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Old 2013-05-12, 14:16   #1
Yamato
 
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Default An Egyptian Fraction

Write \frac{99}{10001} as a sum of distinct unit fractions (all numerators are 1), where all denominators are smaller than 10001.
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Old 2013-05-12, 15:01   #2
c10ck3r
 
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1/102+1/10517+1/228264101+1/188377806760266318
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Old 2013-05-12, 16:04   #3
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cl10ck3r, that doesn't meet the denominator requirement.

I get 99/10001 = 1/105 + 1/6165 + 1/9198 + 1/9590.

Last fiddled with by CRGreathouse on 2013-05-12 at 16:05
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Old 2013-05-12, 16:11   #4
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Quote:
Originally Posted by CRGreathouse View Post
cl10ck3r, that doesn't meet the denominator requirement.

I get 99/10001 = 1/105 + 1/6165 + 1/9198 + 1/9590.
Ah, glanced over that part.
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Old 2013-05-12, 19:35   #5
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Many solutions. Here's just a few:
v=[3,14,15];sum(k=1,#v,1/(73*v[k])+1/(137*v[k]))
v=[3,10,42,70];sum(k=1,#v,1/(73*v[k])+1/(137*v[k]))
v=[5,6,14,30];sum(k=1,#v,1/(73*v[k])+1/(137*v[k]))
v=[7,10,14,15,21,35,70];sum(k=1,#v,1/(73*v[k])+1/(137*v[k]))

Last fiddled with by Batalov on 2013-05-12 at 19:50 Reason: (add a long one)
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Old 2013-05-12, 21:15   #6
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Quote:
Originally Posted by Batalov View Post
Many solutions. Here's just a few:
v=[3,14,15];sum(k=1,#v,1/(73*v[k])+1/(137*v[k]))
v=[3,10,42,70];sum(k=1,#v,1/(73*v[k])+1/(137*v[k]))
v=[5,6,14,30];sum(k=1,#v,1/(73*v[k])+1/(137*v[k]))
v=[7,10,14,15,21,35,70];sum(k=1,#v,1/(73*v[k])+1/(137*v[k]))
Using (99/10001)/(1/73+1/137) = 33/70. Convenient, but not good if you're trying to minimize terms (as I was). Also interesting would be minimizing the maximum term, maximizing the minimum term, and minimizing the maximum ratio (or difference) between terms.
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Old 2013-05-12, 21:41   #7
Batalov
 
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Yes, indeed. There was a P.Euler problem in similar vein. (more than one, actually)

Can we count the number of (distinct) solutions of OP?
(with d1 < d2 < ... < dn < 10001)
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Old 2013-05-13, 05:11   #8
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Good question!

It seems very hard, unless the upper bound on the number of terms can be lowered substantially from the naive 98. I don't see why it should be, though.
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Old 2013-05-16, 15:50   #9
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Quote:
Originally Posted by CRGreathouse View Post
Using (99/10001)/(1/73+1/137) = 33/70. Convenient, but not good if you're trying to minimize terms (as I was). Also interesting would be minimizing the maximum term, maximizing the minimum term, and minimizing the maximum ratio (or difference) between terms.
Thx for your postet solutions, this is an interesting approach to solve this problem.

The "generic" method (which also works for prime denominators) is to choose a smooth number k which factors into small primes and write the numerator in (99k)/(10001k) as a sum of factors of the denominator. This leads to a unit fraction expansion for 99/10001, e.g. 1/247 + 1/292 + 1/949 + 1/1898 + 1/2603 + 1/3562 + 1/5548.

An interesting question is to how to find the expansion, where all denominators are bounded by a number as small as possible.
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