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2005 AMC 12A Problems/Problem 23

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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}...

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}^{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)!}= 25 \cdot 24 = 600, so we get a probability of \frac{62}{600} = \frac{31}{300}\ \mathrm{(B)}.

See also

2005 AMC 12A (ProblemsResources)
Preceded by
Problem 22
Followed by
Problem 24
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