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

Revision as of 15:17, 9 September 2007

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 $a = 2^x$ , $\displaystyle b = 2^y$ ; then $\left(2^x\right)^z = 2^{xz}= 2^y$ , so $\displaystyle x|y$ . 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 $\displaystyle \sum_{x=1}^{25} \lfloor\frac {25}x-1\rfloor = 24 + 11 + 7 + 5 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 = 62$ .

The total number of ways to pick two distinct numbers is $\frac{25!}{(25-2)!} \displaystyle = 25 \cdot 24 = 600$ , so we get a probability of $\frac{62}{500} = \frac{31}{300}\ \mathrm{(B)}$ .

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

@@ Line 10: / Line 10: @@
 == See also ==
-{{AMC12 box|year=2007|ab=A|num-b=22|num-a=24}}
+{{AMC12 box|year=2005|ab=A|num-b=22|num-a=24}}
 [[Category:Introductory Combinatorics Problems]]

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

Revision as of 15:17, 9 September 2007

Problem

Solution

See also