Difference between revisions of "1999 AHSME Problems/Problem 17"
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<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>. | ||
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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
Contents
[hide]Problem
Let be a polynomial such that when is divided by , the remainder is , and when is divided by , the remainder is . What is the remainder when is divided by ?
Solution 1
According to the problem statement, there are polynomials and such that .
From the last equality we get .
The value 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 , from which . This means that is a root of the polynomial . In other words, there is a polynomial such that .
Substituting this into the original formula for we get
Therefore when is divided by , the remainder is .
Solution 2
Since the divisor is a quadratic, the degree of the remainder is at most linear. We can write in the form where is the remainder. By the Remainder Theorem, plugging in and gives us a system of equations.
Solving gives us and , thus, our answer is
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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