2010 AMC 10A Problems/Problem 21

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 $\boxed{\textbf{(A)}}$ .

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2010 AMC 10A Problems/Problem 21

Problem

Solution