An Egyptian Fraction
Write [TEX]\frac{99}{10001}[/TEX] as a sum of distinct unit fractions (all numerators are 1), where all denominators are [B]smaller[/B] than 10001. :smile:

[SPOILER]1/102+1/10517+1/228264101+1/188377806760266318[/SPOILER]

cl10ck3r, that doesn't meet the denominator requirement.
I get [spoiler]99/10001 = 1/105 + 1/6165 + 1/9198 + 1/9590[/spoiler]. 
[QUOTE=CRGreathouse;340147]cl10ck3r, that doesn't meet the denominator requirement.
I get [spoiler]99/10001 = 1/105 + 1/6165 + 1/9198 + 1/9590[/spoiler].[/QUOTE] Ah, glanced over that part. 
Many solutions. Here's just a few:
[SPOILER]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]))[/SPOILER] 
[QUOTE=Batalov;340166]Many solutions. Here's just a few:
[SPOILER]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]))[/SPOILER][/QUOTE] Using [spoiler](99/10001)/(1/73+1/137) = 33/70[/spoiler]. 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. 
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 d[SUB]1[/SUB] < d[SUB]2[/SUB] < ... < d[SUB]n[/SUB] < 10001) 
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. 
[QUOTE=CRGreathouse;340173]Using [spoiler](99/10001)/(1/73+1/137) = 33/70[/spoiler]. 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.[/QUOTE]
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. 
All times are UTC. The time now is 15:49. 
Powered by vBulletin® Version 3.8.11
Copyright ©2000  2020, Jelsoft Enterprises Ltd.