11.3. Does there exist a polynomial $p(x)$ of degree 13 with the coefficient of the leading term equal to $\frac{1}{1001}$, which takes integer values at all integer points? Justify your answer.
Solution: For example, the polynomial $p(x)=\frac{x(x-1)(x-2) \ldots(x-12)}{1001}$ works. Let's show this. When expanding the brackets, the highest power of $x$ will be 13, and the coefficient of this term will be $\frac{1}{1001}$. Since $1001=7 \cdot 11 \cdot 13$, it remains to show that for any integer $x$, the number $A=x(x-1)(x-2) \ldots(x-12)$ is divisible without remainder by each of the numbers 7, 11, and 13. However, each of these divisibilities is obvious, since the number $A$ represents the product of 13 consecutive integers.

Answer: exists.

Recommendations for checking:

| included in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| An example of a polynomial satisfying the condition is given, but divisibility by 1001 for all integers $x$ is not proven | 5 points |
| The idea (not developed into a solution) to use a product of the form $x(x+1)(x+2) \ldots$ for constructing the polynomial | 3 points |
| Incorrect answer or answer without justification, as well as examples of polynomials that do not meet the conditions of the problem | 0 points |