2024 AMC 10 A Problem 25

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To help you master 2024 AMC 10 A Problem 25, 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

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2024 AMC10 A Problem 25

Solution

Solution 1

Inside-Outside Coloring and Run Classification Method

Represent the loop as the boundary between inside and outside cells. The middle-row condition then severely restricts the possible inside-outside pattern in that row, leaving four families that can be counted.

Assign status $1$ to a cell inside the loop and status $0$ to a cell outside. A shared side is covered precisely when the two cells have different statuses. Cells beyond the rectangular grid are outside and therefore have status $0$ .

Step 1: Encode the loop by inside and outside cells

Let the status of the middle cell in column $i$ be $x_{i}$ , and let the statuses of the cells above and below it be $u_{i}$ and $d_{i}$ . The imaginary middle-row cells just beyond the left and right ends are outside. Each disagreement across a side means that side is part of the loop.

x_{i}, u_{i}, d_{i} \in {0, 1}

x_{0} = x_{9} = 0

[u_{i} \neq = x_{i}] + [d_{i} \neq = x_{i}] + [x_{i - 1} \neq = x_{i}] + [x_{i + 1} \neq = x_{i}] = 1

Step 2: Determine what happens in a constant part of the middle row

Suppose a middle cell has the same status as both horizontal neighbors. Neither vertical side of that cell is on the loop, so exactly one of the cells above and below must have the opposite status. Thus a flat outside column contains exactly one inside cell, while a flat inside column contains the middle cell and exactly one of the top or bottom cells.

x_{i - 1} = x_{i} = x_{i + 1} ⟹ [u_{i} \neq = x_{i}] + [d_{i} \neq = x_{i}] = 1

x_{i} = 0 ⟹ (u_{i}, d_{i}) = (1, 0) or (0, 1)

x_{i} = 1 ⟹ (u_{i}, d_{i}) = (1, 0) or (0, 1)

Step 3: Determine what happens next to a change

If the middle-row status changes across one side of a cell, that side already supplies the cell’s one loop edge. Its top and bottom neighbors must therefore have the same status as the cell. Consequently, the outside endpoint of a change is an entirely outside column, while the inside endpoint is an entirely inside column. A middle cell cannot have changes on both horizontal sides because that would already give it two covered sides.

[x_{i - 1} \neq = x_{i}] + [x_{i} \neq = x_{i + 1}] = 1 ⟹ u_{i} = d_{i} = x_{i}

x_{i} = 0 ⟹ (u_{i}, x_{i}, d_{i}) = (0, 0, 0)

x_{i} = 1 ⟹ (u_{i}, x_{i}, d_{i}) = (1, 1, 1)

[x_{i - 1} \neq = x_{i}] + [x_{i} \neq = x_{i + 1}] \leq 1

Step 4: Classify all possible middle-row patterns

The cells inside a simple loop form one connected region. At a change between an outside run and an inside run, the outside endpoint is an empty column. If there were another outside column farther along that run, it would contain an inside top or bottom cell and would be separated from the main inside region by the empty column. That is impossible. Therefore an outside prefix or suffix can have length at most one. An internal outside run cannot occur: it cannot have length one because that middle cell would have two horizontal changes, and a longer run would disconnect the inside cells. Hence either every middle cell is outside, or there is one inside run with at most one outside cell at each end.

00000000

11111111

01111111

11111110

01111110

Step 5: Count the all-outside middle row

Each column must contain one inside cell, either in the top row or the bottom row. For all inside cells to be connected, every choice must use the same row. These give the loops surrounding the entire top row or the entire bottom row.

N_{00000000} = 2

Step 6: Count the all-inside middle row

The first and last columns are next to the outside of the board, so each is forced to be entirely inside. In each of the six remaining columns, either the top cell or the bottom cell joins the middle cell. These choices are independent.

