2025 AMC 10 B Problem 21

2024 AMC 10 B Problem 12

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Minimum excluded value:

$S$ is the set of Sprague-Grundy values of all positions reachable in one move. The value of a wall is defined by applying this operation to all positions that can result from removing one brick or two adjacent bricks.

• Single-wall Sprague-Grundy recurrence:

using all legal integer values of $i$. $g(n)$ is the value of a wall of length $n$. In the first set, $0\le i\le n-1$; in the second set, $0\le i\le n-2$. The two sets represent removing one brick and two adjacent bricks, respectively. Every legal move leaves at most two walls, and the lengths of those walls determine the resulting game position.

• Sprague-Grundy XOR theorem:

The game must be finite, both players must have the same legal moves from any position, and the player making the last move must win. All walls follow the same removal rules, gaps make separate walls independent, and every move removes at least one brick, so the game is finite.

• Zero-XOR response strategy:

This strategy is guaranteed by the Sprague-Grundy theorem for a nonempty zero-XOR position. Choice B has initial total XOR $0$, so Beth can restore zero after every move by Arjun and eventually make the final move.

Key Takeaway: Remember This Strategy for Next Time

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