Difference between revisions of "2005 AMC 10A Problems/Problem 13"
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The second half: <math>n^{100}>2^{200}</math>, or <math>n^{100}>4^{100}</math>, which means <math>n>4</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 <math>125</math> values or choice <math>\boxed{\textbf{(E) } 125}</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 | ~JH. L |
Latest revision as of 06:18, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.