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1999 AHSME Problems/Problem 17

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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 1

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 $\boxed{-x + 118}$.

Solution 2

Since the divisor $(x-19)(x-99)$ is a quadratic, the degree of the remainder is at most linear. We can write $P(x)$ in the form \[P(x) = Q(x)(x-19)(x-99) + cx+d\] where $cx+d$ is the remainder. By the Remainder Theorem, plugging in $19$ and $99$ gives us a system of equations. \[99c+d = 19\] \[19c+d = 99\]

Solving gives us $c=-1$ and $d = 118$, thus, our answer is $\boxed{ \mathrm{(C) \ }-x+118}$

See also

1999 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AHSME Problems and Solutions

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