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

Revision as of 16:01, 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 $1$ .

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

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

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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 16: / Line 16: @@
 <math>x^{100}=(x-2)(x-1)Q(x)+ax+b</math>,
-<math>2^{100}=(2-2)(2-1)Q(x)+2a+b</math>,
+<math>2^{100}=(0)Q(2)+2a+b</math>,
 <math>2^{100}=2a+b</math>.

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

Revision as of 16:01, 10 July 2015

Problem

Solution

See also