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Difference between revisions of "2010 AMC 10A Problems/Problem 21"

(solution to amc 10 a #21)
(→‎Solution)
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==Solution==
 
==Solution==
 
By Vieta's Formulas, we know that <math>a</math> is the sum of the three roots of the polynomial <math>x^3-ax^2+bx-2010</math>.  
 
By Vieta's Formulas, we know that <math>a</math> is the sum of the three roots of the polynomial <math>x^3-ax^2+bx-2010</math>.  
−
Also, 2010 factors into <math>2*3*5*67</math>. But, since there are only three roots to the polynomial, two of the four prime factors must be multiplied so that we are left with two roots. To minimize <math>a</math>, <math>2</math> and <math>3</math> should be multiplied, which means <math>a</math> will be <math>6+5+67=78</math> and the answer is <math>\boxed{\textbf{(D)}}</math>.
+
Also, 2010 factors into <math>2*3*5*67</math>. But, since there are only three roots to the polynomial, two of the four prime factors must be multiplied so that we are left with three roots. To minimize <math>a</math>, <math>2</math> and <math>3</math> should be multiplied, which means <math>a</math> will be <math>6+5+67=78</math> and the answer is <math>\boxed{\textbf{(D)}}</math>.

Revision as of 19:59, 19 November 2010

Problem

The polynomial $x^3-ax^2+bx-2010$ has three positive integer zeros. What is the smallest possible value of $a$?

$\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118$

Solution

By Vieta's Formulas, we know that $a$ is the sum of the three roots of the polynomial $x^3-ax^2+bx-2010$. Also, 2010 factors into $2*3*5*67$. But, since there are only three roots to the polynomial, two of the four prime factors must be multiplied so that we are left with three roots. To minimize $a$, $2$ and $3$ should be multiplied, which means $a$ will be $6+5+67=78$ and the answer is $\boxed{\textbf{(D)}}$.

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