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 2013-05-12, 14:16 #1 Yamato     Sep 2005 Berlin 2×3×11 Posts 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.
 2013-05-12, 15:01 #2 c10ck3r     Aug 2010 Kansas 547 Posts 1/102+1/10517+1/228264101+1/188377806760266318
 2013-05-12, 16:04 #3 CRGreathouse     Aug 2006 2·29·103 Posts 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
2013-05-12, 16:11   #4
c10ck3r

Aug 2010
Kansas

547 Posts

Quote:
 Originally Posted by CRGreathouse 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.

 2013-05-12, 19:35 #5 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×3×5×311 Posts 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)
2013-05-12, 21:15   #6
CRGreathouse

Aug 2006

2·29·103 Posts

Quote:
 Originally Posted by Batalov 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.

 2013-05-12, 21:41 #7 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·3·5·311 Posts 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)
 2013-05-13, 05:11 #8 CRGreathouse     Aug 2006 2·29·103 Posts 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.
2013-05-16, 15:50   #9
Yamato

Sep 2005
Berlin

1028 Posts

Quote:
 Originally Posted by CRGreathouse 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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