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amilapsn

  • one year ago

Hey guy this is a fun question: Given any nine integers whose prime factors lie in the set {3,7,11}, prove that there must be two whose product is a square....

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  1. amilapsn
    • one year ago
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    @rational @AngusV

  2. rational
    • one year ago
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    pigeonhole principle ?

  3. amilapsn
    • one year ago
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    gotchcha

  4. amilapsn
    • one year ago
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    you're in the right path....

  5. anonymous
    • one year ago
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    Oh so it's 3,7 and 11 without their respective powers ?

  6. anonymous
    • one year ago
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    Damn that makes things a lot easier.

  7. anonymous
    • one year ago
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    The box of Dirichlet.

  8. amilapsn
    • one year ago
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    hey @rational not only 1 and 0 there can be many

  9. rational
    • one year ago
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    i read the question wrong haha, let me erase

  10. anonymous
    • one year ago
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    It's a good answer though. This could pose for a nice 5'th grade question or something.

  11. anonymous
    • one year ago
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    Ah, got it!

  12. anonymous
    • one year ago
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    We study the powers and the cases for odd and even. Any such number will have the form of 3^a * 7*b * 11^c. The cases for (a,b,c) are : o o o o o e o e o o e e e e e e e o e o e e o o Where o=odd and e=even. Any combination of two numbers that set will of yield an odd number at least at one of the powers of either 3,7 or 11. However, when we add in a ninth number, regardless of the (odd,even) combination at the powers of 3,7 and 11 - this combination will be found in that list of 8 numbers. Which means that when we multiply those two numbers whose (odd,even) combination at the powers is the same the resulting combination at the powers will be (e e e) and thus a perfect square.

  13. rational
    • one year ago
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    Nice!

  14. amilapsn
    • one year ago
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    you nailed it!

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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