 # GeekDad Puzzle of the Week – Progressive Triplets

This week’s Puzzle of the Week is simply stated, but could use some explanation (or at least translation from math to English.) Simply stated, this week’s puzzle is to find the non-unique, two-digit cumulative differences between the squares of descending triplets of arithmetic sequences. An arithmetic progression is a sequence of numbers in order that have a common difference; for example, the numbers 4, 7, 10, 13, 16, etc., are an arithmetic progression starting with 4 having a common difference of 3. One cumulative difference between the squares of descending triplets of this arithmetic sequence would be (132 – 102 – 72) = 169 – 100 – 49 = 20; another would be (162 – 132 – 102) = 256 – 169 – 100 = -13 and a third would be (102 – 72 – 42) = 100 – 49 – 16 = 35.

For all such arithmetic progressions, some of these cumulative differences between the squares of descending triplets are positive, some are negative and some are exactly zero. Some of these cumulative differences, like 115, can be arrived at through two different sequences (1442 – 1152 – 862 and 302 – 232 – 162), and some like 124 can only be arrived at in one sequence (782 – 622 – 462).

For your chance at this week’s \$50 ThinkGeek Gift Certificate, tell me how many of these cumulative differences between the squares of descending triplets in arithmetic progressions have two digits (i.e., 10-99) and can be arrived at through two or more sequences? Additionally, what are they? (They are finite, and can be listed.)

As always, please send your submissions to GeekDad Central by end of day Friday for consideration. Feel free to show your work for partial credit!

## 1 thought on “GeekDad Puzzle of the Week – Progressive Triplets”

1. Chad says:

No solution submission here, but a note that 63,37,21 is not an arithmetic progression, there being differences if 16 and 26 involved. 63,42,21 would be a sequence.