March 2016

[Xyzzy]

https://www.research.ibm.com/haifa/p...March2016.html

[science_man_88]

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.

[axn]

Using brute force exhaustive search (O(10^8)), I was able to find two solutions for n=15.
Here are the smaller solutions:

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.

[science_man_88]

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:

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)

[R. Gerbicz]

From Ibm Ponder This:

Update (07/03):
A solution with more than 20 digits will earn you a '**'.

[LaurV]

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:

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

[henryzz]

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.

[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.

[henryzz]

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.

[LaurV]

Some supermod should put your post in SPOILER ALERT.
You actually disclosed the method...

[ramshanker]

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.