5. The polynomial $P(x)$, of degree 42, takes the value 0 at the first 21 odd prime numbers and at their reciprocals (it is recalled that the reciprocal of a positive integer $n$ is the rational number $1 / n$). What is the value of the ratio $P(2) / P(1 / 2)$?
(A) 0
(B) 1
(C) $2^{21}$
(D) $3^{21}$
(E) $4^{21}$
5. The answer is $\mathbf{( E )}$. Observe that the expression $Q(x)=P(x)-x^{42} P(1 / x)$ is a polynomial, since the monomial $x^{42}$ simplifies the denominator of $P(1 / x)$. Furthermore, it has a degree of at most

42, and if $r$ is one of the first 21 odd prime numbers, $Q(x)$ vanishes at $r$ and at $1 / r$: indeed, we have

$$
Q(r)=P(r)-r^{42} P(1 / r)=0 \quad Q(1 / r)=P(1 / r)-(1 / r)^{42} P(r)=0,
$$

where we have used the fact that $P(r)=P(1 / r)=0$ by hypothesis. Finally, $Q(x)$ vanishes at 1, because $Q(1)=P(1)-P(1)=0$. Since $Q(x)$ vanishes for at least 43 distinct values of $x$ but has a degree of at most 42, we obtain that $Q(x)$ is the constant zero polynomial, hence $P(x)=x^{42} P(1 / x)$ for every $x$. Therefore, $\frac{P(2)}{P(1 / 2)}=\frac{2^{42} P(1 / 2)}{P(1 / 2)}=2^{42}=4^{21}$.