# 1969 AHSME Problems/Problem 34

## 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 $1$.

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

$2^{100}=(0)Q(2)+2a+b$,

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

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 terms, we see that the answer is $\fbox{B}$.