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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==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. | ||
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+ | == 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
Contents
[hide]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.