2025 AMC 8 Problem 21

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To help you master 2025 AMC 8 Problem 21, 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

Want more practice? Scroll down to download Think Academy’s AMC 8 practice packs and keep building your AMC 8 skills.

2025 AMC8 Problem 21

Solution

Solution 1

Extreme Values and Local Constraints Method

Since consecutive grades cannot be on connected pods, we can identify which pods are missing the most connections. The extreme grades (1 and 7) only need one missing connection, which helps us place them immediately.

Step 1: Interpret the connection rule

The problem states that connected pods must have grades that differ by 2 or more. This means that if two grades differ by exactly 1 (they are consecutive), the pods holding them cannot have a walkway between them.

$$|x-y|=1⟹\text{No Walkway}$$

Step 2: Find the missing walkways

Let’s count how many missing walkways each pod has by looking at the diagram. Pod C has 5 walkways (to A, B, D, E, and F); the only pod it does not connect to is G. Pod F has 5 walkways (to A, B, C, E, and G); the only pod it does not connect to is D.

$$C\text{misses}G$$

$$F\text{misses}D$$

Step 3: Place the extreme grades

Any middle grade (2, 3, 4, 5, or 6) must not connect to the grade right below it AND the grade right above it. This requires at least two missing walkways. Because C and F only have one missing walkway each, they cannot hold a middle grade. They must hold the extreme grades.

$$C,F\in \{1,7\}$$

$$C+F=8$$

Step 4: Place the adjacent grades

The grade 1 cannot connect to 2. Whichever pod gets 1 (either C or F) must use its single missing walkway for 2. Therefore, the pods missing from C and F must hold the grades 2 and 6.

$$G,D\in \{2,6\}$$

Step 5: Use local connections to place the remaining grades

We have three pods left (A, B, and E) and three grades left. Looking at the diagram, pods A and B have a walkway connecting them. Thus, their grades cannot be consecutive.

$$\{A,B,E\}=\{3,4,5\}$$

Step 6: Determine E

The only way to choose two non-consecutive grades from our remaining options for A and B is to pick the smallest and largest available. This leaves exactly one grade remaining for pod E.

$$\{A,B\}=\{3,5\}$$

$$E=4$$

Step 7: Calculate the final sum

We now have the grades for all three requested pods, allowing us to find their sum.

$$C+E+F=1+4+7=12$$

Correct answer: A. $12$
The sum is 12, which matches answer choice A. We can check the full assignment (C=1, G=2, B=3, E=4, A=5, D=6, F=7) and verify that every connected pair in the diagram differs by at least 2.

Solution 2

The Missing Walkways Path Method (Complement Graph)

Instead of focusing on the walkways that exist, we can draw a new path using only the ‘missing’ walkways. The sequence of consecutive grades 1 through 7 must form a continuous chain along these missing connections.

Step 1: Translate to a path of missing walkways

Because consecutive grades must not be connected by a walkway, the sequence of grades from 1 to 7 must form a continuous path of ‘missing walkways’ passing through all seven pods.

Step 2: Identify the endpoints of the path

As found earlier, C and F each have exactly one missing walkway. In a path, the endpoints are the only locations with just one connection. Therefore, the path of consecutive grades must start at C and end at F (or vice versa).

$$\text{Path endpoints:}C\text{and}F$$

Step 3: Trace the path from the endpoints

Let’s trace the path from F. Since F only misses D, the path must go from F to D. Next, we look at pod A. The diagram shows A misses only D and E. Since D is already next to F, A must connect to D on one side and E on the other.

$$F\text{connects to}D$$

$$A\text{connects to}D\text{and}E$$

$$\text{Partial path:}E-A-D-F$$

Step 4: Identify the exact middle of the path

The entire path has 7 pods. Because the path ends with E-A-D-F, we can count backwards from F (which is position 7) to find the position of E. The sequence is position 7 (F), position 6 (D), position 5 (A), and position 4 (E).

$$F=7⟹D=6⟹A=5⟹E=4$$

Step 5: Sum the requested grades

We know the endpoints C and F hold the grades 1 and 7, and E holds the exact middle grade 4.

$$C+E+F=1+4+7=12$$

Correct answer: A. $12$
The sum is 12. Both tracing the exact sequence of grades and using logical constraints independently lead to C+F=8 and E=4.

Similar Problems in Real Exams

The following five AMC 8 problems are closely related to AMC 8 2025 Question 21. These real AMC 8 problems test the same key concept—Coloring, Tiling & Covering. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2025 AMC 8 Problem 11

2024 AMC 8 Problem 7

2011 AMC 8 Problem 3

2010 AMC 8 Problem 7

2009 AMC 8 Problem 18

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Complementary Graph (Missing Walkways):

Applies to restriction problems where items are separated by a minimum difference (e.g., difference $\ge 2$ means difference cannot be $1$). It transforms a negative rule ('cannot be next to') into a positive rule ('must form a path'), making it much easier to trace a sequence.

• Extreme Values Analysis:

Applies when assigning a sequence of integers $1,2,\dots ,n$ to constrained positions. Middle numbers conflict with two other numbers (e.g., 4 conflicts with 3 and 5), requiring two available spaces. Extreme numbers (1 and 7) only conflict with one, locking them into the most restricted positions.

• Pigeonhole Logic for Consecutive Integers:

Given set $\{n,n+1,n+2\}$, the only non-consecutive pair is $\{n,n+2\}$. Applied to pods A and B, which were connected and had to share the remaining grades 3, 4, and 5.

Key Takeaway: Remember This Strategy for Next Time

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