Let m ≥ 5 be an odd integer, and let D(m) denote the number of quadruples (a_1, a_2, a_3, a_4) of distinct integers with 1 ≤ a_i ≤ m for all i such that m divides a_1 + a_2 + a_3 + a_4. There is a polynomial q(x) = c_3x^3 + c_2x^2 + c_1x + c_0 such that D(m) = q(m) for all odd integers m ≥ 5. What is c_1?

2021 AMC 12A Fall Problem 25 — Solution

2021 AMC 12A Fall Problem 25

Below is the professionally curated solution for Problem 25 of the 2021 AMC 12A Fall, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Fall solutions, or check the answer key.

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Concepts:modular arithmetic roots of unity polynomial

Difficulty rating: 2650

25.

Let $m \ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1 \le a_i \le m$ for all $i$ such that $m$ divides $a_1 + a_2 + a_3 + a_4.$ There is a polynomial $q(x) = c_3x^3 + c_2x^2 + c_1x + c_0$ such that $D(m) = q(m)$ for all odd integers $m \ge 5.$ What is $c_1?$

$-6$

$-1$

$4$

$6$

$11$

Solution:

Counting ordered quadruples of distinct residues with sum $\equiv 0 \pmod m$ (via a roots-of-unity filter, using that $m$ is odd) gives $D(m) = (m - 1)(m - 2)(m - 3).$ Direct computation confirms $D(5) = 24,$ $D(7) = 120,$ $D(9) = 336,$ matching this cubic.

Expanding, $D(m) = m^3 - 6m^2 + 11m - 6,$ so $c_1 = 11.$

Thus, the correct answer is E.

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