2020 AMC 10A Problem 17

$P(n) = 0$ exactly when $n$ is one of the $100$ squares $1^2, 2^2, \dots, 100^2$.

For other integers, $P(n) < 0$ when an odd number of the factors $n - k^2$ are negative. If $k^2 < n < (k+1)^2$, the negative factors are those with roots above $n$, namely $(k+1)^2, \dots, 100^2$: that is $100 - k$ factors. This is odd exactly when $k$ is odd. For $n < 1$ all $100$ factors are negative (even, positive product), and for $n > 100^2$ none are.

So $P(n) < 0$ precisely for integers strictly between $k^2$ and $(k+1)^2$ with $k \in \{1, 3, 5, \dots, 99\}$. Each such open gap contains $(k+1)^2 - k^2 - 1 = 2k$ integers, giving
$$\sum_{k \text{ odd},\ 1 \le k \le 99} 2k = 2(1 + 3 + \cdots + 99) = 2 \cdot 50^2 = 5000.$$

Adding the $100$ zeros: $5000 + 100 = 5100$.

The answer is $\boxed{\textbf{(E)}\ 5100}$.

A product of linear factors changes sign at each root, so a sign chart alternates across the sorted roots. With $100$ roots, the sign is positive on both ends and negative on every other gap, starting with the gap $(1, 4)$. Counting integers in each gap is just the spacing of consecutive squares. Remember that "$\le 0$" includes the roots.

• Forgetting the 100 integers where P(n) = 0 and answering 5000, or choosing the gaps with even k instead of odd k.
• Counting $2k + 1$ integers per gap (including an endpoint) and getting $5050$.