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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
avatar
+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
avatar+2362 
+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 = 20 = 30 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
edited by Rom  Oct 6, 2018
edited by Rom  Oct 6, 2018
edited by Rom  Oct 6, 2018
 #3
avatar+93644 
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That is an interesting question and good answers from both guest and Rom.

Thanks :)

Melody  Oct 6, 2018

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