2025 AMC 10B Problem 17

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 $\boxed{\textbf{(B)}\ 1014}$.

Conditions stated about running averages are conditions about $S_k$, and once $S_k$ is a formula in $k$ the terms come back by a single difference. That is the whole problem: the constraint looks like a recursion on averages but is really a closed form. After that, "greatest possible $n$" is a positivity bound on the forced terms, with one piece of care at the boundary, since the first three terms are not determined and the answer is only valid if they can be chosen to fit above $x_4$. Here they can, with room to spare, so the bound is attained.

Avoid the sum formula by working with the averages directly. Writing $x_k = S_k - S_{k-1} = kA_k - (k-1)A_{k-1}$ and using $A_k = A_{k-1} - 1$ gives
$$x_k = k(A_{k-1} - 1) - (k-1)A_{k-1} = A_{k-1} - k = (2029 - k) - k = 2029 - 2k,$$
the same forced tail. This also makes the structure visible: each new term sits $k$ below the previous average, so the terms fall twice as fast as the averages do.

• Reading 'the average drops by 1' as 'each term drops by 1' and running the sequence down from around 2025 to 1, which points at choice (E).
• An arithmetic slip producing $x_k = 2028 - 2k$, which caps $n$ at $1013$, choice (A).
• Stopping at the bound $n \le 1014$ without checking that $x_1 > x_2 > x_3 > 2021$ can be arranged with sum $6075$; the bound is only an answer once it is achieved.