Counting the unusually anal way.

[lavalamp]

If you were counting up through the positive integers, perhaps to pass the time, and you started at 1, but you skipped out numbers that were squares, cubes and quints, and planned on passing a LOT of time. What number would be at place number 10^30?

[cheesehead]

Originally Posted by lavalamp

If you were counting up through the positive integers, perhaps to pass the time, and you started at 1, but you skipped out numbers that were squares, cubes and quints

By "quints", you mean fifth powers, I presume, even though I find no such usage in a quick search. Would "septs" be next?

[fivemack]

Originally Posted by lavalamp

If you were counting up through the positive integers, perhaps to pass the time, and you started at 1, but you skipped out numbers that were squares, cubes and quints, and planned on passing a LOT of time. What number would be at place number 10^30?

You can't start at 1, it's a square.

I think in place 1000000 you will find 1001101, and in place 10^30 you will find 1000000000000001000010000898910. I love and cherish the inclusion-exclusion principle.

[cheesehead]

Originally Posted by fivemack

You can't start at 1, it's a square.

I think in place 1000000 you will find 1001101,

< snip >

I love and cherish the inclusion-exclusion principle.

Number of squares below 1002001 = 1000. Number of cubes below 1030301 = 100. Number of fifth powers below 1048576 = 15. Number of sixth powers below 1771561 = 10. Number of tenth powers below 1048576 = 3. Number of fifteenth powers below 14348907 = 2. Number of thirtieth powers below 2^30 = 1.

1000000 + 1000 + 100 + 15 - 10 - 3 - 2 + 1 = 1001101

(I had to write it all out to find my initial mistake: forgetting tenth powers.)

[lavalamp]

Originally Posted by cheesehead

By "quints", you mean fifth powers, I presume, even though I find no such usage in a quick search. Would "septs" be next?

I do, and I guess they would.

Originally Posted by fivemack

You can't start at 1, it's a square.

Good point, I was just trying to get across that you don't start at 0, but then I suppose 0 is a perfect square and cube too, oh well.

I guess this puzzle didn't take too long to crack then, but perhaps it could be extended.

I've tried to word this in the most specific way I know how, so hopefully the meaning is clear.

For a prime p, find the positive integer number in 10^p# place, where p# is the primorial function, when you skip all prime powers up to and including p.

I don't think there can be a forumla for this, since at some point you're going to end up skipping so many numbers there's another square number to account for skipping and then some more and then a cube and so on. However surely there is some algorithm that can output the correct answer.

[R.D. Silverman]

Originally Posted by lavalamp

<snip>

I don't think there can be a forumla for this

And you would be wrong. Such a formula exists. It is in the form
of a product.

Hint: Look up Mertens' Thm.

[Uncwilly]

Originally Posted by cheesehead

Number of squares below 1002001 = 1000. Number of cubes below 1030301 = 100.

How many of those squares are also cubes or ..... ?

[fivemack]

Originally Posted by Uncwilly

How many of those squares are also cubes or ..... ?

All of those squares which are also cubes are sixth powers, which is why he subtracts off the number of sixth powers.

Except that some of those sixth powers are thirtieth powers, so he adds the number of thirtieth powers.

[cheesehead]

Originally Posted by cheesehead

1000000 + 1000 + 100 + 15 - 10 - 3 - 2 + 1 = 1001101

I should have mentioned

http://mathworld.wolfram.com/Inclusi...Principle.html

and

http://en.wikipedia.org/wiki/Inclusi...sion_principle

[R.D. Silverman]

Originally Posted by fivemack

All of those squares which are also cubes are sixth powers, which is why he subtracts off the number of sixth powers.

Except that some of those sixth powers are thirtieth powers, so he adds the number of thirtieth powers.

Look at the fraction that are NOT 2nd, 3rd, 5th powers etc.
This fraction is given by a simple product formula as I stated.

Look up Mertens' Theorem.