AMC 10 · 2025 · #17

Grade 6 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation
sequences-arithmeticmean-median-mode-rangelinear-equations-one-var convert-to-algebra ↑ Prerequisites: sequences-arithmetic
📏 Medium solution 💡 3 insights
Problem
We have a strictly decreasing list of positive whole numbers x₁ ＞ x₂ ＞ ... ＞ x_n. The average of the first 3 numbers is 2025. From the 4th number on, adding the next number drops the running average by exactly 1. We want the longest such list, that is, the biggest n.

Pick an answer.

(A)
1013
(B)
1014
(C)
1016
(D)
2016
(E)
2025

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

The question asks for the greatest n, so the answer sits at a boundary: the last spot where a term is still positive. That is the Extreme Principle. To find that boundary I first turn each average into a sum, spot the pattern that every term drops by 2, then push n until a term would go negative.

1STEP 1

Turn averages into sums

Sum equals average times count, so the first 3 terms give S₃ = 3 × 2025 = 6075.

S₃ = 3 × 2025 = 6075
2STEP 2

Write the sum of k terms

Each later average drops 1 below 2025, so the first k terms average 2028 - k and sum to k(2028 - k).

avg of first k = 2028 - k, S_k = k(2028 - k)
3STEP 3

Recover a single term

One term is the jump in the running sum, so S_k minus S_(k-1) simplifies to x_k = 2029 - 2k — terms fall by 2.

x_k = S_k - S_k-1 = k(2028-k) - (k-1)(2029-k) = 2029 - 2k
4STEP 4

Push to the boundary

Terms stay positive only while 2029 - 2n ＞ 0, that is n ＜ 1014.5, so n = 1014 and its last term is 1.

2029 - 2n ＞ 0 → n ＜ 1014.5 → n ≤ 1014
5STEP 5

Confirm the list works

Starting 2027, 2025, 2023 sums to 6075 and still beats x₄ = 2021, so a list of length 1014 really exists.

2027 + 2025 + 2023 = 6075, 2023 ＞ 2021 = x₄
Answer
1014
The terms follow x_k = 2029 - 2k, an arithmetic list going 2021, 2019, 2017, ... down by 2. It hits 1 exactly at k = 1014 and would turn negative at k = 1015, so 1014 is the natural cutoff. The far-larger options 2016 and 2025 ignore that the terms drop by 2 each step, which roughly halves how far you can go, so 1014 (near 2028/2) is the sensible size.
💡Key takeaway

Turn each average into a total, watch the terms fall by 2 each step, and stop counting the moment a term would drop below 1.

  • Turn averages into sums
  • Write the sum of k terms
  • Recover a single term
  • Push to the boundary
  • Confirm the list works

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