My puzzle of the week: Fill a 3x4 grid so that triangular numbers read across and down. The answer is unique. If you'd like a tougher challenge, Fermat made the claim that any number could be represented as the sum of at most 3 triangular numbers. Euler tried, unsuccessfully, to prove this (I did say it was tougher). Gauss proved it on July 10, 1796. The entry in his diary was "EUREKA! num = D+D+D." With a bit of Mathematica code, I identified all the ways 3 triangular numbers could sum to 2001. {{2001,2,9,62}, {2001,6,20,59}, {2001,6,44,44}, {2001,7,34,52}, {2001,9,35,51}, {2001,9,39,48}, {2001,10,10,61}, {2001,10,28,55}, {2001,11,14,60}, {2001,12,17,59}, {2001,14,24,56}, {2001,14,30,53}, {2001,14,41,45}, {2001,16,25,55}, {2001,20,30,51}, {2001,21,39,44}, {2001,24,36,45}, {2001,25,34,46}, {2001,34,37,37}, {2001,35,35,38}}. Do any large numbers have unique representations?

4x3 (unique):
3 3 2 1
5 0 5 0
1 0 3 5

5x4 (maybe unique):
5 8 3 1 1
8 7 9 9 0
8 7 1 5 3
6 8 6 3 5

5x5 (not unique, but this one uses all digits):
1 5 9 3 1
1 0 2 9 6
7 4 6 9 1
8 0 6 0 1
1 3 5 3 0

7x4 (unique, I think):
1 3 5 5 4 8 1
6 4 2 4 3 2 4
5 8 5 6 7 5 3
3 6 3 0 1 6 5

7x5 (uses all digits):
1 2 7 1 2 1 5
1 4 9 3 8 5 6
7 9 0 6 2 7 6
8 7 0 9 0 5 1
1 6 3 5 3 3 6

8x3 (not unique, but may be smallest using all digits):
5 2 3 4 1 7 9 6
6 1 7 9 9 4 0 3
1 0 8 6 0 1 3 0

8x5 (maybe unique):
3 6 1 1 2 2 5 1
2 6 0 2 4 5 0 5
3 0 8 8 5 8 7 0
8 6 7 8 3 7 2 5
5 6 8 0 1 8 1 1

--
Roger Phillips

My solution for the triangular grid is:

3321
5050
1035

Let me suggest to add two more decompositions of the number 2001:

(2001,0,18,60) say 2001=0+171+1830;
(2001,0,21,59) say 2001=0+231+1770.

In fact we NEED to accept 0 as a triangular number, else the theorem of
Gauss would fail e.g. for the decomposition of 0, 1, 2, 4, 6, 11, 20 and 29.

Could you give some references for the theorem of Gauss? Thanks

[Ed -- I'm looking for it]

Claudio Baiocchi

So...triangles:

1) dink around by hand to find nothing
2) whip out Perl and dink around for about 10 minutes

3321
5050
1035

and find it unique.

Start thinking about variations, but not too hard yet...

Neat! Thanks!

--
Michael A. Rios

Ed,
You can reduce the search for the Triangular Number Grid by noting that the bottom row is limited to integers
1,3,6,0,5,8
Of 96 four digit triangular numbers, 17 use only those digits.
Top row can have no zeroes.

3 3 2 1
5 0 5 0
1 0 3 5

Also, Sum(x) - Sum(y) factors all Integers.
Dick Saunders Jr.

Hi Ed,
Well, this puzzle is made much easier than I at first thought because as
it happens the correct final four-digit number is the very first you try
if, as I did, you start with the lowest.
So, while I didn't bother to prove that this is unique, I offer:

3321
5050
1035

Best wishes
Chris Lusby Taylor

Hi Ed:

Here are the answers for various grids of triangular numbers:

3*4:
one solution
3321
5050
1035

2*4:
6 solutions
5151
5050

1431
5565

4465
5565

5565
5565

6441
6555

6555
6555

1*4:
no solutions

2*5:
one solution
22791
81810

3*5:
four solutions
87153
68265
10011

14535
53956
35511

69378
69006
60031

42778
91806
60031

It is odd how the number of solutions hovers between about zero and four
(except for square grids, which are a lot easier). Generally, with this sort
of problem, the number of solutions either quickly climbs, or quickly
decreases as the size of the puzzle is increased.

4*5:
1 solution:
58311
87990
87153
68635

3*3:
21 solutions

4*4:
75 solutions

5*5:
247 solutions

2*6:
2 solutions
161596
565516

617716
608856

3*6:
1 solution
153
528
595
990
630
153

4*6:
3 solutions

5*6:
3 solutions

Cherio,
dave clark

I tried a couple of methods for doing the week puzzle, but the only way i could find the answer was with a little program, helding the answer
3321
5050
1035
Juan Montalvo Bressi