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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>
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==Cheap Solution that most people probably used==
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Led <math>b=0.1</math>. Then <math>a/100 = -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 11:40, 5 November 2019

Problem 17

If $a$ is a nonzero integer and $b$ is a positive number such that $ab^2=\log_{10}b$, what is the median of the set $\{0,1,a,b,1/b\}$?

$\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ a \qquad \mathrm{(D)}\ b \qquad \mathrm{(E)}\ \frac{1}{b}$

Solution

Note that if $a$ is positive, then, the equation will have no solutions for $b$. This becomes more obvious by noting that at $b=1$, $ab^2 > \log_{10} b$. The LHS quadratic function will increase faster than the RHS logarithmic function, so they will never intersect.

This puts $a$ as the smallest in the set since it must be negative.

Checking the new equation: $-b^2 = \log_{10}b$

Near $b=0$, $-b^2 > \log_{10} b$ but at $b=1$, $-b^2 < \log_{10} b$

This implies that the solution occurs somewhere in between: $0 < b < 1$

This also implies that $\frac{1}{b} > 1$

This makes our set (ordered) $\{a,0,b,1,1/b\}$

The median is $b \Rightarrow \mathrm {(D)}$

Cheap Solution that most people probably used

Led $b=0.1$. Then $a/100 = -1,$ giving $a=-100$. Then the ordered set is $\{-100, 0, 0.1, 1, 10\}$ and the median is $0.1=b,$ so the answer is $\mathrm {(D)}$.

See Also

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AMC 12 Problems and Solutions

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