2025 AMC 8 Problem 15

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To help you master 2025 AMC 8 Problem 15, 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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2025 AMC8 Problem 15

Solution

Solution 1

Algebraic Modeling Method

By translating the problem into a system of equations, we can find a direct relationship between the different types of pairs. This reveals exactly how maximizing one type limits the others.

Step 1: Identify the total number of squares and pairs

The grid is $6 \times 6$ , which means it has a total of 36 unit squares. We are given that 13 are silver, leaving the rest to be gold. When folded in half, the 36 squares merge into pairs.

36 - 13 = 23 gold squares

36 \div 2 = 18 total pairs

Step 2: Set up variables for the different types of pairs

Every overlapping pair must fall into one of three categories based on the colors of its two squares.

x = number of Silver-Silver pairs

y = number of Gold-Gold pairs

z = number of Silver-Gold pairs

Step 3: Translate the conditions into mathematical equations

We can write three equations representing the total pairs, the total silver squares, and the total gold squares. Notice that a Silver-Silver pair uses 2 silver squares, while a Silver-Gold pair uses only 1.

x + y + z = 18 (Total pairs)

2 x + z = 13 (Total silver squares)

2 y + z = 23 (Total gold squares)

Step 4: Find a direct relationship for the Gold-Gold pairs

We want to find the possible values for $y$ . We can eliminate $z$ and express $y$ entirely in terms of $x$ by subtracting the total silver squares equation from the total pairs equation.

(x + y + z) - (2 x + z) = 18 - 13

y - x = 5

y = x + 5

Step 5: Determine the least possible number of Gold-Gold pairs ( $m$ )

Since $x$ is a physical count of pairs, it cannot be negative ( $x \geq 0$ ). To make $y$ as small as possible, we should make $x$ as small as possible. The smallest integer value for $x$ is 0. We can check that if $x = 0$ , then $z = 13$ , which is valid.

m = 0 + 5 = 5

Step 6: Determine the greatest possible number of Gold-Gold pairs ( $M$ )

To make $y$ as large as possible, we must make $x$ as large as possible. However, the number of Silver-Gold pairs ( $z$ ) must also be zero or greater. We can use the silver squares equation to find the maximum limit for $x$ .

z = 13 - 2 x \geq 0

2 x \leq 13

x \leq 6.5

Step 7: Calculate the greatest value and the final answer

Since $x$ must be a whole number, the maximum possible value for $x$ is 6. Plugging this into our equation for $y$ gives the maximum number of Gold-Gold pairs. Finally, we add $m$ and $M$ together.

M = 6 + 5 = 11

m + M = 5 + 11 = 16

Correct answer: C. $16$
The value of

m + M

is 16, matching answer choice C. We can verify the extreme case

M = 11

: 11 Gold-Gold pairs use 22 gold squares. The remaining 1 gold square forms 1 Silver-Gold pair, leaving exactly 12 silver squares to form 6 Silver-Silver pairs. The total is

11 + 1 + 6 = 18

pairs, perfectly using all given resources.

Solution 2

Greedy Pairing Method

Instead of algebra, we can use a logical ‘greedy’ strategy. To maximize Gold-Gold pairs, we group as many gold squares together as possible. To minimize them, we force gold squares to pair with silver squares.

Step 1: Find the greatest number of Gold-Gold pairs ( $M$ )

To maximize Gold-Gold pairs, we greedily pair gold squares with each other. Each Gold-Gold pair requires 2 gold squares. With 23 gold squares available, we divide by 2 to find the maximum possible pairs.

23 \div 2 = 11 pairs with a remainder of 1 gold square

Step 2: Verify the maximum case works

If we form 11 Gold-Gold pairs, we use 22 gold squares. The 1 left-over gold square must pair with a silver square, forming 1 Silver-Gold pair. The remaining $13 - 1 = 12$ silver squares can form 6 Silver-Silver pairs. The total number of pairs is exactly 18, so 11 is a valid maximum.

M = 11

Step 3: Find the least number of Gold-Gold pairs ( $m$ )

To minimize Gold-Gold pairs, we must greedily use up gold squares by pairing them with silver squares instead. There are 13 silver squares, so we can form a maximum of 13 Silver-Gold pairs. This consumes 13 gold squares. The remaining gold squares have no silver partners left, so they are forced to pair with each other.

23 - 13 = 10 remaining gold squares

10 \div 2 = 5 Gold-Gold pairs

Step 4: Verify the minimum case works and find the sum

We formed 13 Silver-Gold pairs and 5 Gold-Gold pairs. This perfectly accounts for all 18 pairs and uses all 13 silver and 23 gold squares. Therefore, 5 is a valid minimum. Finally, calculate the sum $m + M$ .

m = 5

m + M = 5 + 11 = 16

Correct answer: C. $16$
The sum is 16. Since we successfully built valid grids for both extreme cases without violating any rules, we know the bounds are exact. This confirms answer choice C.

Adapted Problems to Practice

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What Formulas Do You Need for This Problem?

System of Linear Equations for Counting:

Assign variables to mutually exclusive categories (like the three possible types of pairs) and use the given totals (total pairs, total squares) to create independent equations.

Applies when a whole group is divided into non-overlapping subgroups, and you know properties (like size or value) of those subgroups. In this problem, every folded pair is exactly one of three types, and every unit square belongs to exactly one pair, allowing us to perfectly model the constraints using linear equations.

Extreme Value Construction (Greedy Algorithm):

To find the maximum or minimum possible value of an arrangement, push the assignments as far as possible in the desired direction until you run out of resources, then verify the remaining items can form a valid state.

Useful for finding the 'least', 'greatest', 'minimum', or 'maximum' in discrete combinatorial problems. We can directly find $M$ by grouping as many gold squares together as possible, and $m$ by grouping gold and silver squares together as much as possible, provided we verify the leftover squares can still be paired.

Key Takeaway: Remember This Strategy for Next Time

A fixed pool of items (squares) is being divided into different categories of groups (pairs).

Try: Define variables for the number of groups in each category and write equations for the total items and the total groups.

Why: Writing equations organizes all the constraints automatically. You can then substitute one variable into another to clearly see what limits the minimum and maximum values.

A combinatorial problem asks for the 'greatest' or 'least' possible count of something.

Try: Try a greedy logical approach first. To maximize a combination, make as many of them as mathematically possible until you run out of a resource. Then, verify that the leftovers still satisfy all rules.

Why: Discrete pairing is a zero-sum game. Maximizing one type of pair naturally minimizes the alternatives. As long as the leftover pieces fit perfectly into the remaining empty slots, the extreme bound is achievable.

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2025 AMC 8 Problem 15

In This Page:

2025 AMC8 Problem 15