Difference between revisions of "2014 AMC 12B Problems/Problem 20"

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==Problem==
 
==Problem==
 
 
For how many positive integers <math>x</math> is <math>\log_{10}(x-40) + \log_{10}(60-x) < 2</math> ?
 
For how many positive integers <math>x</math> is <math>\log_{10}(x-40) + \log_{10}(60-x) < 2</math> ?
  
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<cmath>x \not = 50</cmath>
 
<cmath>x \not = 50</cmath>
 
Hence, we have integers from 41 to 49 and 51 to 59. There are <math>\boxed{\textbf{(B)} 18}</math> integers.
 
Hence, we have integers from 41 to 49 and 51 to 59. There are <math>\boxed{\textbf{(B)} 18}</math> integers.
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{{AMC12 box|year=2014|ab=B|num-b=19|num-a=21}}
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{{MAA Notice}}

Revision as of 12:30, 21 February 2014

Problem

For how many positive integers $x$ is $\log_{10}(x-40) + \log_{10}(60-x) < 2$ ?

$\textbf{(A) }10\qquad \textbf{(B) }18\qquad \textbf{(C) }19\qquad \textbf{(D) }20\qquad \textbf{(E) }$ infinitely many$\qquad$

Solution

The domain of the LHS implies that \[40<x<60\] Begin from the left hand side \[\log_{10}[(x-40)(60-x)]<2\] \[-x^2+100x-2500<0\] \[(x-50)^2>0\] \[x \not = 50\] Hence, we have integers from 41 to 49 and 51 to 59. There are $\boxed{\textbf{(B)} 18}$ integers.

2014 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 12 Problems and Solutions

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