Difference between revisions of "2013 AMC 10B Problems/Problem 5"
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Positive integers <math>a</math> and <math>b</math> are each less than <math>6</math>. What is the smallest possible value for <math>2 \cdot a - a \cdot b</math>? | Positive integers <math>a</math> and <math>b</math> are each less than <math>6</math>. What is the smallest possible value for <math>2 \cdot a - a \cdot b</math>? | ||
<math>\textbf{(A)}\ -20\qquad\textbf{{(B)}}\ -15\qquad\textbf{{(C)}}\ -10\qquad\textbf{{(D)}}\ 0\qquad\textbf{{(E)}}\ 2</math> | <math>\textbf{(A)}\ -20\qquad\textbf{{(B)}}\ -15\qquad\textbf{{(C)}}\ -10\qquad\textbf{{(D)}}\ 0\qquad\textbf{{(E)}}\ 2</math> | ||
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| + | == Solution 1 == | ||
| + | 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}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:01, 19 February 2025
Problem
Positive integers 
and 
are each less than 
. What is the smallest possible value for 
?
Solution 1
Factoring the equation gives 
. From this we can see that to obtain the least possible value, 
should be negative, and should be as small as possible. To do so, should be maximized. Because is negative, we should maximize the positive value of as well. The maximum values of both and are 
, so the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
