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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 [[numbers]] a and b are chosen randomly from the set {<math>2, 2^2, 2^3, ..., 2^{25}</math>}. What is the probability that log_a b is an [[integer]]?
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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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<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>
  
 
== Solution ==
 
== Solution ==
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{{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>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>.
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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{(B) }\frac{31}{300}}</math>.
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== See also ==
 
== See also ==
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* [[2005 AMC 12A Problems/Problem 22 | Previous problem]]
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{{AMC12 box|year=2005|ab=A|num-b=22|num-a=24}}
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* [[2005 AMC 12A Problems/Problem 24 | Next problem]]
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* [[2005 AMC 12A Problems]]
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[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 19:35, 7 November 2013

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 $\frac{62}{600} = \boxed{\textbf{(B) }\frac{31}{300}}$.

See also

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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All AMC 12 Problems and Solutions

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