<s>[INST] Given a polynomial $P$ with integer coefficients, find $P(2)$ Question:
$P(x)$ be a polynomial with non-negative integer coefficients such that $P(0)=33$ , $P(1)=40$ and $P(9)=60000$.
Find $P(2)$.
My attempt:
Suppose $P(x) = a_nx^n + \dotsb + a_1x + a_0$.
Now, $P(0) = 33$ implies $a_0 = 33$.
$P(1)=40$ implies that the sum of the coefficients other than constant term is $7$. Also,
$$P(9)-P(0)=3(2^5\times5^4-11)=3^2a_{1}+3^4a_{2}+\dotsb+a_n3^{2n}.$$
Dividing both sides by $3^3$ we can easily infer $a_{1}=3$.  After solving it further, I ended up with
$$a_{2}+3a_{3}+\dotsb a_{n}3^{2n-3}=740,$$ 
but I can't find $P(2)$.
There are few more coefficients remaining other than $a_{1}$ whose sum is $4$, and I think casework can give a solution... like assuming consecutive $4$ coefficients $1$, $1$, $1$, $1$; $1$, $2$, $1$ etc., but I don't want to go that way.
 [/INST] The polynomial can be at most of degree $4$ because 
$9^5=59409$
so
$p(x)=33+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5$
and 
$a_1+a_2+a_3+a_4+a_5=7$
It is clear that $a_5=1$ otherwise 
if $a_5=0$ then $p(9)<60000$
while if $a_5>1$ then $a_59^5>60000$
If $a_4\neq 0$  then $p(9)>60000$ so $a_4=0$. If $a_3=0$ we have 
$a_1+a_2=6$ 
$9a_1+81a_2=918$
that it has not solutions
So We must suppose $a_3\neq 0$ and We  must compute
$a_1+a_2+a_3=6$
$9a_1+81a_2+729a_3=918$
This means $a_3=1$ and so 
$a_1+a_2=5$ 
So is clear that We must have $a_1=3$ and $a_2=2$. So the polinomial is 
$p(x)=33+3x+2x^2+x^3+x^5$
</s>