2013 AMC 10B Problem 5

Attempt Problem 5 of the 2013 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 10B solutions, or check the answer key.

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5.

Positive integers aa and bb are each less than 6.6. What is the smallest possible value for 2aab?2 \cdot a - a \cdot b?

20 -20

15 -15

10 -10

0 0

2 2

Answer: B
Concepts:factoringoptimization
Difficulty rating: 870
Solution:

The expression is 2aab=a(2b)2a-ab=a(2-b).

To make it as small as possible, choose bb as large as possible so that 2b2-b is most negative, and choose aa as large as possible. Since a,b<6a,b<6, take a=b=5a=b=5.

The minimum value is 5(25)=155(2-5)=-15.

Thus, the correct answer is B .

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