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ParthKohli

  • 2 years ago

Why is my answer wrong? Consider the function on the integers given by \(f(x,y)=x^2y\). How many ordered pairs of integers satisfying \(−16\le x,y\le16\) is \(f(x,y)=f(y,x)\)?

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  1. ParthKohli
    • 2 years ago
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    \[f(x,y)=f(y,x) \Rightarrow y^2x = x^2 y \]I observed that \(x\) and \(y\) cannot be negative here. So I am left with numbers ranging from \(0\) to \(16\). That gives me \(17\).

  2. ParthKohli
    • 2 years ago
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    Are there any other ordered pairs where \(x \ne y\)?

  3. ParthKohli
    • 2 years ago
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    ZOMG, wait.

  4. ParthKohli
    • 2 years ago
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    \((-4)^2 \times -4 = (-4)^2 \times -4\)

  5. ParthKohli
    • 2 years ago
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    Facepalm. They can be negative.

  6. electrokid
    • 2 years ago
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    PING and the light turns on!

  7. electrokid
    • 2 years ago
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    :D

  8. ParthKohli
    • 2 years ago
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    lol yeah. That fetches me \(17 + 16 = 33\)

  9. ParthKohli
    • 2 years ago
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    Is 33 correct?

  10. electrokid
    • 2 years ago
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    no

  11. ParthKohli
    • 2 years ago
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    Aww.

  12. electrokid
    • 2 years ago
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    you have 33+16+16=?

  13. ParthKohli
    • 2 years ago
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    Why is that so?

  14. electrokid
    • 2 years ago
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    that many ordered pairs (0,?) and (?,0)

  15. ParthKohli
    • 2 years ago
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    Oh, I'm stupid. Thanks man!

  16. electrokid
    • 2 years ago
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    so, it'd be 33+32+32 ordered pairs

  17. electrokid
    • 2 years ago
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    how many ways can you have 1) x=0 2) y=0 3) x=y

  18. ParthKohli
    • 2 years ago
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    (1,1),(2,2)...(16,16) is 16. (-1,-1)...(-16,-16) is 16. (0,?) is 33. (?,0) is 33. 32 + 33 + 33 = 98 We overcounted a (0,0), so 32 + 32 + 33 = 97. Right?

  19. oldrin.bataku
    • 2 years ago
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    Is this from brilliant.org?

  20. ParthKohli
    • 2 years ago
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    @oldrin.bataku Yeah, practicing stuff :-)

  21. electrokid
    • 2 years ago
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    this way you counted (0,0) twice

  22. ParthKohli
    • 2 years ago
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    Am I not allowed to ask questions for practice?

  23. ParthKohli
    • 2 years ago
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    @electrokid I subtracted it in the end.

  24. electrokid
    • 2 years ago
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    right.

  25. oldrin.bataku
    • 2 years ago
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    I'm not sure if you're allowed to ask for help on brilliant.org questions, at least publicly... :-p

  26. perl
    • 2 years ago
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    yes you are allowed

  27. perl
    • 2 years ago
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    and i am god

  28. ParthKohli
    • 2 years ago
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    @oldrin.bataku But I think I'm just practicing techniques.

  29. perl
    • 2 years ago
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    brilliant! 97 is the correct answer

  30. electrokid
    • 2 years ago
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    o'course it'd be lol

  31. perl
    • 2 years ago
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    u wbhi dheu dhee hee:)

  32. electrokid
    • 2 years ago
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    @perl what?

  33. ParthKohli
    • 2 years ago
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    @perl ?

  34. perl
    • 2 years ago
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    i said, i enjoyed the problem

  35. perl
    • 2 years ago
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    now i must go , there are babies to punish

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