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john !!
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ab=25!
0 < a/b < 1

how many solutions a,b are there to the above constraints?
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dvs
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2^23?
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J.F.N
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(Original post by dvs)
2^23?
Yes, I thought so too.. that for n! the number of solutions is 2^(n-2).. but for n=6, 2^(6-2)=16 but there are only 15 solutions in integers.. similarly for n=7, 2^(7-2)=32 but there are only 30 solutions in integers.
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dvs
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(Original post by J.F.N)
Yes, I thought so too.. that for n! the number of solutions is 2^(n-2).. but for n=6, 2^(6-2)=16 but there are only 15 solutions in integers.. similarly for n=7, 2^(7-2)=32 but there are only 30 solutions in integers.
Yeah.. I just noticed.

Perhaps 2^(n-2) - (n-5) instead? It'd be pretty neat if that was true, since factorials have this nice bond to 2 & 5.
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J.F.N
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Nope, for n=8 (yes, I have nothing better to do!) the number of solutions is 48.
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dvs
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(Original post by J.F.N)
Nope, for n=8 (yes, I have nothing better to do!) the number of solutions is 48.
48?! It's useless to try looking for decent patterns now..
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J.F.N
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Yes, I'd venture to guess that the only way to do it is by brute force. Have fun with 25!, mik1w.
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john !!
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trick. there are infintely many solutions. modified question:

ab=25!
0 < a/b < 1
a and b are relatively prime

how many solutions a,b are there to the above constraints?
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J.F.N
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(Original post by mik1w)
trick. there are infintely many solutions. modified question:

ab=25!
0 < a/b < 1
a and b are relatively prime

how many solutions a,b are there to the above constraints?
Its obvious that there are infinitely many real solutions. I thought you were asking for integer solutions.

I dont believe the additional constraint of (a,b)=1 is anymore helpful for finding the general number of solutions for n!. These are the number of solutions satisfying the above constraints up to n=7

2!=1
3!=2
4!=2
5!=4
6!=4
7!=9

No visible pattern, if you ask me.
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john !!
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25! cannot have in its prime factorisation any prime higher than 25.
so in the prime factorisation there are 9 distinct primes: 2,3,5,7,11,13,17,19,23.

when building a fraction from these primes, you can put a prime in either the numerator or denominator- but you can't split prims between them because then the fraction would be reducable. this makes 2^9 choices because each of the 9 primes could go in one of two choices.

however half of these 2^9 possibilities have a numerator larger than the denominator, so the number of solutions is 2^8.

to find a general solution would be much harder...
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