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anonymous

  • 5 years ago

Fool's problem of the day ( on request of asnaseer), probably easy , I haven't thought much, Let \(A = \{a_1, a_2 \cdots a_k\} \) be any set of \( k \) composite numbers such that \(1 \le a_i \le 120 \) for all \(i\) such that \(1 \le i \le k\). Find the least value of \(k\) such that there exists at least one pair \( (a_i, a_j) \), \(1\le i, j \le k\) in \(A\) which is not co prime ?

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  1. anonymous
    • 5 years ago
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    I should go to bed now, Enjoy guys :)

  2. asnaseer
    • 5 years ago
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    good night FFM

  3. asnaseer
    • 5 years ago
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    and thanks for the post

  4. anonymous
    • 5 years ago
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    You are always welcome :)

  5. Mr.Math
    • 5 years ago
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    I guess k=4.

  6. asnaseer
    • 5 years ago
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    isn't it k=2 --> {1,2} are 1 and 2 considered to be co prime?

  7. Mr.Math
    • 5 years ago
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    2 is not composite.

  8. asnaseer
    • 5 years ago
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    ah! - of course - thx for pointing that out Mr.Math

  9. Mr.Math
    • 5 years ago
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    Neither is 1.

  10. asnaseer
    • 5 years ago
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    I thought 1 is coprime?

  11. asnaseer
    • 5 years ago
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    but not composite

  12. Mr.Math
    • 5 years ago
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    Two integers a and b are said to be coprime (also spelled co-prime) or relatively prime if the only positive integer that evenly divides both of them is 1.

  13. Mr.Math
    • 5 years ago
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    What about \(A=\{4,6\}?\), that's k=2.

  14. asnaseer
    • 5 years ago
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    but 2 divides 4 and 6?

  15. Mr.Math
    • 5 years ago
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    Yep, so they are NOT coprime.

  16. Mr.Math
    • 5 years ago
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    As the question asked.

  17. asnaseer
    • 5 years ago
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    note to self: must learn to read the question properly!

  18. asnaseer
    • 5 years ago
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    then your answer seems to be correct.

  19. Mr.Math
    • 5 years ago
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    I think so, but that was too easy. Congrats on being a moderator, they couldn't have chosen any better :-)

  20. asnaseer
    • 5 years ago
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    thx Mr.Math - I am very humbled to have been chosen. FFM did say this is quite easy, but you are right - it seems TOO easy :-)

  21. Mr.Math
    • 5 years ago
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    Plus, he always says that even when it's TOO difficult. So I never trust his judgement. :D

  22. asnaseer
    • 5 years ago
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    he he - yes - I concur. I'll have to mull over this one tomorrow. it's quite late here so I need to get some sleep. bye for now...

  23. Mr.Math
    • 5 years ago
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    Good night.

  24. anonymous
    • 5 years ago
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    Oh well not that easy ;) Does \( K=2 \) or\( K=4 \) works for any co-prime in\( 1\le a_i \le 120 \) ?

  25. anonymous
    • 5 years ago
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    Whenever I say that the problem is easy, it means that there exists a very short solution for that problem which may or may not use some well known theorems or results, and if I remember correctly I have posted only one too difficult problem here.

  26. anonymous
    • 5 years ago
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    and now I have solved this one, I would say this is a easy problem ;)

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