mersenneforum.org March 2016
 Register FAQ Search Today's Posts Mark Forums Read

 2016-03-02, 15:48 #1 Xyzzy     "Mike" Aug 2002 2·5·11·71 Posts March 2016
2016-03-02, 18:09   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by Xyzzy https://www.research.ibm.com/haifa/p...March2016.html
I haven't narrowed it enough to do anything useful with the question.

 2016-03-02, 18:16 #3 axn     Jun 2003 12AD16 Posts Using brute force exhaustive search (O(10^8)), I was able to find two solutions for n=15. Here are the smaller solutions: Code: n=5 65766 n=5 69714 n=7 6317056 n=9 169605719 n=11 96528287587 There were none for n=13. I haven't looked for patterns with even number of digits.
2016-03-02, 18:26   #4
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by axn Using brute force exhaustive search (O(10^8)), I was able to find two solutions for n=15. Here are the smaller solutions: Code: n=5 65766 n=5 69714 n=7 6317056 n=9 169605719 n=11 96528287587 There were none for n=13. I haven't looked for patterns with even number of digits.
the patterns I see that should hold regardless of length are related to the first n must be the end of a square reversed and the rest must match up to give those digits ( so the last has to be a number mod 10 that allows a square to end in the first digit hence the 1 and 0 combination (9^2)%10==1)

 2016-03-06, 16:21 #5 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 2·709 Posts From Ibm Ponder This: Update (07/03): A solution with more than 20 digits will earn you a '**'.
2016-03-07, 10:31   #6
LaurV
Romulan Interpreter

Jun 2011
Thailand

22·7·11·29 Posts

Quote:
 Originally Posted by axn Using brute force exhaustive search (O(10^8)), I was able to find two solutions for n=15. Here are the smaller solutions: Code: n=5 65766 n=5 69714 n=7 6317056 n=9 169605719 n=11 96528287587 There were none for n=13. I haven't looked for patterns with even number of digits.
n=8 90899553

 2016-03-12, 12:14 #7 henryzz Just call me Henry     "David" Sep 2007 Cambridge (GMT/BST) 34·71 Posts Found the two n=15s n=1 1 n=1 5 n=1 6 n=3 963 n=4 9,867 n=5 65,766 n=5 69,714 n=7 6,317,056 n=8 90,899,553 n=9 169,605,719 n=10 4,270,981,082 n=11 96,528,287,587 n=12 465,454,256,742 n=12 692,153,612,536 They seem to be getting rarer. Has anyone a clue of how many we should expect to find of a certain size? Running a search upto 18 digits. This should take a while. 20 digits is out of my reach so far.
 2016-03-13, 10:25 #8 NBtarheel_33     "Nathan" Jul 2008 Maryland, USA 3×7×53 Posts Is there an elegant way to solve this problem or is writing a program and cycling through candidates the only way to go? Sometimes it is difficult to tell when brute-force is actually the intended solution method. I started playing around with a few examples (and even built a grammar-school 15-digit by 15-digit multiplication problem to build up the product line-by-line in Excel) but things get unwieldy fairly quickly. As in GIMPS, knowing the position and size of potential carries becomes vital...but there is no easy way to really do this here.
2016-03-13, 14:31   #9
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT/BST)

131678 Posts

Quote:
 Originally Posted by NBtarheel_33 Is there an elegant way to solve this problem or is writing a program and cycling through candidates the only way to go? Sometimes it is difficult to tell when brute-force is actually the intended solution method. I started playing around with a few examples (and even built a grammar-school 15-digit by 15-digit multiplication problem to build up the product line-by-line in Excel) but things get unwieldy fairly quickly. As in GIMPS, knowing the position and size of potential carries becomes vital...but there is no easy way to really do this here.
Quadratic residues reduce the candidate pool massively. I will explain my method in full once the deadline is up. Found an 18 digit one last night.
I think I may have hit on the way to get to 20 digits today. The squaring was the slowest bit for me. I should be able to go from one square to another though. n^2-(n-x)^2=2*x*n-x^2 which is much easier to calculate and add than do another square.

Last fiddled with by henryzz on 2016-03-13 at 14:31

 2016-03-13, 14:53 #10 LaurV Romulan Interpreter     Jun 2011 Thailand 893210 Posts Some supermod should put your post in SPOILER ALERT. You actually disclosed the method...
 2016-03-16, 01:57 #11 ramshanker     "Ram Shanker" May 2015 Delhi 2516 Posts Are there no 14 & 18 digit answer to this? My program didn't turn-up anything. I am having doubt whether I am doing it right or not. Though it did reproduce the 10 & 12 digit ones posted above. My code is for even digits only. Last fiddled with by ramshanker on 2016-03-16 at 01:58

 Similar Threads Thread Thread Starter Forum Replies Last Post Xyzzy Puzzles 2 2018-04-08 13:45 R. Gerbicz Puzzles 1 2017-04-03 10:57 Xyzzy Puzzles 15 2015-04-02 02:19 Prime95 Lounge 20 2006-03-21 04:35 Mystwalker GMP-ECM 4 2006-02-01 12:00

All times are UTC. The time now is 16:42.

Sun Nov 29 16:42:42 UTC 2020 up 80 days, 13:53, 4 users, load averages: 1.06, 1.08, 1.08