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anonymous
 4 years ago
Fool's problem of the day,
How many ordered pair of \((x,y)\) are there such that \(x, y \in \mathbb{Z} \) and \( \frac 2 x  \frac 3 y = \frac 1 4 \)
Good luck!
anonymous
 4 years ago
Fool's problem of the day, How many ordered pair of \((x,y)\) are there such that \(x, y \in \mathbb{Z} \) and \( \frac 2 x  \frac 3 y = \frac 1 4 \) Good luck!

This Question is Closed

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Oops, maybe this involves some trial and error :P

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0What is \(\mathbb{Z}\) in here?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0This works with 8 and 6

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0If b is 8 then a will be 16

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I see a lot of solutions to this :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not many, less than 25 I believe ;)

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.024 or 23 or 22.. I guess

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0The minimum of b is I guess 36 maximum is 84

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Don't guess, try to form a analytic approach :)

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Oh wait, it goes further.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Your problem is hard.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0How do we do it by the way?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's problem of the day, it's supposed to be interesting. However my approach takes less than 10 seconds ;) I will post it later though.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is the answer 5? @FoolForMath ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay, we can get after simplification.. x=8y/(y+12) as x is an integer, 8y must be a multiple of y+12 8y=2*2*2*y so, here are possible equations y+12=1 y+12=2 y+12=4 y+12=8 y+12=2y y+12=4y y+12=8y as y +12 is a factor of 8y all the equations except the last one will satisfy for an integral value of y so, the answer is.. 6 ordered pairs

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry, there will be other options too if RHS is negative y+12=1 y+12=2 ................. and so on.. for y to be an integer, there are another 5 solutions so, total 11

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the equation is equivalent to (x8)*(y+12)=96 so the answer is 11

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is all the way wrong or just the 11?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I m pretty sur it's 22

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is 24 correct answer.?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0No, you are close by the way

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I was able to find 11 ordered pairs that are solutions: (7, 84) (6,36) (5,20) (4,12) (2,4) (4,4) (8,6) (16,8) (24,9) (40,10) (88,11) I think there might be more, but im missing an idea.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0HINT: There are 23 of them.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Real challenge is to find a quick analytic approach :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i just found the other 12, making 23.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Most people probably got that:\[8y12xxy = 0\]There is a way you can factor this to get an idea of how many solutions there are...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's right joe! Congrats man!
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