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A Puzzle

  1. Mar 10, 2008 #1
    What relationship do 5 and 6 have that make them unlike any other pair of distinct positive integers under 1000? (1000 was as far as I tested.)

    WARNING: Don't read the rest of this thread if you want to solve the puzzle for yourself. It's not that hard--my dad got something rather close to the answer pretty quickly, and he's a computer programmer, not a mathematician. You can PM me if you want a hint.
    Last edited: Mar 10, 2008
  2. jcsd
  3. Mar 10, 2008 #2
    is the only semi-prime number between twin prime numbers under 1000?
  4. Mar 10, 2008 #3
    Please don't put any more answers in this thread.

    Since you answer was wrong, I may as well respond to it. Both 4 and 6 are semiprime numbers that have prime numbers on both sides. Besides, you were supposed to find a relationship between the two numbers.
  5. Mar 10, 2008 #4
    you're right, there are one more case... but seems to be the only 2 cases... (edited because I put some wrong statements)

    if you show that thoses cases are not the only cases, then you'll be famous, because this could be a proof of the twin primes conjecture :smile:

    ps: why we cannot post here???
    Last edited: Mar 10, 2008
  6. Mar 10, 2008 #5
    I don't want anyone to post the answer here so that people can work out the answer for themselves.

    But on second thought, maybe I will just insert a disclaimer in the first post.
  7. Mar 10, 2008 #6
    well... ok... I think your puzzle is a little bit "generic", I mean, at least for me, as a first look, the puzzle can have multiple answers...

    is some information missing?
  8. Mar 10, 2008 #7
    No. I'm sorry it looks generic. Why don't you tell me all the answers you can think of, and I'll tell you whether you got the one I thought of.

    Also, keep in mind that your answer has to express a unique relationship *between* the numbers. So just saying "6 is unique because x, and 5 is unique because y" is not going to cut it. The relationship I am talking about does not exist between any other pair of distinct numbers under 1000.
  9. Mar 12, 2008 #8
    What about this:

    The two positive integers x, y we are looking for satisfy

    x^2-y^2 = (x-y)(x+y) = 11.

    x=6, y=5 is clearly the unique solution.

    There is no need for the restriction that the integers be less than 1000.
    Last edited: Mar 12, 2008
  10. Mar 12, 2008 #9


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    That is the only (unordered) pair of distinct integers lying between 4 and 7. :tongue:
  11. Mar 12, 2008 #10
    Umm ... they're the only two sequential integers with the ratio 5/6? :surprised
  12. Mar 12, 2008 #11
    The only pair of integers that equals (5,6) ....:smile:
  13. Mar 12, 2008 #12


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    Let W be the set of Wilson primes, and let [itex]S=W\cup\{n:2n+1\in W\}[/itex].

    Then (5, 6) is the only pair of consecutive elements in S up to 500 million.
  14. Mar 12, 2008 #13


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    4 is feeling pretty sad right about now. (4 and 6 are also kind of lonely: no company under 100 million.)
  15. Mar 12, 2008 #14
    Guys, thanks for all the responses. The precondition can be stated in terms of all variables - no numbers necessary. (I guess I should have said this sooner, but it seemed like a bit of a hint.) That means *no* coefficients, only multiplication and addition of variables. But the answer can also be stated in plain english in a way that an intelligent middle schooler could understand. So you shouldn't try to figure it out using variables. Just think about what there could be, and keep coefficients and other numbers out of your answer.

    Formally stated, the answer to the problem would go vaguely like this: 5 and 6 are the only numbers A and B such that [insert some equations involving constant variables A, B, C, D, E, and F involving NO COEFFICIENTS OR NONVARIABLE CONSTANTS].

    There, I've given a good number of hints. Hopefully someone will PM me with the reply soon.
  16. Mar 12, 2008 #15
    What do you mean by "constant variables" and "nonvariable constants"...?? And what don't you like about the sugestions presented so far?
    Last edited: Mar 12, 2008
  17. Mar 13, 2008 #16
    A^2 - B^2 = A + B, which is the same Mr. Callahan said yesterday.
  18. Mar 13, 2008 #17
    It is not the same because what you suggest only implies A-B = 1.
  19. Mar 13, 2008 #18
    Something cool.

    The sum of the divisors of 5 plus the sum of the divisors of 6 divides the product of the divisors of 5 times the product of the divisors of 6.

    [tex](1 + 5 ) + (1+6+2+3)=18[/tex]
    [tex](1 * 5 ) * (1*6*2*3)=180[/tex]
  20. Mar 13, 2008 #19
    How about, 5 and 6 are the only consecutive numbers that are the sum and product (respectivley) of the same two prime numbers.
  21. Mar 13, 2008 #20


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    Good call. This is my favorite property so far on the thread. It works unordered, too.
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