If $d_0=0$ then $x \mid d_9x^9+\cdots+d_0$ If $d_k=d_{9-k}$ for all $k$ then $x+1 \mid d_9x^9+\cdots+d_0$ For everything else, we can pair $\overline{d_9\ldots d_0}$ with $\overline{d_0\ldots d_9}$ as a polynomial is irreducible iff its reverse is.

IAmTheHazard wrote: If $d_0=0$ then $x \mid d_9x^9+\cdots+d_0$ If $d_k=d_{9-k}$ for all $k$ then $x+1 \mid d_9x^9+\cdots+d_0$ For everything else, we can pair $\overline{d_9\ldots d_0}$ with $\overline{d_0\ldots d_9}$ as a polynomial is irreducible iff its reverse is. felt a bit easy for a p3 but i enjoyed the quick solution

ooks, again with the difficulty

If $d_0 = 0$, then the polynomial is reducible (divisible by $x$) If $d_9, d_9, \dots d_0$ is palindromic, then $-1$ is a root so the polynomial is reducible. Otherwise, we can reverse the sequence to get a new distinct legal sequence. The polynomial formed here is simply the previous polynomial evaluated at $\frac{1}{x}$ multiplied by $x^9$, so if the original polynomial is irreducible, then the new polynomial is as well. Therefore, there is a bijection between the irreducible polynomials, so $N$ is even.

Very standard problem. If $d_0=0$, then $0$ is a root. If the number is a palindrome, then $-1$ is a root. Otherwise, it is well known that reversing the coefficients preserves irreducibility.

Denote the above polynomial to be $P(x)$., and for any polynomial $F(x)$ for which $x$ is not a factor, call its reciprocal polynomial $F'(x)$. (We define the *reciprocal polynomial* of a polynomial not divisible by $x$ to be the polynomial with its coefficients in reverse order of the original polynomial; in particular, the roots of the reciprocal polynomial are the reciprocals of the roots of the original polynomial) Note that if $d_0 = 0$, then $P$ is reducible as $x$ is a factor. Henceforth consider all $P$ with $d_0 \ne 0$. The main claim is as follows: Claim: If $P$ is irreducible, so is $P'$ and vice-versa. Proof: Assume $P'$ is indeed reducible as $Q(x)R(x)$. Denote $Q(x) = K(x-a_1)\cdots(x-a_m), R(x) = M(x-b_1)\cdots(x-b_n)$, where $a_1, \cdots, a_m, b_1, \cdots, b_n \in \mathbb{C}, K, M \in \mathbb{Q}$. (This is possible due to the fundamental theorem of arithmetic) Then $Q'(x) = K' \left(x-\frac{1}{a_1}\right) \cdots \left(x-\frac{1}{a_m}\right), R'(x) = M' \left(x-\frac{1}{b_1}\right) \cdots \left(x-\frac{1}{b_n}\right)$, for some $K', M' \in \mathbb{Q}$. But now $Q'(x)R'(x) = K'M'\left(x-\frac{1}{a_1}\right)\cdots\left(x-\frac{1}{a_m}\right)\left(x-\frac{1}{b_1}\right)\cdots'\left(x-\frac{1}{b_n}\right) = \frac{K'M'}{c}P(x)$, where $c$ is the leading coefficient of $P(x)$ (this is due to the definition of reciprocal polynomials). But now $P(x) = \left(\frac{cQ(x)}{M'K'}\right) \cdot R(x)$, so $P$ is reducible, contradiction. Now observe that due to the claim, unless $P(x) = P'(x)$, we can pair up $P(x)$ and $P'(x)$ to get two polynomials. But if indeed $P(x) = P'(x)$, then $P(-1) = 0$, so $(x+1)$ is a factor, contradicting the irreducibility of $P$. Thus, we can pair up the irreducible polynomials as asked in the problem, so $N$ is indeed even. $\square$

This problem was easy for a cmo p3 but I didn't solve in contest smh