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

Consider a decreasing sequence of $n$ positive integers
$x_1 > x_2 > \cdots > x_n$
that satisfies the following conditions:

The average of the first $3$ terms in the sequence is $2025$ .
For all $4 \le k \le n$ , the average of the first $k$ terms is $1$ less than the average of the first $k-1$ terms.

What is the greatest possible value of $n$ ?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: We have a strictly decreasing list of positive whole numbers x_1 > x_2 > ... > 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.

Givens: The sequence is strictly decreasing: x_1 > x_2 > ... > x_n.; Every term is a positive integer.; The average of x_1, x_2, x_3 is 2025.; For each k from 4 to n, the average of the first k terms equals the average of the first k-1 terms minus 1.

Unknowns: The greatest possible number of terms, n.

Understand

Plan

Primary tool: #14 Extreme Principle

Secondary: #5 Look for a Pattern, #13 Convert to Algebra, #6 Guess and Check

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.

Execute — Answer: B

#13 Convert to Algebra 6.SP.B.5 Step 1

Turn averages into sums

An average is a sum divided by how many numbers there are, so sum = average times count.
The first 3 terms average 2025, so their sum is S_3 = 3 * 2025 = 6075.

$$S_3 = 3 \times 2025 = 6075$$

💡 An average is just the total shared out equally, so multiplying the average by the count gives the total back.

#5 Look for a Pattern 6.SP.A.3 Step 2

Write the sum of k terms

The average of the first 3 terms is 2025, and each later average is 1 smaller.
So the average of the first k terms is 2025 minus (k - 3), which simplifies to 2028 - k.
Multiplying by k gives the sum of the first k terms: S_k = k(2028 - k).

$$\text{avg of first } k = 2028 - k, \qquad S_k = k(2028 - k)$$

💡 Each step lowers the running average by exactly 1, so the average is a simple straight-line count downward from 2025.

#13 Convert to Algebra 6.EE.A.3 Step 3

Recover a single term

Any one term is the change in the running sum: x_k = S_k - S_{k-1}.
Subtracting k(2028 - k) minus (k-1)(2029 - k) and simplifying gives x_k = 2029 - 2k.
Check it: x_4 = 2029 - 8 = 2021 and x_5 = 2029 - 10 = 2019, so each new term is 2 less than the one before.
That keeps the list strictly decreasing.

$$x_k = S_k - S_{k-1} = k(2028-k) - (k-1)(2029-k) = 2029 - 2k$$

💡 The gap between two running totals is exactly the newest number added, so subtracting the totals uncovers that term.

#14 Extreme Principle 6.EE.B.5 Step 4

Push to the boundary

Every term must be a positive integer, so we need x_n > 0.
Using x_n = 2029 - 2n, set 2029 - 2n > 0, which gives n < 1014.5.
The largest whole number that fits is n = 1014, where x_1014 = 2029 - 2028 = 1 (still positive).
One step further, x_1015 = 2029 - 2030 = -1, which is not allowed.

$$2029 - 2n > 0 \;\Rightarrow\; n < 1014.5 \;\Rightarrow\; n \le 1014$$

💡 The list can only grow as long as the shrinking terms stay above zero; the last positive term marks the hard stop.

#6 Guess and Check 6.SP.B.5 Step 5

Confirm the list works

We still need three starting terms that average 2025 (sum 6075), stay above x_4 = 2021, and keep the list decreasing.
Choosing 2027, 2025, 2023 works: they sum to 6075, each is bigger than 2021, and 2023 > 2021 continues the decrease.
So a valid list of length 1014 truly exists, and the greatest possible value of n is 1014, which is answer (B).

$$2027 + 2025 + 2023 = 6075, \quad 2023 > 2021 = x_4$$

💡 A boundary answer only counts if you can actually build one example that reaches it.

[1] #13 6.SP.B.5 An average is a sum divided by how many numbers there are, so sum = average time

[2] #5 6.SP.A.3 The average of the first 3 terms is 2025, and each later average is 1 smaller. S

[3] #13 6.EE.A.3 Any one term is the change in the running sum: x_k = S_k - S_{k-1}. Subtracting

[4] #14 6.EE.B.5 Every term must be a positive integer, so we need x_n > 0. Using x_n = 2029 - 2n

[5] #6 6.SP.B.5 We still need three starting terms that average 2025 (sum 6075), stay above x_4

Review

Reasonableness: 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.

Alternative: Instead of the term formula, notice the running sum S_k = k(2028 - k) = 2028k - k^2 is a downward parabola peaking at k = 1014. The sum stops growing exactly where the added terms stop being positive, pointing to the same n = 1014.

CCSS standards used (min grade 6)

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures (Converting an average into a total (sum = average times count) for the first 3 and first k terms.)
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Reading the shrinking averages as a straight-line count to write the average of the first k terms as 2028 - k.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Simplifying S_k - S_{k-1} to the clean term formula x_k = 2029 - 2k.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Solving 2029 - 2n > 0 to find the largest n that keeps every term positive.)

⭐ 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.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Turn averages into sums

Write the sum of k terms

Recover a single term

Push to the boundary

Confirm the list works

Review

CCSS standards used (min grade 6)

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