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 12A Problems/Problem 23"

m (→‎Solution: typo 5->6)
(→‎Solution)
Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
−
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}^{25} \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}\ \mathrm{(B)}</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}\ \mathrm{(B)}</math>.

Revision as of 21:57, 20 February 2012

Problem

Two distinct numbers a and b are chosen randomly from the set $\{2, 2^2, 2^3, ..., 2^{25}\}$. What is the probability that $\mathrm{log}_a b$ is an integer?

$\mathrm{(A)}\ \frac{2}{25}\qquad \mathrm{(B)}\ \frac{31}{300}\qquad \mathrm{(C)}\ \frac{13}{100}\qquad \mathrm{(D)}\ \frac{7}{50}\qquad \mathrm{(E)}\ \frac 12$

Solution

Let $\log_a b = z$, so $a^z = b$. Define $a = 2^x$, $b = 2^y$; then $\left(2^x\right)^z = 2^{xz}= 2^y$, so $x|y$. Here we can just make a table and count the number of values of $y$ per value of $x$. The largest possible value of $x$ is 12, and we get $\sum_{x=1}^{12} \lfloor\frac {25}x-1\rfloor = 24 + 11 + 7 + 5 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 = 62$.

The total number of ways to pick two distinct numbers is $\frac{25!}{(25-2)!}= 25 \cdot 24 = 600$, so we get a probability of $\frac{62}{600} = \frac{31}{300}\ \mathrm{(B)}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_23&oldid=44918"