AMC12 2011 A

AMC12 2011 A · Q20

AMC12 2011 A · Q20. It mainly tests Linear equations, Quadratic equations.

Let $f(x)=ax^2+bx+c$, where $a$, $b$, and $c$ are integers. Suppose that $f(1)=0$, $50<f(7)<60$, $70<f(8)<80$, $5000k<f(100)<5000(k+1)$ for some integer $k$. What is $k$?

设 $f(x)=ax^2+bx+c$，其中 $a$、$b$ 和 $c$ 是整数。假设 $f(1)=0$，$50<f(7)<60$，$70<f(8)<80$，且对某个整数 $k$ 有 $5000k<f(100)<5000(k+1)$。求 $k$。

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (C)

正确答案：（C）

Solution

From $f(1) = 0$, we know that $a+b+c = 0$. From the first inequality, we get $50 < 49a+7b+c < 60$. Subtracting $a+b+c = 0$ from this gives us $50 < 48a+6b < 60$, and thus $\frac{25}{3} < 8a+b < 10$. Since $8a+b$ must be an integer, it follows that $8a+b = 9$. Similarly, from the second inequality, we get $70 < 64a+8b+c < 80$. Again subtracting $a+b+c = 0$ from this gives us $70 < 63a+7b < 80$, or $10 < 9a+b < \frac{80}{7}$. It follows from this that $9a+b = 11$. We now have a system of three equations: $a+b+c = 0$, $8a+b = 9$, and $9a+b = 11$. Solving gives us $(a, b, c) = (2, -7, 5)$ and from this we find that $f(100) = 2(100)^2-7(100)+5 = 19305$ Since $15000 < 19305 < 20000 \to 5000(3) < 19305 < 5000(4)$, we find that $k = 3 \rightarrow \boxed{\textbf{(C)}\ 3}$.

由 $f(1) = 0$ 可知 $a+b+c = 0$。由第一个不等式得 $50 < 49a+7b+c < 60$。两边减去 $a+b+c = 0$ 得 $50 < 48a+6b < 60$，从而 $\frac{25}{3} < 8a+b < 10$。由于 $8a+b$ 必为整数，故 $8a+b = 9$。同理，由第二个不等式得 $70 < 64a+8b+c < 80$。两边减去 $a+b+c = 0$ 得 $70 < 63a+7b < 80$，即 $10 < 9a+b < \frac{80}{7}$。因此 $9a+b = 11$。于是得到方程组 $a+b+c = 0$，$8a+b = 9$，$9a+b = 11$。解得 $(a, b, c) = (2, -7, 5)$，从而 $f(100) = 2(100)^2-7(100)+5 = 19305$。因为 $15000 < 19305 < 20000 \to 5000(3) < 19305 < 5000(4)$，所以 $k = 3 \rightarrow \boxed{\textbf{(C)}\ 3}$。

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