Question

设 $P(x)$ 是一个多项式，当 $P(x)$ 除以 $x - 19$ 时余数为 $99$ ，当 $P(x)$ 除以 $x - 99$ 时余数为 $19$ 。当 $P(x)$ 除以 $(x - 19)(x - 99)$ 时，余数是多少？

Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x - 19,$ the remainder is $99,$ and when $P(x)$ is divided by $x - 99,$ the remainder is $19.$ What is the remainder when $P(x)$ is divided by $(x - 19)(x - 99)?$

$-x + 80$

$x + 80$

$-x + 118$

$x + 118$

$0$

答案：C

知识点：多项式方程组

难度评级：1680

解答：

由余数定理， $P(19) = 99$ ， $P(99) = 19$ 。设则 $\begin{aligned} &P(x) = (x - 19)(x - 99)Q(x) \\ &\quad {}+ ax + b. \end{aligned}$ $19a + b = 99, \qquad 99a + b = 19.$

相减得 $80a = -80$ ，所以 $a = -1$ ，进而 $b = 118$ 。余数为 $-x + 118$ 。

所以正确答案是 C。

By the Remainder Theorem, $P(19) = 99$ and $P(99) = 19.$ Write $\begin{aligned} &P(x) = (x - 19)(x - 99)Q(x) \\ &\quad {}+ ax + b. \end{aligned}$ Then $19a + b = 99, \qquad 99a + b = 19.$

Subtracting gives $80a = -80,$ so $a = -1$ and $b = 118.$ The remainder is $-x + 118.$

Thus, the correct answer is C.

1999 AMC 12 第 17 题

答案：C

难度评级：1680

解答：

其他年份的第 17 题