Posers and Puzzles

Posers and Puzzles

  1. Standard memberMathurine
    sorozatgyilkos
    leölés ellenfeleim
    Joined
    15 Jul '06
    Moves
    40507
    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.
  2. Joined
    25 Aug '06
    Moves
    0
    02 Jan '07 20:07
    Does "cycle" mean that the last 2 digits of the last number are the first 2 digits of the first number?
  3. Standard memberXanthosNZ
    Cancerous Bus Crash
    p^2.sin(phi)
    Joined
    06 Sep '04
    Moves
    25076
    02 Jan '07 23:29
    Interesting 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.
  4. Standard memberXanthosNZ
    Cancerous Bus Crash
    p^2.sin(phi)
    Joined
    06 Sep '04
    Moves
    25076
    03 Jan '07 08:29
    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?
    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.
  5. Standard memberMathurine
    sorozatgyilkos
    leölés ellenfeleim
    Joined
    15 Jul '06
    Moves
    40507
    04 Jan '07 17:47
    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.
    Very clever, XanthosNZ !!

    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)



    🙂
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