Difference between revisions of "2007 AMC 12B Problems/Problem 17"
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The median is <math>b \Rightarrow \mathrm {(D)}</math> | The median is <math>b \Rightarrow \mathrm {(D)}</math> | ||
==Cheap Solution that most people probably used== | ==Cheap Solution that most people probably used== | ||
− | Led <math>b=0.1</math>. Then <math>a | + | Led <math>b=0.1</math>. Then <math>a\cdot0.01 = -1,</math> giving <math>a=-100</math>. Then the ordered set is <math>\{-100, 0, 0.1, 1, 10\}</math> and the median is <math>0.1=b,</math> so the answer is <math>\mathrm {(D)}</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2007|ab=B|num-b=16|num-a=18}} | {{AMC12 box|year=2007|ab=B|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:42, 5 November 2019
Problem 17
If is a nonzero integer and is a positive number such that , what is the median of the set ?
Solution
Note that if is positive, then, the equation will have no solutions for . This becomes more obvious by noting that at , . The LHS quadratic function will increase faster than the RHS logarithmic function, so they will never intersect.
This puts as the smallest in the set since it must be negative.
Checking the new equation:
Near , but at ,
This implies that the solution occurs somewhere in between:
This also implies that
This makes our set (ordered)
The median is
Cheap Solution that most people probably used
Led . Then giving . Then the ordered set is and the median is so the answer is .
See Also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 12 Problems and Solutions |
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