How many different rational numbers between 1/1000 and 1000 can be written either as a power of 2 or as a power of 3, where the exponent is a (possibly negative) integer?

I got 31; is this correct?

mathbum
Oct 6, 2018

#1**+1 **

\(2^9=512\)

\(3^6=729\)

Those are the largest positive powers of 2 and 3 under 1000.

\(2^{-9}=0.001953.....\)

\(3^{-6}=0.0013717.....\)

These are the smallest negative powers greater than \(\frac{1}{1000}\)

Therefore, we have \(9-(-9)+1=19\) different powers of 2 between \(\frac{1}{1000}\) and 1000.

We also have \(6-(-6)+1=13\) different powers of 3 between \(\frac{1}{1000}\) and 1000.

This gives us \(19+13=32\) total integers...

32 is the answer I got but someone is going to have to check over my work :b

Guest Oct 6, 2018

#2**+1 **

The problem is symmetric in x vs. 1/x so we can find the number of integers between 0 and 1000 that are either a power of two or 3 and just double the answer.

Note that any non-zero power of 2 is even and any non-zero power of 3 is odd so these two sets are disjoint.

There are 10 powers of 2, 0-9, and 6 non-zero powers of 3 that are less than 100.

We use non-zero power for 3 since we only want to count 1 = 2^{0} = 3^{0} once

That gets us 16 from 0 to 1000. Doubling this we get 32 but we don't want to count \(1 = \dfrac{1}{1}\)

twice so we subtract 1 from this getting 31 as you found.

Rom
Oct 6, 2018