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

(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