N_{11111111} = 2^{6} = 64

Step 7: Count a single outside end

For a pattern with one outside cell at an end, that end column is entirely outside. The neighboring inside column and the inside column at the other boundary are entirely inside. The five inside columns between them each have two choices. The left-end and right-end cases are symmetric.

N_{01111111} = 2^{5} = 32

N_{11111110} = 2^{5} = 32

Step 8: Count outside cells at both ends

Both end columns are entirely outside, and their neighboring inside columns are entirely inside. Each of the four remaining middle columns has two choices for whether its top or bottom cell is included.

N_{01111110} = 2^{4} = 16

Step 9: Verify that every counted assignment gives one legal loop

In every counted pattern, the inside cells are connected: either they form one complete outer row, or they are joined by the consecutive inside middle cells. Every outside top or bottom cell touches the exterior through the corresponding outer edge, so no enclosed hole is created. The permitted column types also avoid a diagonal checkerboard meeting at a vertex, so the boundary does not cross or touch itself. Thus each assignment produces exactly one legal simple closed loop.

N = 2 + 64 + 32 + 32 + 16

N = 146

Correct answer: C. $146$
The number of toothpick arrangements is

146

, corresponding to choice

C

. As a check, the five classified middle-row patterns do not overlap, and their counts total

2 + 64 + 32 + 32 + 16 = 146

. Every legal pattern is included because any additional outside run would either give a numbered cell two covered sides or disconnect the region inside the loop.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Inside-outside boundary encoding:

A grid edge is part of the loop exactly when the cells on its two sides have different inside-outside statuses.

Use status $1$ for an inside cell and $0$ for an outside cell. A cell beyond the grid is outside. This applies because the toothpicks form a simple closed loop along grid edges. A simple closed loop separates the plane into an inside and an outside, and its edges are precisely the boundaries between those two regions.

One-boundary-side condition:

[u_{i} \neq = x_{i}] + [d_{i} \neq = x_{i}] + [x_{i - 1} \neq = x_{i}] + [x_{i + 1} \neq = x_{i}] = 1

$x_{i}$ is the status of middle cell $i$ , while $u_{i}$ and $d_{i}$ are the statuses above and below it. The notation $[P]$ equals $1$ when statement $P$ is true and $0$ otherwise. Each disagreement corresponds to one covered side of the numbered cell, and every numbered cell requires exactly one covered side.

Multiplication principle:

If

k

independent positions each have two choices, then they produce

2^{k}

assignments.

The choices must be independent, and every resulting assignment must satisfy the required conditions. Once the forced columns are fixed, each remaining inside middle-row column independently chooses whether its top or bottom cell is outside.

Addition principle for disjoint cases:

If valid objects are divided into non-overlapping, complete cases, the total is the sum of the case counts.

The cases must not overlap and must include every legal arrangement. The five middle-row status patterns are distinct, and the connectivity argument proves that no other pattern is possible.

Connectivity requirement for a simple loop:

The cells inside one simple closed loop must form one connected region and cannot be split into separate groups by an entirely outside column.

The loop is required to be a single non-self-intersecting closed curve. Separate inside groups would require separate boundary components or a disconnected interior, rather than one simple loop.

Key Takeaway: Remember This Strategy for Next Time

A grid problem describes one simple closed loop and gives the number of loop edges around cells.

Try: Color cells as inside or outside and translate covered edges into disagreements between neighboring statuses.

Why: The loop is exactly the boundary separating inside cells from outside cells.

A local condition creates an entirely empty or entirely full column at every status change.

Try: Use connectivity to rule out long or internal runs that would separate the inside region.

Why: A single simple loop must enclose one connected region.

After structural restrictions are found, several columns still have independent top-or-bottom choices.

Try: Count those choices with powers of two, then add the non-overlapping pattern cases.

Why: Each free column contributes two choices, while the classified cases cover all legal arrangements exactly once.

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2024 AMC 10 A Problem 25

In This Page:

2024 AMC10 A Problem 25