Difference between revisions of "2005 AMC 10A Problems/Problem 13"
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How many positive integers <math>n</math> satisfy the following condition: | How many positive integers <math>n</math> satisfy the following condition: | ||
− | <math> (130n)^{50} > n^{100} > 2^{200} </math> | + | <math> (130n)^{50} > n^{100} > 2^{200}\ ?</math> |
− | <math> \ | + | <math> \textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125 </math> |
==Solution== | ==Solution== | ||
− | <math> (130n)^{50} > n^{100} > 2^{200} </math> | + | We're given <math> (130n)^{50} > n^{100} > 2^{200} </math>, so |
− | <math> \sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}} </math> | + | <math> \sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}} </math> (because all terms are positive) and thus |
<math> 130n > n^2 > 2^4 </math> | <math> 130n > n^2 > 2^4 </math> | ||
− | <math> 130n | + | <math> 130n > n^2 > 16 </math> |
− | Solving each part | + | Solving each part separately: |
− | <math> n^2 > 16 </math> | + | <math> n^2 > 16 \Longrightarrow n > 4 </math> |
− | <math> n | + | <math> 130n > n^2 \Longrightarrow 130 > n </math> |
− | <math> | + | So <math> 4 < n < 130 </math>. |
− | <math> 130 < | + | 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 | + | ==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== | |
+ | {{AMC10 box|year=2005|ab=A|num-b=12|num-a=14}} | ||
− | + | [[Category:Introductory Number Theory Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 07:18, 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
which contains the same number of elements as
which clearly contains
values or choice
.
~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 |
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