Each of the triangular numbers that are below 1000 is written on a separate card. Some of these cards are then placed in a circle so that, viewed from the center, the right-hand digit of each card matches the left-hand digit of the card to its right. No tricks are used such as turning a card upside-down.

What's the maximum number of digits that can appear on such a subset of the cards?

There are a total of 44 triangular numbers less than 1000 (41 with 2+ digits in length). 
No triangular number ends with a 2, 4, 7 or 9, thus we can eliminate those numbers that begin with a 2, 4, 7 or 9.  And, since, all triangular numbers that begin with a 6 also end with a 6 (which would greatly limit the ring's size as it could only include three numbers {6, 66, 666}), those numbers and all other numbers ending in 6 can be eliminated. In addition, with the assumption leading zeroes are not permitted, numbers ending in 0 can also be removed from consideration.  This leaves 14 (12 w/2+ digits) numbers:
[1, 3,] 15, 55, 105, 153, 171, 325, 351, 378, 528, 561, 595, and 861.

As only one of the remaining numbers begin with an 8 and two other remaining numbers end in an 8, only one of the two ending in the digit 8 can be included.  And, as there are three numbers (15, 105, and 325) that remain that end in 5 that do not also begin with a 5, while there are only two (561, and 528) that begin with a 5 that do not also end with a 5, only two of the three (15, 105, or 325) can be included in the ring.  This leaves the maximum ring size to be 12 (10 w/2+ digits) triangular numbers.

Permitting single digit numbers where the right-hand digit is therefore the left-hand digit, the ring will then be 30 digits using 12 triangular numbers (8 distinct digits):

1 - 15 - 561 - 105 - 55 - 595 - 528 - 861 - 171 - 153 - 3 - 351

If each number must be at least two digits, 1 and 3 are removed from the ring giving a solution of 28 digits using 10 triangular numbers.

Edited on January 10, 2008, 10:46 am

Posted by Dej Mar on 2008-01-10 05:45:11