- 02 Jan '07 15:23
**PART 1:**

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 exactly one of the following types (with each of the 5 types being represented exactly once): Square, Cube, Triangular, Prime, Fibonacci.**The solution is unique.**

**PART 2:**

Find a cycle of six 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 6 numbers must be exactly one of the following types (with each of the 6 types being represented exactly once): Square, Cube, Triangular, Prime, Fibonacci, Power-of-Two.**The solution is unique.** - 02 Jan '07 23:29Interesting puzzles. I think a good place to start is finding the Fibonacci (and/or the power of two) as there are only four of each with four digits.

A program could be written to check if each of those was possible (could be contained in the cycle) and then leave you with a much smaller sample to search for the cycle. - 03 Jan '07 08:29

I guess it does. The solution isn't unique if we don't have a cycle formed (first two equal last two).*Originally posted by David113***Does "cycle" mean that the last 2 digits of the last number are the first 2 digits of the first number?**

Part 1:

2850, 5041, 4181, 8117, 1728

Or any of the other four shifted versions of this order.

Part 2:

9261, 6133, 3364, 6441, 4181, 8192

Or as above.

I approached this using Matlab. - 04 Jan '07 17:47

Very clever, XanthosNZ !!*Originally posted by XanthosNZ***I guess it does. The solution isn't unique if we don't have a cycle formed (first two equal last two).**

Part 1:

2850, 5041, 4181, 8117, 1728

Or any of the other four shifted versions of this order.

Part 2:

9261, 6133, 3364, 6441, 4181, 8192

Or as above.

I approached this using Matlab.

PART 1: (8117, 1728, 2850, 5041, 4181) (P, C, T, S, F)

PART 2: (6133, 3364, 6441, 4181, 8192, 9261) (P, S, T, F, W, C) (W = power of two)