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Difference between revisions of "1999 AHSME Problems/Problem 17"

(New page: == Problem == Let <math>P(x)</math> be a polynomial such that when <math>P(x)</math> is divided by <math>x-19</math>, the remaidner is <math>99</math>, and when <math>P(x)</math> is divide...)
 
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== Problem ==
 
== Problem ==
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Let <math>P(x)</math> be a polynomial such that when <math>P(x)</math> is divided by <math>x-19</math>, the remaidner is <math>99</math>, and when <math>P(x)</math> is divided by <math>x - 99</math>, the remainder is <math>19</math>. What is the remainder when <math>P(x)</math> is divided by <math>(x-19)(x-99)</math>?
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Let <math>P(x)</math> be a polynomial such that when <math>P(x)</math> is divided by <math>x-19</math>, the remainder is <math>99</math>, and when <math>P(x)</math> is divided by <math>x - 99</math>, the remainder is <math>19</math>. What is the remainder when <math>P(x)</math> is divided by <math>(x-19)(x-99)</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>
 
<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>

Revision as of 19:10, 14 March 2010

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 $\boxed{-x + 118}$.

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
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