Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ParthKohli

  • 3 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 \(f(x,y)=f(y,x)\)?

  • This Question is Closed
  1. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[f(x,y)=f(y,x) \Rightarrow y^2x = x^2 y \]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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Are there any other ordered pairs where \(x \ne y\)?

  3. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ZOMG, wait.

  4. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \((-4)^2 \times -4 = (-4)^2 \times -4\)

  5. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Facepalm. They can be negative.

  6. electrokid
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    PING and the light turns on!

  7. electrokid
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    :D

  8. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    lol yeah. That fetches me \(17 + 16 = 33\)

  9. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Is 33 correct?

  10. electrokid
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    no

  11. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Aww.

  12. electrokid
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you have 33+16+16=?

  13. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Why is that so?

  14. electrokid
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    that many ordered pairs (0,?) and (?,0)

  15. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Oh, I'm stupid. Thanks man!

  16. electrokid
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    so, it'd be 33+32+32 ordered pairs

  17. electrokid
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    how many ways can you have 1) x=0 2) y=0 3) x=y

  18. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    (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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is this from brilliant.org?

  20. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @oldrin.bataku Yeah, practicing stuff :-)

  21. electrokid
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    this way you counted (0,0) twice

  22. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Am I not allowed to ask questions for practice?

  23. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @electrokid I subtracted it in the end.

  24. electrokid
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    right.

  25. oldrin.bataku
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm not sure if you're allowed to ask for help on brilliant.org questions, at least publicly... :-p

  26. perl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes you are allowed

  27. perl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and i am god

  28. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @oldrin.bataku But I think I'm just practicing techniques.

  29. perl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    brilliant! 97 is the correct answer

  30. electrokid
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    o'course it'd be lol

  31. perl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    u wbhi dheu dhee hee:)

  32. electrokid
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @perl what?

  33. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @perl ?

  34. perl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i said, i enjoyed the problem

  35. perl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now i must go , there are babies to punish

  36. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy