Difference between revisions of "1969 AHSME Problems/Problem 34"
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== Solution == | == Solution == | ||
− | <math>\fbox{B}</math> | + | Let the polynomial <math>Q(x)</math> be the quotient when <math>x^100</math> is divided by <math>x^2-3x+2</math>, and let <math>R</math> the remainder be <math>ax+b</math>, for some real <math>a</math> and <math>b</math>. Then we can write: <math>x^100=(x^2-3x+2)Q(x)+ax+b</math>. Since it is hard to deal with <math>Q(x)</math> (it is of degree 98!), we factor <math>x^2-3x+2</math> as <math>(x-2)(x-1)</math> so we can eliminate <math>Q(x)</math> by plugging in <math>x</math> values of <math>2</math> and <math>1</math>. |
+ | |||
+ | <math>x^100=(x-2)(x-1)Q(x)+ax+b</math> | ||
+ | <math>2^100=(2-2)(x-1)Q(x)+ax+b</math> | ||
+ | <math>2^100=2a+b</math>. | ||
+ | |||
+ | Similarly, <math>1^100=a+b</math>. | ||
+ | Solving this system of equations gives <math>a=2^100-1</math> and <math>b=2-2^100</math>. Thus, <math>R=ax+b=(2^100-1)x+(2-2^100)</math>. Expanding and combining <math>x</math> terms, we see that the answer is <math>\fbox{B}</math>. | ||
== See also == | == See also == |
Revision as of 15:52, 10 July 2015
Problem
The remainder obtained by dividing by is a polynomial of degree less than . Then may be written as:
Solution
Let the polynomial be the quotient when is divided by , and let the remainder be , for some real and . Then we can write: . Since it is hard to deal with (it is of degree 98!), we factor as so we can eliminate by plugging in values of and .
.
Similarly, . Solving this system of equations gives and . Thus, . Expanding and combining terms, we see that the answer is .
See also
1969 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 33 |
Followed by Problem 35 | |
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All AHSME Problems and Solutions |
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