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Difference between revisions of "2005 AMC 12A Problems/Problem 23"

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== Problem ==
 
== Problem ==
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Two distinct [[number]]s a and b are chosen randomly from the [[set]] <math>\{2, 2^2, 2^3, ..., 2^{25}\}</math>. What is the [[probability]] that <math>\mathrm{log}_a b</math> is an [[integer]]?
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Two distinct numbers <math>a</math> and <math>b</math> are chosen randomly from the set <math>\{2, 2^2, 2^3, \ldots, 2^{25}\}</math>. What is the probability that <math>\log_{a}b</math> is an integer?
  
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<math>\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</math>
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<math>\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{1}{2}</math>
  
 
== Solution ==
 
== Solution ==
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Let <math>\log_a b = z</math>, so <math>a^z = b</math>. Define <math>a = 2^x</math>, <math>\displaystyle b = 2^y</math>; then <math>\left(2^x\right)^z = 2^{xz}= 2^y</math>, so <math>\displaystyle 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>\displaystyle \sum_{x=1}^{25} \lfloor\frac {25}x-1\rfloor = 24 + 11 + 7 + 5 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 = 62</math>.  
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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 \mid 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 <math>12</math>, and so we get <math>\sum_{x=1}^{12} \left\lfloor\frac{25}{x}-1\right\rfloor = 24 + 11 + 7 + 5 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 = 62</math>.  
  
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The total number of ways to pick two distinct numbers is <math>\frac{25!}{(25-2)!} \displaystyle = 25 \cdot 24 = 600</math>, so we get a probability of <math>\frac{62}{500} = \frac{31}{300}\ \mathrm{(B)}</math>.
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The total number of ways to pick two distinct numbers (where the order matters) is <math>\frac{25!}{(25-2)!}= 25 \cdot 24 = 600</math>, so we get a probability of <math>\frac{62}{600} = \boxed{\textbf{(B) }\frac{31}{300}}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 15:11, 1 July 2025

Problem

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

$\mathrm{(A)}\ \frac{2}{25}\qquad \mathrm{(B)}\ \frac{31}{300}\qquad \mathrm{(C)}\ \frac{13}{100}\qquad \mathrm{(D)}\ \frac{7}{50}\qquad \mathrm{(E)}\ \frac{1}{2}$

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 $x \mid y$. 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 so we get $\sum_{x=1}^{12} \left\lfloor\frac{25}{x}-1\right\rfloor = 24 + 11 + 7 + 5 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 = 62$.

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

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

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