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2021 AMC 10B Problem 24

2021 AMC 10B Problem 24AMC 10 Step by Step

Name: Date: September 13, 2026

Games & Processes · difficulty 5 of 5 · about 10 min · position P21-25

Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move: $(3,2),(2,1,2),(4),(4,1),(2,2),$ or $(1,1,2).$

Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?

A)
$(6,1,1)$
B)
$(6,2,1)$
C)
$(6,2,2)$
D)
$(6,3,1)$
E)
$(6,3,2)$

Answer

Key insight

Each wall has a nim-value (mex of reachable values): walls 1–6 give 1,2,3,1,4,3; Beth wins exactly when the XOR is 0, only for (6,2,1).

Solution

This is a sum of independent games (the walls), so use nim-values. Assign each wall size $k$ a value $g(k)$ : the smallest nonnegative integer not among the values of positions reachable from a single wall of $k$ , where a multi-wall position's value is the XOR ( $\oplus$ ) of its walls' values. The player to move loses exactly when the total XOR is $0$ .

Compute, listing the positions reachable from one wall:

$k$	reachable positions	their values	$g(k)$
1	$()$	$0$	$1$
2	$(1), ()$	$1, 0$	$2$
3	$(2), (1,1), (1)$	$2, 0, 1$	$3$
4	$(3), (1,2), (2), (1,1)$	$3, 3, 2, 0$	$1$
5	$(4), (1,3), (2,2), (3), (1,2)$	$1, 2, 0, 3, 3$	$4$
6	$(5), (1,4), (2,3), (4), (1,3), (2,2)$	$4, 0, 1, 1, 2, 0$	$3$

Beth, the second player, wins when the starting XOR is $0$ :

$(6,1,1)$ : $3 \oplus 1 \oplus 1 = 3$ .
$(6,2,1)$ : $3 \oplus 2 \oplus 1 = 0$ . Beth wins.
$(6,2,2)$ : $3 \oplus 2 \oplus 2 = 3$ .
$(6,3,1)$ : $3 \oplus 3 \oplus 1 = 1$ .
$(6,3,2)$ : $3 \oplus 3 \oplus 2 = 2$ .

Only $(6,2,1)$ is a losing position for the first player.

The answer is $\boxed{\textbf{(B)}\ (6,2,1)}$ .

Why this works

A position is losing for the mover if every move leads to a winning position, and winning if some move leads to a losing one. For games that split into independent piles, the Sprague–Grundy theorem packages this recursion: each pile gets a "mex" value and piles combine by XOR, with XOR $0$ meaning the mover loses. The mirror strategy on symmetric positions like $(2,2)$ is the special case "equal values XOR to $0$ ." Learn to compute small nim-value tables; they settle many take-away games in minutes.

Alternative approach

Without nim-values one can verify (B) by hand: every Arjun move from $(6,2,1)$ has a Beth reply into a known losing position, using that mirror-image positions and $(3,2,1)$ and $(4,1)$ are losing for the mover. For instance $(5,2,1) \to (3,2,1)$ , $(4,2,1) \to (4,1)$ , $(6,1,1) \to (1,4,1,1)$ , $(6,2) \to (1,3,2)$ , $(6,1) \to (4,1)$ , $(1,4,2,1) \to (1,2,2,1)$ , $(2,3,2,1) \to (2,1,2,1)$ . Quick eliminations: in $(6,2,2)$ Arjun removes the middle two bricks of the $6$ to leave $(2,2,2,2)$ and mirrors.

The trap

Treating the game as ordinary Nim (using wall sizes 6, 2, 1 directly) even though removing a middle brick splits a wall into two.

Common mistakes

Treating the game as ordinary Nim (using wall sizes 6, 2, 1 directly) even though removing a middle brick splits a wall into two.
Forgetting the split moves when listing what a wall of $4$ , $5$ or $6$ can reach, which corrupts the nim-value table (e.g. getting $g(6) \ne 3$ ).

Techniques

Define states/recurrence and iterate · Exploit symmetry to reduce work or pair up objects