Difference between revisions of "1969 AHSME Problems/Problem 34"
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− | 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 the remainder <math>R=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>. | + | 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 the remainder <math>R=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>. |
Revision as of 16:02, 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 , 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 | |
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 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
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