1999 AMC 12 Problem 17

Below is the professionally curated solution for Problem 17 of the 1999 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 12 solutions, or check the answer key.

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Concepts:polynomial system of equations

Difficulty rating: 1680

17.

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 $(x - 19)(x - 99)?$

$-x + 80$

$x + 80$

$-x + 118$

$x + 118$

$0$

Solution:

By the Remainder Theorem, $P(19) = 99$ and $P(99) = 19.$ Write $P(x) = (x - 19)(x - 99)Q(x) + ax + b.$ 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.

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