2026 AMC 8 Problem 22: Solution

Work through 2026 AMC 8 Problem 22 on medians and optimization with a written solution, hints, and a visual explanation.

Number TheoryLevel 5 · Final stretch

The integers from 1 through 25 are arbitrarily separated into five groups of 5 numbers each. The median of each group is identified. Let M equal the median of the five medians. What is the least possible value of M?

  • A. 9
  • B. 10
  • C. 12
  • D. 13
  • E. 14
Hints
  1. In any group of 5 numbers, exactly 3 of them are less than or equal to that group's median.
  2. Think about how many numbers total must be ≤ M if M is to be the median of the five medians — how many groups need a small median, and how many numbers does each contribute?
  3. Once you have a lower bound for M, try to actually construct five groups of 1–25 that achieve it.
Read the step-by-step solution

Answer: A · 9

  1. Lower bound: M ≥ 9

    In a group of 5, three numbers are ≤ its median. For M to be the median of the five medians, at least three groups must have median ≤ M. Those three groups contribute 3 × 3 = 9 distinct numbers that are ≤ M, so M ≥ 9.

    3 × 3 = 9 ⇒ M ≥ 9

  2. Construct groups reaching M = 9

    Pair small numbers with the largest ones to keep three medians small: {1,2,3,24,25}, {4,5,6,22,23}, {7,8,9,20,21}, {10,11,12,13,14}, {15,16,17,18,19}. Exactly the 9 numbers 1–9 land in the three small-median groups.

  3. Take the median of the medians

    The five medians are 3, 6, 9, 12, 17. Sorted, the middle one is 9, so M = 9 — matching the bound. The least possible value is 9.

    median(3, 6, 9, 12, 17) = 9

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