2025 AMC 10B Problem 17

P16-20~6 minPrint

Sequences & Series Number Properties Ratios, Percents & Averages Bound the quantity above/below or estimate to pin it down Set up the equation/formula and compute; no special trick needed Consider the largest/smallest element or boundary case

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

$\qquad\bullet$ The average (arithmetic mean) of the first $3$ terms in the sequence is $2025.$

$\qquad\bullet$ For all $4 \leq k \leq n,$ the average of the first $k$ terms in the sequence is $1$ less than the average of the first $k - 1$ terms in the sequence.

What is the greatest possible value of ${n?}$

Show solution without answering

Key insight

The average of the first k terms is 2028 − k, so their total is k(2028 − k). Subtracting totals shows later terms drop by 2.

Solution

Start with the numbers. The average of the first $3$ terms is $2025$ , so their total is $3 \times 2025 = 6075$ . Each later average is $1$ less than the one before.

First $k$ terms	Average	Total
$3$	$2025$	$6075$
$4$	$2024$	$8096$
$5$	$2023$	$10115$
$6$	$2022$	$12132$

2. Get each term from the totals. A new term is the new total minus the old total:
$x_4 = 8096 - 6075 = 2021, \quad x_5 = 10115 - 8096 = 2019, \quad x_6 = 12132 - 10115 = 2017.$
The terms drop by $2$ each time, not by $1$ .

3. Check the pattern in general. The average of the first $k$ terms is $2025 - (k - 3) = 2028 - k$ , so their total is $k(2028 - k)$ . Then
$x_k = k(2028 - k) - (k-1)(2029 - k).$
Multiplying out, $k(2028-k) = 2028k - k^2$ and $(k-1)(2029-k) = 2030k - k^2 - 2029$ . Subtracting,
$x_k = 2029 - 2k \qquad \text{for } k \ge 4.$
So from the fourth term on, every term is fixed: $2021, 2019, 2017, \ldots$

Upper limit. All terms must be positive whole numbers, so $2029 - 2n \ge 1$ , which gives $2n \le 2028$ and $n \le 1014$ . At $n = 1014$ the last term is exactly $1$ .

5. Can $n = 1014$ really happen? The first three terms are not fixed; they just need to add to $6075$ and be bigger than $x_4 = 2021$ , in decreasing order. For example, $2028 + 2025 + 2022 = 6075$ works. So the full sequence
$2028,\ 2025,\ 2022,\ 2021,\ 2019,\ 2017,\ \ldots,\ 3,\ 1$
is decreasing, positive, and follows both rules.

So the greatest possible $n$ is $1014$ .

The answer is $\boxed{\textbf{(B)}\ 1014}$ .

Why this works

A rule about averages is really a rule about totals, since total $=$ average $\times$ count. Once you have a formula for the total of the first $k$ terms, subtracting two neighboring totals gives each term. For a "greatest possible" answer, finish both halves: find the bound (terms must stay positive) and show the bound is reached (the first three terms can be chosen to fit).

Alternative approach

Think about what a new term does to the average. Adding $x_k$ to $k - 1$ terms with average $A$ gives a new average $A - 1$ only if the new term is $k$ less than the old average: $x_k = A - k$ . With $A = 2029 - k$ (the average of the first $k-1$ terms), this gives $x_k = 2029 - 2k$ again.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let $S_k = x_1 + \dots + x_k$ and $A_k = S_k / k$ . The conditions say $A_3 = 2025$ and $A_k = A_{k-1} - 1$ for $k \ge 4$ , so the averages drop by one at every step:
$A_k = 2025 - (k - 3) = 2028 - k \qquad (k \ge 3),$
and therefore
$S_k = k(2028 - k) \qquad (k \ge 3).$

Recover the individual terms. For $k \ge 4$ ,
$x_k = S_k - S_{k-1} = k(2028-k) - (k-1)\bigl(2029 - k\bigr).$
Expanding, $k(2028-k) = 2028k - k^2$ and $(k-1)(2029-k) = 2030k - k^2 - 2029$ , so
$x_k = 2029 - 2k.$

So from the fourth term on nothing is free: the sequence must read $x_4 = 2021$ , $x_5 = 2019$ , $x_6 = 2017, \dots$ , dropping by $2$ each time. These are automatically decreasing, but they must stay positive integers:
$2029 - 2n \ge 1 \iff n \le 1014.$

It remains to check that $n = 1014$ is achievable, which requires the first three terms to exist. They must satisfy $x_1 > x_2 > x_3 > x_4 = 2021$ and $x_1 + x_2 + x_3 = 3 \cdot 2025 = 6075$ . The smallest possible sum of three distinct integers greater than $2021$ is $2024 + 2023 + 2022 = 6069 \le 6075$ , so there is room; for instance
$2028,\ 2025,\ 2022$
works. With those and $x_k = 2029 - 2k$ for $4 \le k \le 1014$ , the sequence is strictly decreasing, entirely positive (the last term is $x_{1014} = 1$ ), and satisfies both conditions.

The answer is $\boxed{\textbf{(B)}\ 1014}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Number Theory, Chapter 1: Integers: The Basics
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)