View source for 1969 AHSME Problems/Problem 34

== Problem ==

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


<math>\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)</math>

== Solution ==
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 ==
{{AHSME 35p box|year=1969|num-b=33|num-a=35}}  

[[Category: Intermediate Algebra Problems]]
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