1999 AHSME Problems/Problem 17

Problem

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

$\mathrm{(A) \ } -x + 80 \qquad \mathrm{(B) \ } x + 80 \qquad \mathrm{(C) \ } -x + 118 \qquad \mathrm{(D) \ } x + 118 \qquad \mathrm{(E) \ } 0$

Solution

According to the problem statement, there are polynomials $Q(x)$ and $R(x)$ such that $P(x) = Q(x)(x-19) + 99 = R(x)(x-99) + 19$ .

From the last equality we get $Q(x)(x-19) + 80 = R(x)(x-99)$ .

The value $x=99$ is a root of the polynomial on the right hand side, therefore it must be a root of the one on the left hand side as well. Substituting, we get $Q(99)(99-19) + 80 = 0$ , from which $Q(99)=-1$ . This means that $99$ is a root of the polynomial $Q(x)+1$ . In other words, there is a polynomial $S(x)$ such that $Q(x)+1 = S(x)(x-99)$ .

Substituting this into the original formula for $P(x)$ we get $P(x) = Q(x)(x-19) + 99 = (S(x)(x-99) - 1)(x-19) + 99 =$ $= S(x)(x-99)(x-19) - (x-19) + 99$

Therefore when $P(x)$ is divided by $(x-19)(x-99)$ , the remainder is $\boxed{-x + 118}$ .

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1999 AHSME Problems/Problem 17

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