Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1999 AHSME Problems/Problem 17"
(→‎Solution)
 
Line 4: Line 4:
 
<math> \mathrm{(A) \ } -x + 80 \qquad \mathrm{(B) \ } x + 80 \qquad \mathrm{(C) \ } -x + 118 \qquad \mathrm{(D) \ } x + 118 \qquad \mathrm{(E) \ } 0</math>
 
<math> \mathrm{(A) \ } -x + 80 \qquad \mathrm{(B) \ } x + 80 \qquad \mathrm{(C) \ } -x + 118 \qquad \mathrm{(D) \ } x + 118 \qquad \mathrm{(E) \ } 0</math>
  
−
== Solution ==
+
== Solution 1==
  
 
According to the problem statement, there are polynomials <math>Q(x)</math> and <math>R(x)</math> such that <math>P(x) = Q(x)(x-19) + 99 = R(x)(x-99) + 19</math>.
 
According to the problem statement, there are polynomials <math>Q(x)</math> and <math>R(x)</math> such that <math>P(x) = Q(x)(x-19) + 99 = R(x)(x-99) + 19</math>.
Line 17: Line 17:
  
 
Therefore when <math>P(x)</math> is divided by <math>(x-19)(x-99)</math>, the remainder is <math>\boxed{-x + 118}</math>.
 
Therefore when <math>P(x)</math> is divided by <math>(x-19)(x-99)</math>, the remainder is <math>\boxed{-x + 118}</math>.
 +
 +
== Solution 2==
 +
Since the divisor <math>(x-19)(x-99)</math> is a quadratic, the degree of the remainder is at most linear. We can write <math>P(x)</math> in the form
 +
<cmath>P(x) = Q(x)(x-19)(x-99) + cx+d</cmath>
 +
where <math>cx+d</math> is the remainder.
 +
By the Remainder Theorem, plugging in <math>19</math> and <math>99</math> gives us a system of equations.
 +
<cmath>99c+d = 19</cmath>
 +
<cmath>19c+d = 99</cmath>
 +
 +
Solving gives us <math>c=-1</math> and <math>d = 118</math>, thus, our answer is <math>\boxed{ \mathrm{(C) \ }-x+118}</math>
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1999|num-b=16|num-a=18}}
 
{{AHSME box|year=1999|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 04:01, 7 February 2016

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1999_AHSME_Problems/Problem_17&oldid=75657"