A2. The coefficients of the polynomial $p$ of degree 3 are single-digit natural numbers, and it is also given that $p(\sqrt{10})=$ $12+34 \sqrt{10}$. What is $p(10)$?
(A) 46
(B) 352
(C) 2022
(D) 3142
(E) 3494
A2. Let's write $p(x)=a x^{3}+b x^{2}+c x+d$, where $a, b, c$ and $d$ are one-digit natural numbers. Then,

$$
p(\sqrt{10})=a \cdot 10 \sqrt{10}+b \cdot 10+c \cdot \sqrt{10}+d=(10 b+d)+(10 a+c) \sqrt{10}
$$

Since $a, b, c$ and $d$ are one-digit natural numbers, i.e., digits, we have

$$
p(\sqrt{10})=\overline{b d}+\overline{a c} \sqrt{10}=12+34 \sqrt{10}
$$

It follows that $b=1, d=2, a=3$ and $c=4$. Therefore, $p(x)=3 x^{3}+x^{2}+4 x+2$ and thus $p(10)=$ $3000+100+40+2=3142$.