Bryce Herdt Triangle Puzzle

Hello Ed,

I like Bryce Herdt's puzzle because it is small enough to do by hand and 
appears
to me to have only one solution.  Clearly one could write a program that could
solve it, but fortunately some math smarts about triangular numbers helps a 
lot.
Here is my reasoning, or skip to the bottom for my answer(s).

A triangular number cannot end in 2, 4, 7, 9.  Is there a proof of this?  Must
be.  Therefore the bottom 4 digit number cannot have any of these digits, so
looking through the 4 digit triangular numbers there are only 9 possibilities
for the bottom row.  In addition, the downward 2 digit number cannot start with
2, 4, 7, or 9, and looking at the possibilities it must then end in either 0, 5
or 6.  Thus the number in the bottom row cannot contain a 2, 4, 7 or 9, and the
3rd digit must be a 0, 5 or 6.  Thus there are only six possibilities for the
bottom row: 1653, 1953, 3001, 5151, 5356 and 8001.

Working through each of these six cases individually, there are only a few
possibilites in each case for the vertical 4 digit number.  Assuming that none
of the digits in the vertical 4 digit number can be 0, I get only one solution:

6
3 6
2 3 1
8 0 0 1

However if one considers "06" as a valid 2 digit number there is also the
solution:

1
0 6
8 6 1
1 6 5 3

It appears to me that these are the only solutions possible that do not repeat
any triangular numbers.  But I did this pretty quickly and I may have missed
another solution.

-George Bell
------------------------------------------------

What a nice puzzle! A blank grid and all the information you need to get a 
unique solution. (Reminds me of Juha Hyvonen's puzzles.)
6
36
231
8001
 
Best wishes
 
John Gowland
------------------------------
Finally found a moment to work on this.  While I expect there are other
answers, the one I found with pen and paper was...

6
3 6
2 3 1
8 0 0 1

Warren Phillips
---------------------------------
It looks like a crossword and in crosswords, unchecked letters (one-letter 
entries) do not matter. Therefore, I did not think about the one-digit numbers and 
got multiple solutions. Since multiple solutions are not fun, I thought maybe 
each digit 0-9 are used once. No luck.

It took me a while to realize that the one-digit numbers must be triangular, 
too.

6
3 6
2 3 1
8 0 0 1

Solution if the grid is rotated 90 degrees clockwise:

1 4 3 1
6 6 6
5 5
3

And if rotated counter-clockwise:

_ _ _ 1
_ _ 3 6
_ 1 0 5
3 0 0 3

Juha Hyvönen
-----------------------------------------------------
6
36
231
8001

Alan Lemm
-----------------------------------------
A quick program took care of this one. I modified it and also got results for "3x3", "5x5", and "6x6" ones(if the puzzle as it stands is called a "4x4"). The program didn't filter out results with leading zeroes, and after I manually pulled those out, I was left with only one 5x5 and no 6x6es. (Most of the 6x6es had 0's in the first digit of the last row.)
 
Results:
 
"3x3"
 
1
3 6
6 6 6
1
5 5
3 5 1
 
3
5 5
1 5 3
 
And my favorite:
6
6 6
6 6 6
 
"4x4"

1
4 5
8 6 1
5 1 5 1
6
3 6
2 3 1
8 0 0 1
 
6
5 5
5 6 1
5 1 5 1
 
"5x5"
 
 6
 7 8
 1 5 3
 6 1 0 5
 1 5 0 5 1
 
(Note: Since I did leading-zero filtering "by hand" I might've missed a few.)
 
--Ellis M. Eisen


-----------------------------------------
Hi Ed,
I found the solution to Bryce Herdt's problem from June 2.

1. Rows: 6, 36, 231, 8001 (Columns: 6328, 630, 10, 1)  

Thanks for the fun you provide, Matt Sheppeck

BTW, I boiled it down to this before using if-thens...
A					
B	C				
D	E	F			
G	H	I	J		

BC	CEH	DEF	FI	ABDG	GHIJ
21	120	231	10	1225	1653
28	136	325	15	1431	3003
36	153	351	55	1485	5151
45	190	435		1653	5356
55	561	465		1953	8001
66	630	561		3655	
91	666	595		6328	
	820	861		6441	
	861			6555	
----------------------------------------------------
Paul Cleary

Joseph DeVincentis

Andrew Weimholt

George Sicherman

John Dalbec

Bob Kraus

Jaime Rangel-Mondragon

James L Melby

Mark Michell

Kirk Bresniker

DENIS BORRIS