2005 AMC 10A Problems/Problem 13: Difference between revisions
Sevenoptimus (talk | contribs) Improved formatting/explanations and removed Solution 2, since it's really the same as Solution 1 |
Sevenoptimus (talk | contribs) m Fixed solution heading |
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==Solution | ==Solution== | ||
Since <math>n > 0</math>, all <math>3</math> terms of the inequality are positive, so we may take the <math>50</math>th root, yielding | Since <math>n > 0</math>, all <math>3</math> terms of the inequality are positive, so we may take the <math>50</math>th root, yielding | ||
Latest revision as of 16:42, 1 July 2025
Problem
How many positive integers 
satisfy the following condition:
![\[\left(130n\right)^{50} > n^{100} > 2^{200} \ \text{?}\]](http://latex.artofproblemsolving.com/b/3/7/b371e65c189b0ab26e841e29c6668556880e7f6b.png)
Solution
Since 
, all 
terms of the inequality are positive, so we may take the 
th root, yielding
![\begin{align*}&\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}} \\ \iff &130n > n^2 > 2^4 \\ \iff &130n > n^2 > 16.\end{align*}](http://latex.artofproblemsolving.com/8/c/e/8ce506d0c1da0714f2cbf52acf2c37e376cfcd6c.png)
Solving each part separately, while noting that , therefore gives 
and 
.
Hence the solution is 
, and therefore the answer is the number of positive integers in the open 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 | ||
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