Question

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

$78$

$88$

$98$

$108$

$118$

Answer: A

Concepts:Vieta’s Formulas prime factorization optimization

Difficulty rating: 2070

Solution:

Let the roots be positive integers $r\le s\le t.$ By Vieta's formulas, $rst=2010$ and $a=r+s+t.$

Since $2010=2\cdot3\cdot5\cdot67,$ one root must be divisible by $67.$ If that root is larger than $67,$ then it is at least $134,$ which is already worse than the construction below.

Thus take $t=67$ and minimize $r+s$ with $rs=30.$ The factor pair with smallest sum is $5$ and $6,$ so $a=5+6+67=78.$

Thus, A is the correct answer.

2010 AMC 10A Problem 21

Answer: A

Difficulty rating: 2070

Solution:

Problem 21 in Other Years