2010 AMC 10A Problem 21

Let the roots be positive integers $p, q, r$. Expanding $(x - p)(x - q)(x - r)$ and matching coefficients (Vieta) gives $p + q + r = a$ and $pqr = 2010$. So we want three positive integers with product $2010$ and the smallest possible sum.

Factor: $2010 = 2 \cdot 3 \cdot 5 \cdot 67$. The prime $67$ must divide one of the roots, so that root is at least $67$. Taking it to be exactly $67$ leaves $30$ to split between the other two; the sum is smallest when the pair is as balanced as possible:
$$30 = 5 \cdot 6 \;\Rightarrow\; 67 + 5 + 6 = 78,$$
compared with $67 + 3 + 10 = 80$, $67 + 2 + 15 = 84$, $67 + 1 + 30 = 98$.

Could the root containing $67$ be larger? Then it is at least $134$, and the sum exceeds $134 > 78$. So the minimum is $78$, attained by roots $5, 6, 67$ (and $b = 5\cdot6 + 5\cdot67 + 6\cdot67 = 425$ is fine).

The answer is $\boxed{\textbf{(A)}\ 78}$.

Vieta translates "integer roots" into "integer factorization of the constant term," and minimizing a sum with fixed product is a balancing problem: factors should be as close together as possible. The large prime $67$ dominates, so the real decision is how to split the cofactor $30$.

• Splitting 2010 into factors that are not as balanced as possible (e.g. 67, 30, 1 or 67, 10, 3) and stopping early.
• Sign confusion with the $-ax^2$ term, leading to a search for a negative or maximal $a$.