Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AMC 10A Problems/Problem 13"

(→‎Solution)
(→‎Solution 2)
 
(2 intermediate revisions by the same user not shown)
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\}</math> which contains the same number of elements as <math>\left\{1,\ 2,\ 3,\ \cdots ,\ 124,\ 125\right\}</math> which clearly contains <math>125</math> values or choice <math>\boxed{\textbf{(E) } 125}</math>.
 +
 +
~JH. L
  
 
==See also==
 
==See also==

Latest revision as of 06:18, 16 July 2022

Problem

How many positive integers $n$ satisfy the following condition:

$(130n)^{50} > n^{100} > 2^{200}\ ?$

$\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125$

Solution

We're given $(130n)^{50} > n^{100} > 2^{200}$, so

$\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}}$ (because all terms are positive) and thus

$130n > n^2 > 2^4$

$130n > n^2 > 16$

Solving each part separately:

$n^2 > 16 \Longrightarrow n > 4$

$130n > n^2 \Longrightarrow 130 > n$

So $4 < n < 130$.

Therefore the answer is the number of positive integers over the interval $(4,130)$ which is $129-5+1 = \boxed{\textbf{(E) }125}$


Solution 2

We're given $\left(130n\right)^{50}>n^{100}>2^{200}$.

Alternatively to solution 1, first deal with the first half: $\left(130n\right)^{50}>\left(n^{2}\right)^{50}$. Because the exponents are equal, we can ignore them and solve for $n$: $130n>n^{2}$, or $n<130$.

The second half: $n^{100}>2^{200}$, or $n^{100}>4^{100}$, which means $n>4$.

Therefore $4<n<130$ and $n=\left\{5,\ 6,\ 7,\ \cdots ,\ 128,\ 129\right\}$ which contains the same number of elements as $\left\{1,\ 2,\ 3,\ \cdots ,\ 124,\ 125\right\}$ which clearly contains $125$ values or choice $\boxed{\textbf{(E) } 125}$.

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_13&oldid=175966"