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Difference between revisions of "1969 AHSME Problems/Problem 34"

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Line 19: Line 19:
  
 
<math>2^{100}=2a+b</math>.
 
<math>2^{100}=2a+b</math>.
 +
  
 
Similarly, <math>1^{100}=a+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>.
 
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>.
  

Revision as of 15:57, 10 July 2015

Problem

The remainder $R$ obtained by dividing $x^{100}$ by $x^2-3x+2$ is a polynomial of degree less than $2$. Then $R$ may be written as:


$\text{(A) }2^{100}-1 \quad \text{(B) } 2^{100}(x-1)-(x-2)\quad \text{(C) } 2^{200}(x-3)\quad\\ \text{(D) } x(2^{100}-1)+2(2^{99}-1)\quad \text{(E) } 2^{100}(x+1)-(x+2)$

Solution

Let the polynomial $Q(x)$ be the quotient when $x^{100}$ is divided by $x^2-3x+2$, and let the remainder $R=ax+b$, for some real $a$ and $b$. Then we can write: $x^100=(x^2-3x+2)Q(x)+ax+b$. Since it is hard to deal with $Q(x)$ (it is of degree 98!), we factor $x^2-3x+2$ as $(x-2)(x-1)$ so we can eliminate $Q(x)$ by plugging in $x$ values of $2$ and $1$.


$x^{100}=(x-2)(x-1)Q(x)+ax+b$,

$2^{100}=(2-2)(2-1)Q(x)+2a+b$,

$2^{100}=2a+b$.


Similarly, $1^{100}=a+b$.

Solving this system of equations gives $a=2^{100}-1$ and $b=2-2^{100}$. Thus, $R=ax+b=(2^{100}-1)x+(2-2^{100})$. Expanding and combining $x$ terms, we see that the answer is $\fbox{B}$.

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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