1728, 2850, 5041, 4181, 8117 If I can add a comment, let me say, first of all, "nice problem!". Then let me point out that one can not solve it without a computer; on the other hand, having a Computer, you can choose between two strategies: - a bad program written in a short time but requiring a very high execution time; - or a program requiring some more time to be constructed, but with a very short execution time. Of course the second strategy is much more exciting... Best regards, Claudio Baiocchi.

1728, 2850, 5041, 4181, 8117 Cube, Triangular, Square, Fibonacci, Prime Woohoo! I think I finally got one right!! Danny Potts

Answer to number cycle problem, submitted by Andrew Ofiesh 8117-1728-2850-5041-4181

number cycle: 5041(s), 4181(f), 8117(p), 1728(c), 2850(t) Nick Baxter

4181 is Fibonacci 8117 is Prime 1728 is a Cube (12^3) 2850 is Triangular (75 * 76 / 2) 5041 is Square (71^2) -Ken Bateman

2850 (triangular) - 5041 (square) - 4181 (fibonacci) - 8117 (prime) - 1728 (cube) -Dan Hennessy and Stacey Benson (bored physics instructors)

This week's answer seems to be 1728, 2850, 5041, 4181, 8117. Best wishes Chris Lusby Taylor

Hi! I tried this on the way to Osaka by Shinkansen super-express. Very nice problem! How did he make or notice this problem...It's very interesting. The Answer of "Number Cycle by Bob Kraus" is: 1728 (Cube) 2850 (Triangular) 5041 (Square) 4181 (Fibonacci) 8117 (Prime) --Koshi Arai

JP Ikäheimonen -- Fibonacci 4181 Prime 8117 Cube 1728 Triangular 2850 Square 5041 Found with a computer program when I didn't find any other alternative to brute force. I thought Planiverse was out of print. I have been searching for a new copy to replace the old paperback version which I've loaned too often to careless friends. I hope this version has a better binding.

> Find a cycle of five 4-digit numbers such that the last 2 digits > of each number are equal to the first 2 digits of the next number > in the cycle. Each of the 5 numbers must be one of the following > types (with all 5 types being represented): Square, Cube, > Triangular, Prime, Fibonacci. The solution is unique. square = 5041 Fibonacci = 4181 prime = 8117 cube = 1728 triangular = 2850 I actually solved this mostly by hand. Having generated a list of 4-digit numbers of each type, I deleted those with 0 as the third digit and a few hundred others which started with a digit pair which no numbers ended with. I then had 755 numbers, of which only four were Fibonacci numbers: 2584, 4181, 6765 and 1597 (also prime). There were only two numbers ending in 25 (both squares) and two ending in 15 (both triangular), so it was easy to eliminate 2584 and 1597. I then tried 4181. There were many primes starting with 81, plus the triangular 8128. There were several primes ending in 41, two triangular numbers (3741 and 6441) and a single square, 5041. Concentrating on the square, there were four numbers ending in 50, all triangular: 2850, 4950, 5050, 7750. I rejected 5050 because it obviously wouldn't advance the cycle. The last number now had to be a cube, as I had a tentative chain of triangular -> square -> Fibonacci -> prime. There were no cubes ending in 49 or 77, so the cube had to be 1728. And the many primes starting with 81 included 8117, so the cycle was complete. -- Roger Phillips