2026 AMC 8 Problem 22

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To help you master 2026 AMC 8 Problem 22, 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.

2026 AMC8 Problem 22

Solution

Solution 1

2D Grid and Bounding Method

By arranging the groups as rows in a sorted grid, we can easily count exactly how many numbers must be smaller than or equal to M.

To rigorously prove the minimum possible value of a quantity, you must accomplish two steps: 1. Bound: Prove logically that the value cannot be less than a certain limit. 2. Construction: Provide a specific, valid example where the value is exactly that limit.

Step 1: Arrange the groups in a sorted 5×5 grid

Imagine the 25 numbers arranged in a grid with 5 rows and 5 columns. Let each row represent one group of 5 numbers. Sort each row from smallest to largest, so the 3rd column contains the median of each group. Then, sort the rows from top to bottom so their medians are in ascending order.

Step 2: Locate M within the grid

The 3rd column contains the 5 medians. The variable M is the median of these 5 medians. Since the rows are sorted vertically by these medians, M is the 3rd median, which means it sits exactly in the 3rd row and 3rd column of our grid.

$$M=m_3$$

Step 3: Count how many numbers are forced to be less than or equal to M

Look at the top-left 3×3 section of our grid. In the 3rd row, the first 3 numbers are less than or equal to M. In the 2nd row, the median is less than M, so the first 3 numbers in the 2nd row are also strictly less than M. Similarly, the first 3 numbers in the 1st row are strictly less than M. This gives us a 3×3 block of numbers that must be less than or equal to M.

$$3\text{rows}\times 3\text{columns}=9\text{numbers}$$

Step 4: Determine the lower bound for M

Because all the integers are distinct and chosen from 1 to 25, the 9 numbers in our 3×3 block must be at least the smallest 9 positive integers (1 through 9). Since M is the largest number in this block, M must be at least 9.

$$M\ge 9$$

Step 5: Construct an example to prove M = 9 is achievable

To complete the proof, we must show that M = 9 is actually possible. We can place the numbers 1 through 9 in the 3×3 block, and fill the rest of the grid with the remaining numbers 10 through 25. Let the groups be: Group 1: 1, 2, 3, 20, 21 (median 3) Group 2: 4, 5, 6, 22, 23 (median 6) Group 3: 7, 8, 9, 24, 25 (median 9) Group 4: 10, 11, 12, 13, 14 (median 12) Group 5: 15, 16, 17, 18, 19 (median 17) The medians are 3, 6, 9, 12, and 17. Their median is indeed 9. All problem rules are satisfied.

Correct answer: A. 9
The least possible value of M is $\boxed{9}$ (Choice A). This passes the boundary-condition check because we systematically proved no smaller integer could be the 9th term, and we verified the answer with a complete, rule-abiding construction.

Solution 2

Multiple-Choice Construction Shortcut

Since 9 is the smallest answer choice available, if we can successfully build a valid configuration where M=9, we immediately know it must be the answer without having to prove a lower bound.

Step 1: Identify the strategy based on the answer choices

The smallest option given in the problem is 9. If we can divide the numbers 1 through 25 into 5 valid groups such that the median of medians is 9, then 9 must be the correct answer. We do not need to check 8 or lower.

Step 2: Plan the necessary medians

To make M = 9, we need two groups to have medians smaller than 9, and two groups to have medians larger than 9. We also need 9 itself to be a median. So we need three ‘small’ medians. Let’s make them 3, 6, and 9.

Step 3: Build the small halves of the first three groups

Each median must be greater than two other numbers in its group. To keep these three medians as small as possible, we use the absolute smallest integers available: 1 through 9. Group 1 begins: 1, 2, 3 Group 2 begins: 4, 5, 6 Group 3 begins: 7, 8, 9

Step 4: Build the remaining two complete groups

We still have two groups left to form, which require 10 numbers total. We can use the next sequence of numbers to make them valid. Group 4: 10, 11, 12, 13, 14 (median is 12) Group 5: 15, 16, 17, 18, 19 (median is 17)

Step 5: Complete the first three groups

We have exactly six numbers remaining: 20, 21, 22, 23, 24, 25. We need to add two large numbers to Groups 1, 2, and 3. Since these are all greater than our medians (3, 6, and 9), adding them will not change the medians. Group 1 gets 20, 21. Group 2 gets 22, 23. Group 3 gets 24, 25. Everything works perfectly, proving that M = 9 is achievable.

Correct answer: A. 9
The correct answer is $\boxed{9}$ (Choice A). Because we successfully matched the smallest answer choice with a valid construction, we can confidently conclude the problem without writing out an algebraic lower-bound proof.

Similar Problems in Real Exams

The following five AMC 8 problems are closely related to AMC 8 2026 Question 22. These real AMC 8 problems test the same key concept—Pigeonhole & Extremal Principles. 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 15

2023 AMC 8 Problem 14

2014 AMC 8 Problem 19

2005 AMC 8 Problem 16

2000 AMC 8 Problem 12

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Minimum/Maximum Proof Structure:

Required in "minimum", "maximum", "least", or "greatest" competition math problems. If you only find a bound, you don't know if the boundary value is actually possible. If you only provide an example, you don't know if an even smaller example exists.

• Median of a Finite List:

Applies to any finite set with an odd number of elements. The problem requires calculating the median of 5 numbers multiple times. In each case, exactly 2 numbers will be smaller than the median, and 2 will be larger.

Key Takeaway: Remember This Strategy for Next Time

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