Major Probability Question 2.

$\textbf{Lemma:}$ $\textit{If p is a prime number and}$ $a_0, a_1, ..., a_{p-1}$ $\textit{are rational numbers satisfying}$ $a_0 + a_1\epsilon + a_2\epsilon^2 + ... + a_{p-1}\epsilon^{p-1} = 0,$ $\textit{where}$ $\epsilon = e^{\frac{2\pi}{p}i},$ $\textit{that is the primitive root of unity, then }$ $a_0 = a_1 = a_2 = ... = a_{p-1}.$

$\textbf{Proof:}$

Observe that $a_0 + a_1x + a_2x^2 + ... + a_{p-1}x^{p-1}$ and $1+x + x^2 + ... + x^{p-1}$ are not relatively prime since they have a common root. Since the latter is irreducible over $\mathbb{Q}$ , $1 + x + x^2 + ... + x^{p-1}$ must divide $a_0 + a_1x + ... + a_{p-1}x^{p-1}$ . which can only happen if $a_0 = a_1 = ... = a_{p-1}$ . We have proven the lemma.

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Denote $a_k^{(j)}$ to be the number of $k$ element subsets such that the sum of the elements is congruent to $j$ modulo $3$ . We will also define $\epsilon = e^{\frac{2\pi}{3}i}.$

It is pretty clear that $\sum_{j = 0}^{2}a_k^{j}\epsilon^{j} = \sum_{1 \le c_1 < c_2 < ... < c_k \le 1986}\epsilon^{c_1 + c_2 + ... + c_k}$ since they are both counting the same thing. However, notice that the right hand side is the $x^{1986 - k}$ coefficient of $A = (x + \epsilon)(x + \epsilon^2)...(x + \epsilon^{1986}).$ We know that $x^3 + 1 = (x + \epsilon)(x + \epsilon^2)(x + \epsilon^3)$ , so we can greatly simplify the product $A = (x^3 + 1)^{662}$ Thus, we can find the $x^{1986 - k}$ coefficient. Notice that if the coefficient of the $x^{1986 - k}$ term is $t$ , then we know that by the lemma, one of the below cases must be fulfilled $\begin{aligned} a_k^{(0)} - t & = a_k^{(1)} = a_k^{(2)} \\ a_k^{(0)} & = a_k^{(1)} - t= a_k^{(2)} \\ a_k^{(0)} & = a_k^{(1)} = a_k^{(2)} - t \end{aligned}$ However, we want to find the $k$ where $a_k^{(0)} = a_k^{(1)} = a_k^{(2)}.$ Clearly, this occurs when the $x^{1986 - k}$ coefficient is zero. Now notice that in $A$ , there only exist terms with exponents divisible by three. That means that the coefficient is zero whenever $3 \nmid k$ and the only value that follows that condition among the answer choices is $\boxed{6002}$ .