The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. 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 1
By [Vieta's Formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas), we know that $a$ is the sum of the three roots of the polynomial $x^3-ax^2+bx-2010$. Again Vieta's Formulas tell us that $2010$ is the product of the three integer roots. Also, $2010$ factors into $2\cdot3\cdot5\cdot67$. But, since the polynomial has only three roots, 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 $\boxed{\textbf{(A)78}}$.
~JimPickens


Note:
If you are feeling unconfident about $78$, you can try to expand the expression $(x-5)(x-6)(x-67)$ which has roots $5$, $6$, and $67$.
\[(x-5)(x-6)(x-67)=(x^{2}-11x+30)(x-67)\]
\[=x\left(x^{2}-11x+30\right)-67\left(x^{2}-11x+30\right)=x^{3}-11x^{2}+30x-67x^{2}+737x-2010=x^{3}-78x^{2}+767x-2010\]
As we can see, $a=78$, and since this is the least answer choice, we can be confident that the right option is $\boxed{\textbf{(A) } 78}$.
~JH. L

Solution 2
We can expand $(x+a)(x+b)(x+c)$ as $(x^2+ax+bx+ab)(x+c)$
$(x^2+ax+bx+ab)(x+c)=x^3+abx+acx+bcx+abx^2+acx^2+bcx^2+abc=x^3+x^2(a+b+c)+x(ab+ac+bc)+abc$
We do not care about $+bx$ in this case, because we are only looking for $a$.  We know that the constant term is $-2010=-(2\cdot 3\cdot 5\cdot 67)$
We are trying to minimize a, such that we have $-ax^2$
Since we have three positive solutions, we have $(x-a)(x-b)(x-c)$ as our factors.  We have to combine two of the factors of $2\cdot 3\cdot 5\cdot 67$, and then sum up the $3$ resulting factors.  Since we are minimizing, we choose $2$ and $3$ to combine together.  We get $(x-6)(x-5)(x-67)$ which gives us a coefficient of $x^2$ of $-6-5-67=-78$
Therefore $-a=-78$ or $a=\boxed{\textbf{(A)}78}$

Solution 3
We want the polynomial $x^3-ax^2+bx-2010$ to have POSITIVE integer roots. That means we want to factor it in to the form $(x-a)(x-b)(x-c).$ We therefore want the prime factorization for $2010$. The prime factorization of $2010$ is $2 \cdot 3 \cdot 5 \cdot 67$. We want the smallest difference of the $3$ roots since by [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas), $a$ is the sum of the $3$ roots.
We proceed to factorize it in to $(x-5)(x-6)(x-67)$. Therefore, our answer is $5+6+67$ = $\boxed{\textbf{(A)78}}$.
~Arcticturn

Suggestion for the author: The variables $a$ and $b$ are already defined as coefficients of the cubic polynomial. So, consider using different variables when describing how we want to factor this polynomial in the second sentence. Thanks!
~Jwarner