Difference between revisions of "2005 AMC 10A Problems/Problem 13"
Dairyqueenxd (talk | contribs) (→Solution) |
Erics son07 (talk | contribs) (→Solution) |
||
Line 24: | Line 24: | ||
Therefore the answer is the number of [[positive integer]]s over the interval <math> (4,130) </math> which is <math>129-5+1 = \boxed{\textbf{(E) }125}</math> | Therefore the answer is the number of [[positive integer]]s over the interval <math> (4,130) </math> which is <math>129-5+1 = \boxed{\textbf{(E) }125}</math> | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | We're given <math>\left(130n\right)^{50}>n^{100}>2^{200}</math>. | ||
+ | |||
+ | Alternatively to solution 1, first deal with the first half: <math>\left(130n\right)^{50}>\left(n^{2}\right)^{50}</math>. Because the exponents are equal, we can ignore them and solve for <math>n</math>: <math>130n>n^{2}</math>, or <math>n<130</math>. | ||
+ | |||
+ | The second half: <math>n^{100}>2^{200}</math>, or <math>n^{100}>4^{100}</math>, which means <math>n>4</math>. | ||
+ | |||
+ | Therefore <math>4<n<130</math> and <math>n=\left\{5,\ 6,\ 7,\ \cdots ,\ 128,\ 129\right\} which contains </math>125<math> values or choice </math>\boxed{\textbf{(E) } 125}$. | ||
+ | |||
+ | ~JH. L | ||
==See also== | ==See also== |
Revision as of 06:16, 16 July 2022
Contents
Problem
How many positive integers satisfy the following condition:
Solution
We're given , so
(because all terms are positive) and thus
Solving each part separately:
So .
Therefore the answer is the number of positive integers over the interval which is
Solution 2
We're given .
Alternatively to solution 1, first deal with the first half: . Because the exponents are equal, we can ignore them and solve for : , or .
The second half: , or , which means .
Therefore and 125\boxed{\textbf{(E) } 125}$.
~JH. L
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.