2013 AMC 10B Problem 5

Below is the professionally curated solution for Problem 5 of the 2013 AMC 10B, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:factoringoptimization

Difficulty rating: 870

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

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 .

Problem 5 in Other Years