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Consider all even natural numbers. Every 4th number has a power of 4 (or $2^2$)

Every 8th number has a power of 8 (or $2^3$)

Every 16th number has a power of 16 (or $2^4$)

What is the average number of power of 2 in any random even number?

Since every 2nd number has a power of 2,and every 4th number is a power of 4 and so on

$\left(2 \times \dfrac{1}{2}\right) + \left(4 \times \dfrac{1}{4}\right) + \left(8 \times \dfrac{1}{8}\right) + \ldots$

This goes all the way to Infinity and hence the answer is $\infty$. Is this correct?

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A multiple of $4$ only has two powers of $2$ in its factorization and a multiple of $8$ only has three. Your sum should therefore be $$\left(1 \times \dfrac{1}{2}\right) + \left(2 \times \dfrac{1}{4}\right) + \left(3 \times \dfrac{1}{8}\right) + \ldots=2$$ because half the evens have exactly one factor of two, one quarter have exactly two factors of two and so on. Although it you can't pick a random natural number, this makes sense as the limit of the average number of factors of two in the even numbers up to $n$ as $n \to \infty$.

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  • $\begingroup$ thanks a lot.I wonder how is the sum 2?the 1st 2 add upto to 1....not sure how to add rest,unable to use GP formula directly $\endgroup$ – Rahul Shah Dec 13 '18 at 17:42
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    $\begingroup$ It is an arithmetico-geometric series. This particular one is solved on this site. Here and here. You can search for arithmetico-geometric $\endgroup$ – Ross Millikan Dec 13 '18 at 17:56
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I think you are double counting the multiples of $4$ since half of them have been counted as multiples of $2$. That multiple counting continues.

I think the sum (corrected as I suggest) for the power of $2$ dividing a "random integer", is $$ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots $$ which sums to $1$. That makes sense for the average power of $2$.

The result for a random even integer would be $2$.

I put "random integer" in quotes since defining a "random integer" is a little subtle.

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  • $\begingroup$ A "little subtle" ... like trisecting the angle. :) $\endgroup$ – John Hughes Dec 9 '18 at 21:50
  • $\begingroup$ @JohnHughes Not quite. I think you can make sense of that expected value as a limit. I didn't want to spend the time. $\endgroup$ – Ethan Bolker Dec 9 '18 at 21:54
  • $\begingroup$ Indeed... I meant that picking a uniform distribution on the integers might be a little problematic. $\endgroup$ – John Hughes Dec 9 '18 at 23:26

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