Difference between revisions of "2014 AMC 12B Problems/Problem 20"
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==Problem== | ==Problem== | ||
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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}} | ||
+ | {{MAA Notice}} |
Revision as of 12:30, 21 February 2014
Problem
For how many positive integers is ?
infinitely many
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.
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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