AMC 10 · 2025 · #14

Grade 6 algebra
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
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

The first three sum to 6075.

S₃ = 3 × 2025 = 6075
2STEP 2

Write the sum of k terms

The first k sum to k times 2028 minus k.

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

Recover a single term

Differencing gives the term 2029 minus two k.

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

Push to the boundary

Positivity forces n at most 1014.

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

Confirm the list works

Such a list really exists, so 1014.

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