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

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\textbf{(D) }20\qquad
 
\textbf{(D) }20\qquad
 
\textbf{(E) }\text{infinitely many}\qquad</math>
 
\textbf{(E) }\text{infinitely many}\qquad</math>
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== Video Solution ==
 
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https://youtu.be/RdIIEhsbZKw?t=1088
 
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~ pi_is_3.14
 
  
 
==Solution==
 
==Solution==
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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.
 +
 +
== Video Solution ==
 +
https://youtu.be/RdIIEhsbZKw?t=1088
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 +
~ pi_is_3.14
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2014|ab=B|num-b=19|num-a=21}}
 
{{AMC12 box|year=2014|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:16, 27 January 2021

Problem

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

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

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.

Video Solution

https://youtu.be/RdIIEhsbZKw?t=1088

~ pi_is_3.14

See also

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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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