Question

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 \(f(x,y)=f(y,x)\)?

Answer 1

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

Answer 2

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

Answer 3

ZOMG, wait.

Answer 4

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

Answer 5

Facepalm. They can be negative.

Answer 6

PING and the light turns on!

Answer 7

:D

Answer 8

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

Answer 9

Is 33 correct?

Answer 10

no

Answer 11

Aww.

Answer 12

you have 33+16+16=?

Answer 13

Why is that so?

Answer 14

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

Answer 15

Oh, I'm stupid. Thanks man!

Answer 16

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

Answer 17

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

Answer 18

(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?

Answer 19

Is this from brilliant.org?

Answer 20

@oldrin.bataku Yeah, practicing stuff :-)

Answer 21

this way you counted (0,0) twice

Answer 22

Am I not allowed to ask questions for practice?

Answer 23

@electrokid I subtracted it in the end.

Answer 24

right.

Answer 25

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

Answer 26

yes you are allowed

Answer 27

and i am god

Answer 28

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

Answer 29

brilliant! 97 is the correct answer

Answer 30

o'course it'd be
lol

Answer 31

u wbhi dheu dhee hee:)

Answer 32

@perl what?

Answer 33

@perl ?

Answer 34

i said, i enjoyed the problem

Answer 35

now i must go , there are babies to punish