Difference between revisions of "2010 AMC 10A Problems/Problem 21"
(Created page with '==Problem== The polynomial <math>x^3-ax^2+bx-2010</math> has three positive integer zeros. What is the smallest possible value of <math>a</math>? <math>\textbf{(A)}\ 78 \qquad \…') |
(solution to amc 10 a #21) |
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<math>\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118</math> | <math>\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118</math> | ||
| + | ==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>. | ||
| + | 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>. | ||
Revision as of 18:57, 18 November 2010
Problem
The polynomial 
has three positive integer zeros. What is the smallest possible value of 
?
Solution
By Vieta's Formulas, we know that is the sum of the three roots of the polynomial .
Also, 2010 factors into 
. 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 , 
and 
should be multiplied, which means will be 
and the answer is 
.
