2010 AMC 10A Problem 21

First connect the roots to the numbers $a$ and $2010$.

1. Roots and coefficients. If the roots are the positive whole numbers $p$, $q$, $r$, the polynomial is $(x - p)(x - q)(x - r)$. Multiplying out,
$$(x - p)(x - q)(x - r) = x^3 - (p + q + r)x^2 + (pq + pr + qr)x - pqr.$$
Match this with $x^3 - ax^2 + bx - 2010$:
- $a = p + q + r$ (the sum of the roots),
- $2010 = pqr$ (the product of the roots).
2. The new question. Find three positive whole numbers that multiply to $2010$ with the smallest possible sum.
3. Factor 2010. $2010 = 2 \times 3 \times 5 \times 67$.
4. The big prime. $67$ is prime, so one of the roots must be a multiple of $67$. Keep it as small as possible: make that root exactly $67$. The other two roots multiply to $2010 \div 67 = 30$.
5. Split 30. Two numbers with a fixed product have the smallest sum when they are close together. List the options:

1. Could a bigger multiple of 67 help? If the root containing $67$ were $134$ or more, the sum alone would be over $134$, far more than $78$. So no.
2. The smallest sum is $78$, from the roots $5$, $6$, $67$.

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

Writing a polynomial as a product of $(x - \text{root})$ shows that the number in front of $x^2$ is minus the sum of the roots and the constant is minus their product. That turns a polynomial question into a factoring question. To keep a sum small when the product is fixed, make the factors as close to each other as you can; a large prime like $67$ decides one factor for you.

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 $\boxed{\textbf{(A)}\ 78}$.