68*. All coefficients of the polynomial $p(x)$ are natural numbers less than 10. It is known that $p(10)=1248$. Find this polynomial.
$\triangle$ Let's write $p(x)$ in the form

$$
p(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}
$$

Let's set $x=10$ here. Then

$$
a_{0} 10^{n}+a_{1} 10^{n-1}+\cdots+a_{n-1} 10+a_{n}=1248
$$

From this,

$$
a_{0} 10^{n}+a_{1} 10^{n-1}+\cdots+a_{n-1} \cdot 10=1240+\left(8-a_{n}\right)
$$

Therefore, the difference $8-a_{n}$ is divisible by 10. But since $a_{n}$ is a natural number less than 10, then

$$
8-a_{n}=0, \quad a_{n}=8
$$

We obtain the equality:

$$
a_{0} \cdot 10^{n-1}+a_{1} \cdot 10^{n-2}+\cdots+a_{n-1}=124
$$

Here, for the same reason, $a_{n-1}=4$. Then

$$
a_{0} \cdot 10^{n-2}+a_{1} \cdot 10^{n-3}+\cdots+a_{n-2}=12
$$

From this, $a_{n-2}=2, a_{n-1}=1$. It makes no sense to continue further.

Answer: $p(x)=x^{3}+2 x^{2}+4 x+8$.