13. Angela has the polynomials $x-1,(x-1)(x-2),(x-1)(x-2)(x-3), \ldots$ up to $(x-1)(x-2) \cdots(x-2017)(x-2018)$, and she divides them into two groups. Let $p(x)$ be the product of the polynomials in the first group and $q(x)$ the product of the polynomials in the second group. Angela notices that the polynomial $p(x)$ divides the polynomial $q(x)$, and that the degree of the quotient $\frac{q(x)}{p(x)}$ is the smallest possible: what is this degree?
13. The answer is 1009. First, observe that Angela can assign the polynomials $(x-1)(x-$ $2),(x-1)(x-2)(x-3)(x-4), \cdots,(x-1)(x-2) \cdots(x-2017)(x-2018)$ (i.e., those of even degree) to the group corresponding to $p(x)$, and the others to the group corresponding to $q(x)$. With this choice, the ratio $p(x) / q(x)$ is equal to

$$
\frac{(x-1)(x-2) \cdot(x-1)(x-2)(x-3)(x-4) \cdots}{(x-1) \cdot(x-1)(x-2)(x-3) \cdots}=(x-2)(x-4) \cdots(x-2018),
$$

which has degree 1009. To show that a smaller degree cannot be achieved, observe that there are exactly 2017 factors $(x-2)$, exactly 2015 factors $(x-4)$, and in general, there is an odd number of factors $(x-k)$ for each even $k$. For the polynomial $q(x)$ to divide the polynomial $p(x)$, the number of factors $x-k$ that appear in $p(x)$ must be greater than or equal to the number of factors $x-k$ that appear in $q(x)$: if the total number of factors $x-k$ is odd, at least one more must appear in $p(x)$ than in $q(x)$. This reasoning shows that in the ratio $p(x) / q(x)$, at least one factor $(x-2)$, at least one factor $(x-4)$, ..., and at least one factor $(x-2018)$ must appear, and thus this ratio cannot have a degree less than 1009.