Problem 11.6. The polynomial $P(x)$ has all coefficients as non-negative integers. It is known that $P(1)=4$ and $P(5)=152$. What is $P(11) ?$
Answer: 1454.

Solution. Suppose the degree of the polynomial $P$ is not less than 4, then its leading coefficient is not less than 1. Since all other coefficients of $P(x)$ are non-negative, then $P(5) \geqslant 5^{4}=625$, which contradicts the condition $P(5)=152$.

Therefore, the degree of the polynomial $P$ is not more than 3, and $P(x)=a x^{3}+b x^{2}+c x+d$ for some non-negative integers $a, b, c, d$. Then

$$
a+b+c+d=P(1)=4, \quad 125 a+25 b+5 c+d=P(5)=152
$$

Suppose $a \geqslant 2$. Then $P(5) \geqslant 125 \cdot 2=250>152$, a contradiction.

Suppose $a=0$. Then $P(5)=25 b+5 c+d \leqslant 25(b+c+d)=25 \cdot 427$, a contradiction.

Suppose $b=0$. Then $27=5 c+d \leqslant 5(c+d)=5 \cdot 3<27$, a contradiction.

Thus, $b=1$. Then

$$
c+d=2, \quad 5 c+d=2
$$

From this, it easily follows that $c=0$ and $d=2$, i.e., $P(x)=x^{3}+x^{2}+2$. Therefore, $P(11)=11^{3}+11^{2}+2=1454$.

Another solution. One can notice that $P(1)=4$ is in any case the sum of the coefficients of the polynomial $P(x)=a_{n} x^{n}+\ldots+a_{1} x+a_{0}$, from which each of them does not exceed 4. Then these coefficients can be considered as digits in the base-5 number system, and the value of the polynomial at the point 5 is the number written by these digits:

$$
152=P(5)=5^{n} a_{n}+\ldots+5 a_{1}+a_{0}
$$

Since $152_{10}=1102_{5}$, and the representations of numbers in the base-5 system are unique, then $P(x)=1 x^{3}+1 x^{2}+0 x+2$.