2010 AMC 10A Problem 21

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.