Difference between revisions of "2014 AMC 12B Problems/Problem 20"
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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. | ||
| + | == 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 12:25, 22 February 2014
Problem
For how many positive integers 
is 
?
Solution
The domain of the LHS implies that ![\[40<x<60\]](http://latex.artofproblemsolving.com/e/6/1/e61ac4e06141f931a998331029b57ad65f67c533.png)
Begin from the left hand side
![\[\log_{10}[(x-40)(60-x)]<2\]](http://latex.artofproblemsolving.com/6/7/4/6749fe9dcc8059ce47ccd96766d2b0d9d1f0eb3a.png)
![\[-x^2+100x-2500<0\]](http://latex.artofproblemsolving.com/8/c/b/8cb07d38ab8be9ccbb57dd02425ab79f4a9da74d.png)
![\[(x-50)^2>0\]](http://latex.artofproblemsolving.com/d/c/7/dc75fe198e27a34d5b4d4876eed002ce2eb24c26.png)
![\[x \not = 50\]](http://latex.artofproblemsolving.com/3/9/c/39c9aefab22bafbe506412f1eeed25a27f7cf9d5.png)
Hence, we have integers from 41 to 49 and 51 to 59. There are 
integers.
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
