Problem 22 · 2026 AMC 8
Stretch
Logic & Word Problems
extremalmedian
The integers 1 through 25 are arbitrarily separated into five groups of 5 numbers each. The median of each group is found, and M is the median of those five medians. What is the least possible value of M?
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Answer: A — 9.
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Hint 1 of 2
Unpack what M even is: it's the median of the five medians, i.e. the 3rd-smallest of them. To push M down, you need three groups whose medians are all small at once.
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Hint 2 of 2
Here's the bottleneck: a group's median needs two strictly smaller numbers sitting below it. Three small medians plus their six ‘below’ numbers eat up a lot of the tiny values — and tiny values are scarce. Count how many you'd need.
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Approach: lower-bound by counting scarce small numbers, then build a matching example
- First decode M: it is the 3rd-smallest of the five group medians. So making M tiny requires three groups to each have a small median simultaneously.
- Now the scarcity argument. Each of those three medians needs two numbers strictly below it inside its group. Suppose M ≤ 8. Then three medians (each ≤ 8) together with their six below-numbers (each smaller still) are 9 distinct values all ≤ 8 — impossible, since only 8 numbers (1–8) are that small. So M ≥ 9.
- Then show 9 is actually reachable: {1, 2, 7, 24, 25}, {3, 4, 8, 22, 23}, {5, 6, 9, 20, 21} have medians 7, 8, 9, while the last two groups {10–14} and {15–19} absorb the big numbers (medians 12, 17). The five medians 7, 8, 9, 12, 17 have median 9.
- Bound met by an example ⇒ the least value is 9.
- Why this transfers: extremal problems are won by pairing a lower bound (a counting/pigeonhole reason it can't be smaller) with a construction (an explicit example hitting that bound). Neither half alone is a proof — you need both jaws of the vise.
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