Difference between revisions of "2005 AMC 10A Problems/Problem 13"
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<math> 130n > n^2 > 16 </math> | <math> 130n > n^2 > 16 </math> | ||
− | Solving each part | + | Solving each part separately: |
<math> n^2 > 16 \Longrightarrow n > 4 </math> | <math> n^2 > 16 \Longrightarrow n > 4 </math> | ||
Line 25: | Line 25: | ||
Therefore the answer is the number of [[positive integer]]s over the interval <math> (4,130) </math> which is <math> 125 \Longrightarrow \mathrm{(E)} </math>. | Therefore the answer is the number of [[positive integer]]s over the interval <math> (4,130) </math> which is <math> 125 \Longrightarrow \mathrm{(E)} </math>. | ||
− | ==See | + | ==See also== |
− | + | {{AMC10 box|year=2005|ab=A|num-b=12|num-a=14}} | |
− | + | [[Category:Introductory Number Theory Problems]] | |
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{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:38, 8 December 2020
Problem
How many positive integers satisfy the following condition:
?
Solution
We're given , so
(because all terms are positive) and thus
Solving each part separately:
So .
Therefore the answer is the number of positive integers over the interval which is .
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.