2010 AMC 10A Problem 21

Below is the professionally curated solution for Problem 21 of the 2010 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 10A solutions, or check the answer key.

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Concepts:Vieta’s Formulasprime factorizationoptimization

Difficulty rating: 2070

21.

The polynomial x3ax2+bx2010x^3-ax^2+bx-2010 has three positive integer roots. What is the smallest possible value of a?a?

7878

8888

9898

108108

118118

Solution:

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

Since 2010=23567,2010=2\cdot3\cdot5\cdot67, one root must be divisible by 67.67. If that root is larger than 67,67, then it is at least 134,134, which is already worse than the construction below.

Thus take t=67t=67 and minimize r+sr+s with rs=30.rs=30. The factor pair with smallest sum is 55 and 6,6, so a=5+6+67=78.a=5+6+67=78.

Thus, A is the correct answer.

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