Easy math puzzles
Poll
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36 members have voted
I would say this now constitutes an interesting open question in elementary number theory -- whether another such N exists ... such questions are hard to come by.Quote: ThatDonGuyQuote: ThatDonGuyQuote: teliotFor any integer N, consider the list of all primes that are in the half-open interval (N/2,N]. Call them p1, p2, p3 ... , pk. The question is, for which N is N!/(p1*p2*p3*...*pk) a perfect square? In other words, divide N! by all the primes greater than N/2 and less than or equal to N (the Bertrand's Postulate primes) and ask if what's leftover is a perfect square.
Any others? Infinitely many? This isn't in the OEIS (at least that I can find).
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I wrote something that can calculate for larger numbers, and have not found any others for any N < 2000.
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Update: no other solutions for N <= 3300. I stopped there as it is taking about 12 seconds to check each further number.
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Quote: teliotThe question about N! not being a perfect square for N >= 2 leads to a question that I don't know the answer to. Here goes.
For any integer N, consider the list of all primes that are in the half-open interval (N/2,N]. Call them p1, p2, p3 ... , pk. The question is, for which N is N!/(p1*p2*p3*...*pk) a perfect square? In other words, divide N! by all the primes greater than N/2 and less than or equal to N (the Bertrand's Postulate primes) and ask if what's leftover is a perfect square.
Here are the five examples I found searching N <= 22 (the largest N for which Excel accurately computes N!).
With N = 6, then we divide 6! by 5 and 6!/5 = 144 is a perfect square.
With N = 7, then we divide 7! by 5*7 = 35 and get 7!/35 = 144, again a perfect square.
With N = 10, then we divide 10! by 7 (the only prime in the interval (5,10]) and get 10!/7 = 518400 = 720^2 is a perfect square.
With N = 11, then we divide 11! by 11*7 and again get 10!/7 = 518400 = 720^2.
With N = 20, then we divide 20! by 19*17*13*11 and get 52672757760000 = 7257600^2.
Any others? Infinitely many? This isn't in the OEIS (at least that I can find).
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There is a paper by Jitsuro Nagura, which proves that for any real number x≥9, there exists a prime p such that x≤p<4x/3. If N≥35, we can apply this theorem with x=(N+1)/4 to obtain that there exists a prime p such that (N+1)/4≤p<(N+1)/3. Then p>N/4 clearly, and 3p<N+1, so 3p≤N, because 3p and N are integers, therefore p≤N/3.The only numbers among 1,2,...,N which are divisible by p are p, 2p and 3p, so N! is divisible by p^3 but not by p^4. After factoring out the primes between N and 2N, the resulting number will still be divisible by p^3, but not by p^4, so it cannot be a square.
This means that if N! divided by the product of primes in (N/2,N] is a square then N≤34. It is easy to check that the only solutions are N=2,3,6,7,10,11,20.
An elegant and simple argument. Many thanks!Quote: GMQuote: teliotThe question about N! not being a perfect square for N >= 2 leads to a question that I don't know the answer to. Here goes.
For any integer N, consider the list of all primes that are in the half-open interval (N/2,N]. Call them p1, p2, p3 ... , pk. The question is, for which N is N!/(p1*p2*p3*...*pk) a perfect square? In other words, divide N! by all the primes greater than N/2 and less than or equal to N (the Bertrand's Postulate primes) and ask if what's leftover is a perfect square.
Here are the five examples I found searching N <= 22 (the largest N for which Excel accurately computes N!).
With N = 6, then we divide 6! by 5 and 6!/5 = 144 is a perfect square.
With N = 7, then we divide 7! by 5*7 = 35 and get 7!/35 = 144, again a perfect square.
With N = 10, then we divide 10! by 7 (the only prime in the interval (5,10]) and get 10!/7 = 518400 = 720^2 is a perfect square.
With N = 11, then we divide 11! by 11*7 and again get 10!/7 = 518400 = 720^2.
With N = 20, then we divide 20! by 19*17*13*11 and get 52672757760000 = 7257600^2.
Any others? Infinitely many? This isn't in the OEIS (at least that I can find).
link to original post
There is a paper by Jitsuro Nagura, which proves that for any real number x≥9, there exists a prime p such that x≤p<4x/3. If N≥35, we can apply this theorem with x=(N+1)/4 to obtain that there exists a prime p such that (N+1)/4≤p<(N+1)/3. Then p>N/4 clearly, and 3p<N+1, so 3p≤N, because 3p and N are integers, therefore p≤N/3.The only numbers among 1,2,...,N which are divisible by p are p, 2p and 3p, so N! is divisible by p^3 but not by p^4. After factoring out the primes between N and 2N, the resulting number will still be divisible by p^3, but not by p^4, so it cannot be a square.
This means that if N! divided by the product of primes in (N/2,N] is a square then N≤34. It is easy to check that the only solutions are N=2,3,6,7,10,11,20.
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This reminds me of a theorem I have long since forgotten, an improvement on Bertrand's Postulate that follows from the PNT, that for any e > 0 there is a positive integer N such that for all n > N, there exists a prime p with n <= p < (1+e)n. The result you quoted was this bound N being explicitly computed when e equals 1/3.
Again fabulous work.
By the way, I tried to look up your reference, Wikipedia pointed to this result by Nagura: there is always a prime between n and 6/5*n for n >= 25. I couldn't find the result in the paper you referenced. But your proof holds, we would just have to run the exhaustive search out a bit further.
Quote: teliot
By the way, I tried to look up your reference, Wikipedia pointed to this result by Nagura: there is always a prime between n and 6/5*n for n >= 25. I couldn't find the result in the paper you referenced. But your proof holds, we would just have to run the exhaustive search out a bit further.
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There is a link on that page to download the whole paper, the statement of the theorem is at the bottom of the penultimate page.
Okay -- found the download button and the theorem. For reference:Quote: GMQuote: teliot
By the way, I tried to look up your reference, Wikipedia pointed to this result by Nagura: there is always a prime between n and 6/5*n for n >= 25. I couldn't find the result in the paper you referenced. But your proof holds, we would just have to run the exhaustive search out a bit further.
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There is a link on that page to download the whole paper, the statement of the theorem is at the bottom of the penultimate page.
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By the way, one consequence of an easy generalization of your argument is that for any integer k there exists an integer N such that for n > N there is a prime p so that p^k exactly divides n!.
Consider a pyramid with a base of a regular N-gon. All other sides are triangles and every edge is of unit length.
1) What is the maximum N?
2) What is the angle between the base and a side?
3) What is the angle between two adjacent sides?
Quote: Gialmere
Consider a pyramid with a base of a regular N-gon. All other sides are triangles and every edge is of unit length.
1) What is the maximum N?
2) What is the angle between the base and a side?
3) What is the angle between two adjacent sides?
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1. 5. At six, the pyramid would be flat. I'm not sure if that counts as a pyramid.
2. I'll bother to answer that if you confirm my answer of 5.
3. Based on a pentagonal base, 108.
I have a feeling I am not understanding the question correctly.
Quote: Gialmere
Consider a pyramid with a base of a regular N-gon. All other sides are triangles and every edge is of unit length.
...
2) What is the angle between the base and a side?
...
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