Difference between revisions of "2005 AMC 12A Problems/Problem 23"
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Let <math>\log_a b = z</math>, so <math>a^z = b</math>. Define <math>a = 2^x</math>, <math>b = 2^y</math>; then <math>\left(2^x\right)^z = 2^{xz}= 2^y</math>, so <math>x|y</math>. Here we can just make a table and count the number of values of <math>y</math> per value of <math>x</math>. The largest possible value of <math>x</math> is 12, and we get <math>\sum_{x=1}^{12} \lfloor\frac {25}x-1\rfloor = 24 + 11 + 7 + 5 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 = 62</math>. | Let <math>\log_a b = z</math>, so <math>a^z = b</math>. Define <math>a = 2^x</math>, <math>b = 2^y</math>; then <math>\left(2^x\right)^z = 2^{xz}= 2^y</math>, so <math>x|y</math>. Here we can just make a table and count the number of values of <math>y</math> per value of <math>x</math>. The largest possible value of <math>x</math> is 12, and we get <math>\sum_{x=1}^{12} \lfloor\frac {25}x-1\rfloor = 24 + 11 + 7 + 5 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 = 62</math>. | ||
− | The total number of ways to pick two distinct numbers is <math>\frac{25!}{(25-2)!}= 25 \cdot 24 = 600</math>, so we get a probability of <math>\frac{62}{600} = \frac{31}{300} | + | The total number of ways to pick two distinct numbers is <math>\frac{25!}{(25-2)!}= 25 \cdot 24 = 600</math>, so we get a probability of <math>\frac{62}{600} = \boxed{\textbf{(C) }\frac{31}{300}}</math>. |
== See also == | == See also == |
Revision as of 20:35, 7 November 2013
Problem
Two distinct numbers a and b are chosen randomly from the set . What is the probability that is an integer?
Solution
Let , so . Define , ; then , so . Here we can just make a table and count the number of values of per value of . The largest possible value of is 12, and we get .
The total number of ways to pick two distinct numbers is , so we get a probability of .
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.