2010 AMC 10A Problem 21

P21-25~4 minPrint

Polynomials Divisibility & Factors Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Organized listing / direct enumeration

The polynomial $x^3 -ax^2 + bx -2010$ has three positive integer roots. What is the smallest possible value of $a$ ?

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Key insight

The roots multiply to 2010 and add to a. One root must contain the prime 67; split the leftover 30 as evenly as possible, 5 and 6.

Solution

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:

Split of 30	Sum of all three roots
$1 \times 30$	$67 + 1 + 30 = 98$
$2 \times 15$	$67 + 2 + 15 = 84$
$3 \times 10$	$67 + 3 + 10 = 80$
$5 \times 6$	$67 + 5 + 6 = 78$

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.
The smallest sum is $78$ , from the roots $5$ , $6$ , $67$ .

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Intermediate Algebra, Chapter 6: Polynomial Division (§6.2, §6.4)
Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Algebra, Chapter 11: Special Factorizations
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)