Consider polynomials $P(x)$ of degree at most 3, each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}.$ How many such polynomials satisfy $P(-1) = -9$ ?

286
165
110
220
143

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Let $P(x) = ax^{3} + bx^{2} + cx + d$ where $a$ , $b$ , $c$ , and $d$ are elements of the set $\{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ . By direct substitution, $P(-1) = -9 = -a +b -c +d$ . Rearrangement leads to $b + d + 9 = a + c$ . This is the constraint that must be satisfied in order to find the correct number of polynomials. We also know that the coefficients must remain in the aforementioned subset. Thus, the maximum value of the expression is $18$ (by maximizing $c + d$ ) and the minimum value is $9$ (by letting $a = b = 0$ ). We have now reduced the problem to finding the number of ways we can satisfy $b + d + 9 = a + c$ within the range $9 \leq a + c \leq 18$ .

We can now do some casework.

Case $1$ : $a + c = 18; a = 9, c= 9$

$b + d = (0, 9), (1, 8), (2, 7) \cdots (7, 2), (8, 1), (9, 0) \implies 10$ possible polynomials.

Case $2$ : $a + c = 17; a = 9, c = 8$ OR $a = 8, c = 9$

$b + d = (0, 8), (1, 7), (2, 6) \cdots (6, 2), (7, 1), (8, 0) \implies 9 * 2 = 18$ possible polynomials.

Case $3$ : $a + c = 16; a = 9, c = 7$ OR $a = 8, c = 8$ OR $a = 7, c = 9$

$b + d = (0, 7), (1, 6), (2, 5) \cdots (5, 2), (6, 1), (7, 0) \implies 8 * 3 = 24$ possible polynomials.

Clearly, we can observe a pattern. Expressing this in summation notation,

$\begin{aligned} & \sum_{n=1}^{10}(11-n)(n) \\ = & \sum_{n=1}^{10} 11n - n^{2} \\ = & 11\sum_{n=1}^{10} n - \sum_{n=1}^{10} n^{2} \\ = & (11)\frac{(10)(11)}{2} - \frac{(10)(11)(21)}{6} \\ = & 605 - 385 = \boxed{220} \end{aligned}$