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Posers and Puzzles

Posers and Puzzles

  1. 25 Jan '07 08:44
    31 is a prime, so is 331 and 3331.
    Prove (or disprove) that every number with threes and ending with a one is prime.
  2. 25 Jan '07 10:05 / 2 edits
    Here are some composite numbers of this form:

    (10 ^ 9 - 7) / 3 = 17 x 19607843
    (10 ^ 10 - 7) / 3 = 673 x 4952947
    (10 ^ 11 - 7) / 3 = 307 x 108577633
    (10 ^ 12 - 7) / 3 = 19 x 83 x 211371803
    (10 ^ 13 - 7) / 3 = 523 x 3049 x 2090353
    (10 ^ 14 - 7) / 3 = 607 x 1511 x 1997 x 18199
    (10 ^ 15 - 7) / 3 = 181 x 1841620626151
    (10 ^ 16 - 7) / 3 = 199 x 16750418760469
  3. 25 Jan '07 10:19
    Originally posted by David113
    333333331 = 17 x 19607843
    Correct.

    31, 331, 3331, 33331, 333331, 3333331, 33333331 are primes, 333333331 is not.

    Once, before computers and mechanical calculators, there was a conjecture that every number started with a number of threes, ending with a one was a prime until, with great effort, 333333331 was factorized, and therefore, the conjecture was disproven.
  4. 25 Jan '07 10:42
    Originally posted by FabianFnas
    Correct.

    31, 331, 3331, 33331, 333331, 3333331, 33333331 are primes, 333333331 is not.

    Once, before computers and mechanical calculators, there was a conjecture that every number started with a number of threes, ending with a one was a prime until, with great effort, 333333331 was factorized, and therefore, the conjecture was disproven.
    Interesting.
  5. 25 Jan '07 12:49 / 3 edits
    I find this hard to believe, since it is easy to prove - without any computer - that this sequence contains a composite number.

    By Fermat's Little Theorem, if p is a prime and x is an integer not divisible by p, then x ^ (p - 1) - 1 is divisible by p.

    Let p = 31, x = 10. Then you get the result - 999999999999999999999999999999 (30 9's) is divisible by 31.

    This means that also this number divided by 3, which is 333333333333333333333333333333, is divisible by 31; and so, 33333333333333333333333333333300 also is divisible by 31; and
    so is 33333333333333333333333300 + 31 = 33333333333333333333333333333331.

    so 33333333333333333333333333333331 is not prime.
  6. Standard member PocketKings
    Banned from edits
    25 Jan '07 14:32
    Ah, the old divisibility rules are still going strong
  7. 25 Jan '07 15:57
    Originally posted by David113
    I find this hard to believe, since it is easy to prove - without any computer - that this sequence contains a composite number.

    By Fermat's Little Theorem, if p is a prime and x is an integer not divisible by p, then x ^ (p - 1) - 1 is divisible by p.

    Let p = 31, x = 10. Then you get the result - 999999999999999999999999999999 (30 9's) is divisible by ...[text shortened]... + 31 = 33333333333333333333333333333331.

    so 33333333333333333333333333333331 is not prime.
    And I thought im geek...
  8. Standard member TheMaster37
    Kupikupopo!
    25 Jan '07 22:24
    I came across this puzzle last week;

    The sum of the digits of the number 37 is 10, wich has sum of digits 1.

    The number 37^2=1369 has sum of digits 19, wich has sum 10, wich results to 1.

    A) Prove or disprove that every power of 37 will end in 1 after taking the sum of the digits repeatedly.

    B) Prove or disprove that repeatedly taking the sum of the digits of 37^n will be 10, before becoming 1, for all integers n>0.

    B implies A, I know, but A is a bit easier then B.
  9. 26 Jan '07 10:19
    A follows from the fact that 37 = 1 (mod 9), so 37 ^ n = 1 (mod 9).