2020 AMC 10A Problem 17

"$P(n) \le 0$" means $P(n) = 0$ or $P(n) < 0$. Count the two kinds separately.

1. Where $P(n) = 0$. A product is zero when a factor is zero, so $n$ is one of $1^2, 2^2, \ldots, 100^2$. That is $100$ numbers.
2. The sign of a product. A product of nonzero numbers is negative when it has an odd number of negative factors.
3. Try some values of $n$.
- $n = 0$: all $100$ factors are negative. $100$ is even, so $P(0) > 0$.
- $n = 2$ (between $1$ and $4$): the factor $n - 1$ is positive and the other $99$ are negative. $99$ is odd, so $P(2) < 0$.
- $n = 5$ (between $4$ and $9$): two factors are positive and $98$ are negative. Even, so $P(5) > 0$.
- In general, each time $n$ passes a square, one more factor turns positive and the sign flips.
4. So $P(n) < 0$ in the gaps between $1$ and $4$, between $9$ and $16$, between $25$ and $36$, and so on: the gap from $k^2$ to $(k+1)^2$ for $k = 1, 3, 5, \ldots, 99$. (Past $100^2$ every factor is positive.)
5. How many whole numbers are in a gap? Strictly between $k^2$ and $(k+1)^2$ there are $(k+1)^2 - k^2 - 1 = 2k$ numbers. For example, between $1$ and $4$ there are $2$ (namely $2, 3$), and between $9$ and $16$ there are $6$.
6. Add these for $k = 1, 3, 5, \ldots, 99$:
$$2 + 6 + 10 + \cdots + 198 = 2(1 + 3 + 5 + \cdots + 99) = 2 \times 50^2 = 5000.$$
(The first $50$ odd numbers add up to $50^2$.)
7. Total. $5000 + 100 = 5100$.

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

A product of factors like $(x - r)$ switches sign every time $x$ passes one of the $r$ values. So the signs alternate from gap to gap, and you only need to find the sign in one gap to know them all. Watch for "$\le$": the zeros count too.

$P(n) = 0$ 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 $\boxed{\textbf{(E)}\ 5100}$.