AMC10 2020 A

AMC10 2020 A · Q17

AMC10 2020 A · Q17. It mainly tests Sequences & recursion (algebra), Basic counting (rules of product/sum).

Define $P(x) = (x - 1^2)(x - 2^2)\cdots(x - 100^2)$. How many integers $n$ are there such that $P(n) \le 0$?

定义$P(x) = (x - 1^2)(x - 2^2)\cdots(x - 100^2)$。有多少个整数$n$使得$P(n) \le 0$？

(A) 4900 4900

(B) 4950 4950

(D) 5050 5050

(E) 5100 5100

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Observe that $P(n)\le 0$ exactly when either one of the factors in the product is zero or an odd number of the factors are negative. Therefore $P(n)\le 0$ exactly when $n$ lies in one of the intervals $[1^2,2^2]$, $[3^2,4^2]$, $\ldots$, $[99^2,100^2]$. The number of integers in the interval $[(2k-1)^2,(2k)^2]$ is $(2k)^2-(2k-1)^2+1=4k$, so the total number of values for $n$ is \[ \sum_{k=1}^{50}4k=4\cdot\frac{50\cdot 51}{2}=5100. \]

答案（E）：注意到当且仅当乘积中的某个因子为零，或负因子的个数为奇数时，$P(n)\le 0$。因此，当且仅当 $n$ 落在区间 $[1^2,2^2]$、$[3^2,4^2]$、$\ldots$、$[99^2,100^2]$ 之一中时，$P(n)\le 0$。区间 $[(2k-1)^2,(2k)^2]$ 中整数的个数为 $(2k)^2-(2k-1)^2+1=4k$，所以 $n$ 的取值总数为 \[ \sum_{k=1}^{50}4k=4\cdot\frac{50\cdot 51}{2}=5100. \]

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