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Posers and Puzzles

Posers and Puzzles

  1. 09 Apr '08 11:07 / 1 edit
    142857 is a curious number:

    1 * 142857 = 142857
    2 * 142857 = 285714
    3 * 142857 = 428571
    4 * 142857 = 571428
    5 * 142857 = 714285
    6 * 142857 = 857142
    Note that the numbers are the same and in the same order circularly

    7 * 142857 however = 999999.

    Explanation, anyone? Or is it just a coincidence?
  2. 09 Apr '08 11:10
    Not an explanation, but I did immediately notice that these are the first six digits of 1/7.
  3. 09 Apr '08 11:49
    http://en.wikipedia.org/wiki/Cyclic_number
  4. Standard member EinsteinMind
    Seeker
    09 Apr '08 17:02
    Originally posted by FabianFnas
    142857 is a curious number:

    1 * 142857 = 142857
    2 * 142857 = 285714
    3 * 142857 = 428571
    4 * 142857 = 571428
    5 * 142857 = 714285
    6 * 142857 = 857142
    Note that the numbers are the same and in the same order circularly

    7 * 142857 however = 999999.

    Explanation, anyone? Or is it just a coincidence?
    Then, if you continue the series,

    142857 * 8 = 1142856 (the 7 on the end splits into a 1 and a 6, the 1 going to the beginning of the number and the six remaining at the end).

    142857 * 9 = 1258713 (the 4 on the end splits into a 1 and a 3).

    Yes, that number is very interesting.
  5. 09 Apr '08 17:31 / 1 edit
    I have sometimes wondered if Cyclical numbers exist in other bases, Base 12 for instance..

    I suspect so, and I suspect 1/7 in Base 12 would share similar properties, because 12 % 7 is 5, which is a prime modulo root, a term I didn't know before but an aspect of number theory I had noticed before.

    (Composite numbers don't have prime modulo roots unless they are the power of odd prime numbers. The main property of a prime modulo root is as follows...

    We define the following:
    N is the base number we want to find a prime modulo root for.
    P is the candidate for prime modulo root.
    f(x) = (P ^ x) mod N

    If we start with x=1 and examine the value of f(x) until we find a value of x where f(x)=1, we find length of the cyclical pattern for f(x).

    If the length of this pattern is equal to the number of relatively prime numbers below N, then P is a prime modulo root of N.
  6. 09 Apr '08 18:07
    Composite numbers don't have prime modulo roots unless they are the power of odd prime numbers.
    Actually, that is not true.

    If n is a positive integer the congruence classes coprime to n (that is a set of all integerts 0 < m < n-1 such that gcd(m, n) = 1) form a group with multiplication modulo n). This group, usually denoted az Z/nZ, is cyclic if and only if n is 1, 2, 4, p^k, or 2p^k where p^k is a power of an odd prime number. If Z/nZ is cyclic, the generators are primitive root modulo n.

    For example, for n = 14, Z/14Z is { 1, 3, 5, 9, 11, 13 } which is isomorphic to C6 (cyclic group with 6 elements), and the generators are 3 and 5. Hence 3 and 5 are primitive roots modulo 14.

    more on this truly interesting topic is here:
    http://en.wikipedia.org/wiki/Primitive_root_modulo_n
    http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n
  7. 09 Apr '08 20:19
    1/7 = .142857142857142857142857142857142857142857142857...
    2/7 = .285714285714285714285714285714285714285714285714...
    3/7 = .428571428571428571428571428571428571428571428571...
    4/7 = .571428571428571428571428571428571428571428571428...
    5/7 = .714285714285714285714285714285714285714285714285...
    6/7 = .857142857142857142857142857142857142857142857142...
    7/7 = .999999999999999999999999999999999999999999999999... = 1

    You get this cycle because when you do 1 divided by 7 there are 6 possible remainders, each of which gets used once. Then you hit a remainder that has already happened before and the cycle repeats (try it).

    There are many other numbers like this; Wikipedia has some information here: http://en.wikipedia.org/wiki/Recurring_decimal
  8. 09 Apr '08 22:05
    Originally posted by 3v1l5w1n
    Actually, that is not true.

    If n is a positive integer the congruence classes coprime to n (that is a set of all integerts 0 < m < n-1 such that gcd(m, n) = 1) form a group with multiplication modulo n). This group, usually denoted az Z/nZ, is cyclic if and only if n is 1, 2, 4, p^k, or 2p^k where p^k is a power of an odd prime number. If Z/nZ is cyclic ...[text shortened]... /Primitive_root_modulo_n
    http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n
    I missed the 2p^k term... My apologies.

    I was mainly thinking about numbers like 15, 21, 35, 33... Composite numbers with at least 2 odd prime divisors.