20160302, 15:48  #1 
"Mike"
Aug 2002
5×7×227 Posts 
March 2016

20160302, 18:09  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20160302, 18:16  #3 
Jun 2003
3×5×17×19 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 
20160302, 18:26  #4  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20160306, 16:21  #5 
"Robert Gerbicz"
Oct 2005
Hungary
1,429 Posts 
From Ibm Ponder This:
Update (07/03): A solution with more than 20 digits will earn you a '**'. 
20160307, 10:31  #6  
Romulan Interpreter
Jun 2011
Thailand
10001111000001_{2} Posts 
Quote:


20160312, 12:14  #7 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5795_{10} 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. 
20160313, 10:25  #8 
"Nathan"
Jul 2008
Maryland, USA
45B_{16} 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 bruteforce is actually the intended solution method. I started playing around with a few examples (and even built a grammarschool 15digit by 15digit multiplication problem to build up the product linebyline 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. 
20160313, 14:31  #9  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
1011010100011_{2} Posts 
Quote:
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(nx)^2=2*x*nx^2 which is much easier to calculate and add than do another square. Last fiddled with by henryzz on 20160313 at 14:31 

20160313, 14:53  #10 
Romulan Interpreter
Jun 2011
Thailand
3^{4}×113 Posts 
Some supermod should put your post in SPOILER ALERT.
You actually disclosed the method... 
20160316, 01:57  #11 
"Ram Shanker"
May 2015
Delhi
45_{8} Posts 
Are there no 14 & 18 digit answer to this? My program didn't turnup 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 20160316 at 01:58 
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