Difference between revisions of "2014 AMC 12B Problems/Problem 20"
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+ | ==Problem== | ||
+ | For how many positive integers <math>x</math> is <math>\log_{10}(x-40) + \log_{10}(60-x) < 2</math> ? | ||
+ | |||
+ | <math>\textbf{(A) }10\qquad | ||
+ | \textbf{(B) }18\qquad | ||
+ | \textbf{(C) }19\qquad | ||
+ | \textbf{(D) }20\qquad | ||
+ | \textbf{(E) }\text{infinitely many}\qquad</math> | ||
+ | |||
+ | ==Solution== | ||
The domain of the LHS implies that <cmath>40<x<60</cmath> Begin from the left hand side | The domain of the LHS implies that <cmath>40<x<60</cmath> Begin from the left hand side | ||
− | <cmath>log_{10}[(x-40)(60-x)]<2</cmath> | + | <cmath>\log_{10}[(x-40)(60-x)]<2</cmath> |
<cmath>-x^2+100x-2500<0</cmath> | <cmath>-x^2+100x-2500<0</cmath> | ||
<cmath>(x-50)^2>0</cmath> | <cmath>(x-50)^2>0</cmath> | ||
<cmath>x \not = 50</cmath> | <cmath>x \not = 50</cmath> | ||
− | Hence, we have integers from 41 to 49 and 51 to 59. There are 18 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 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2014|ab=B|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} |
Revision as of 08:16, 27 January 2021
Contents
Problem
For how many positive integers is ?
Solution
The domain of the LHS implies that Begin from the left hand side Hence, we have integers from 41 to 49 and 51 to 59. There are 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.