2023/24 AHSMC Part 1 Problem 15

If $P(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + \cdots$ then $$\begin{align*} \frac{P(10)+P(-10)}{2} &= 179077 = a_0 + a_2 10^2 + a_4 10^4 + \cdots, \\ \frac{P(10)-P(-10)}{2} &= 55590 = a_1 10 + a_3 10^3 + a_5 10^5 + \cdots. \end{align*}$$ Thus $a_0 = 77$, $a_1 = 59$, $a_2 = 90$, $a_3 = 55$, $a_4 = 17$ and $a_n = 0$ for $n > 4$. Therefore $P(1) = 77+59+90+55+17 = 298$. The answer is (B).