[Guest]

Determine a cycle of five 4-digit positive integers , each with no leading zero, such that the last 2 digits of each number are equal to the first 2 digits of the next number in the cycle with the proviso that each of the 5 numbers must be exactly one of the following types : Square, Cube, Triangular, Prime, Fibonacci (albeit not necessarily in this order). Each of the five types must be represented exactly once.

Edited March 9, 2010 by K Sengupta

[superprismatic]

(1728,2850,5041,4181,8117) which cycles as (Cube, Triangular, Square, Fibonacci, Prime)

[Guest]

8649.|.................|.7841

3249.|.4913.1378.|

1849.|.................|.7867

these are the only solutions

If u generate all of this number in a 4 digits range, u will notice that there is a quite small amount of squares (12)

and Fibonaccis (4, and 2 of them are starting with *5 so they cant be the ending of a prime number), but the solution the 12

squares. u have to check the starting and ending numbers of boundaries.

all of this stuff is about 10 minutes, so quicker than writing a program.

Edited March 11, 2010 by det

[Guest]

my answer is for the Square, Cube, Triangular, Prime, Fibonacci order

[superprismatic]

8649.|.................|.7841

3249.|.4913.1378.|

1849.|.................|.7867

these are the only solutions

If u generate all of this number in a 4 digits range, u will notice that there is a quite small amount of squares (12)

and Fibonaccis (4, and 2 of them are starting with *5 so they cant be the ending of a prime number), but the solution the 12

squares. u have to check the starting and ending numbers of boundaries.

all of this stuff is about 10 minutes, so quicker than writing a program.

I didn't see any Fibonacci number in there nor did I spot a cycle. Please elaborate.

[Guest]

oh..i forgot the Fibonacci numbers

..S........C........T........P.......F

8649 |...................| 7841 | 4181

3249 | 4913 | 1378 |

1849 |...................| 7867 | 6765

Edited March 12, 2010 by det

[superprismatic]

oh..i forgot the Fibonacci numbers

..S........C........T........P.......F

8649 |...................| 7841 | 4181

3249 | 4913 | 1378 |

1849 |...................| 7867 | 6765

OK, but they don't cycle. The last two digits of the last number (in this case Fibonacci) must be the first two digits of the first number (in this case Square). I'm afraid you don't have any solutions to the OP here.

[Guest]

ok, cycle.. i was drunk or i dont know, sorry..