note sk I am not sure if you had a chance to look at my node [id://482867]<p> I guess my approach is same as [Limbic~Region] as I was using subtraction of lower factorial terms with higher ones. <p>I copied the input list from one of your previous examples and I have <c> 45700</c> instead of <c>4570</c> but aside from that the implementation should be easy to understand. <p> I guess if you sort numerator and denominator and subtract them you should be all set. I have not proved this formally but here is a stab at a simple proof that shows sorting and subtracting should work...<p> Let the fraction be <P> <c> X! Y! Z! ------- a! b! c! WLOG let's assume that X > Y > Z and a > b > c. Also let's assume that b > Y , X > a (weaker assumption: Z > c)...<p> NOTE: if we divide X!/a! we will have (X-a) elements <p> To Prove: (X-a) + (Z-c) + (b-Y) is the shortest list one can find or in other words (X-a) + (Z-c) + (b-Y) <= (X-p) + (Z-q) + (r-Y) for any p,q,r in permutation(a,b,c) Proof: From the above equation since b > Y and Z > c, r should be equal to either a or b. If r = b then the solution is trivial<p> If r = a then we get (X-a) + (Z-c) + (b-Y) ?<= (X-b) + (Z-c) + (a-Y) canceling terms -a - c + b ?<= -b -c + a -a + b ?<= -b + a ====> YES since a > b we see that r = a is not the smallest list so r = b<p> Similarly we can also show that (X-a) + (Z-c) + (b-Y) <= (X-a) + (Y-c) + (b-Z) </c> I don't think this is a rigourous proof this method but i sort of feel sorting and subtracting should give us what we need...<p> cheers<p> SK<p> PS: I think there will be 47448 elements and not 47444 as you suggested? as you need to count the first element too..<P> 481987 483172