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Difference between revisions of "2013 AMC 10B Problems/Problem 5"

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==Solution==
 
==Solution==
 
Factoring the equation gives <math>a(2 - b)</math>. From this we can see that to obtain the least possible value, <math>2 - b</math> should be negative, and should be as small as possible. To do so, <math>b</math> should be maximized. Because <math>2 - b</math> is negative, we should maximize the positive value of <math>a</math> as well. The maximum values of both <math>a</math> and <math>b</math> are <math>5</math>, so the answer is <math>5(2 - 5) = \boxed{\textbf{(B)}\ -15}</math>.
 
Factoring the equation gives <math>a(2 - b)</math>. From this we can see that to obtain the least possible value, <math>2 - b</math> should be negative, and should be as small as possible. To do so, <math>b</math> should be maximized. Because <math>2 - b</math> is negative, we should maximize the positive value of <math>a</math> as well. The maximum values of both <math>a</math> and <math>b</math> are <math>5</math>, so the answer is <math>5(2 - 5) = \boxed{\textbf{(B)}\ -15}</math>.
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== See also ==
 +
{{AMC10 box|year=2013|ab=B|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 15:45, 3 July 2013

Problem

Positive integers $a$ and $b$ are each less than $6$. What is the smallest possible value for $2 \cdot a - a \cdot b$?


$\textbf{(A)}\ -20\qquad\textbf{{(B)}}\ -15\qquad\textbf{{(C)}}\ -10\qquad\textbf{{(D)}}\ 0\qquad\textbf{{(E)}}\ 2$

Solution

Factoring the equation gives $a(2 - b)$. From this we can see that to obtain the least possible value, $2 - b$ should be negative, and should be as small as possible. To do so, $b$ should be maximized. Because $2 - b$ is negative, we should maximize the positive value of $a$ as well. The maximum values of both $a$ and $b$ are $5$, so the answer is $5(2 - 5) = \boxed{\textbf{(B)}\ -15}$.

See also

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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