Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2007 AMC 12B Problems/Problem 17"

(→‎Cheap Solution that most people probably used)
(→‎Cheap Solution that most people probably used)
Line 23: Line 23:
 
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*0.01 = -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>.
+
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 11:42, 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 $a\cdot0.01 = -1,$ 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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_17&oldid=110915"