ParthKohli
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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ParthKohli
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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 \(17\).
ParthKohli
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Are there any other ordered pairs where \(x \ne y\)?
ParthKohli
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ZOMG, wait.
ParthKohli
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\((-4)^2 \times -4 = (-4)^2 \times -4\)
ParthKohli
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Facepalm. They can be negative.
electrokid
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PING and the light turns on!
electrokid
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:D
ParthKohli
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lol yeah. That fetches me \(17 + 16 = 33\)
ParthKohli
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Is 33 correct?
electrokid
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no
ParthKohli
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Aww.
electrokid
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you have 33+16+16=?
ParthKohli
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Why is that so?
electrokid
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that many ordered pairs
(0,?)
and
(?,0)
ParthKohli
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Oh, I'm stupid. Thanks man!
electrokid
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so, it'd be
33+32+32
ordered pairs
electrokid
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how many ways can you have
1) x=0
2) y=0
3) x=y
ParthKohli
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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?
oldrin.bataku
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Is this from brilliant.org?
ParthKohli
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@oldrin.bataku Yeah, practicing stuff :-)
electrokid
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this way you counted (0,0) twice
ParthKohli
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Am I not allowed to ask questions for practice?
ParthKohli
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@electrokid I subtracted it in the end.
electrokid
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right.
oldrin.bataku
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I'm not sure if you're allowed to ask for help on brilliant.org questions, at least publicly... :-p
perl
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yes you are allowed
perl
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and i am god
ParthKohli
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@oldrin.bataku But I think I'm just practicing techniques.
perl
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brilliant! 97 is the correct answer
electrokid
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o'course it'd be
lol
perl
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u wbhi dheu dhee hee:)
electrokid
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@perl what?
ParthKohli
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@perl ?
perl
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i said, i enjoyed the problem
perl
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now i must go , there are babies to punish