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!
Oops, maybe this involves some trial and error :P
What is \(\mathbb{Z}\) in here?
This works with -8 and -6
If b is -8 then a will be -16
I see a lot of solutions to this :/
Not many, less than 25 I believe ;)
24 or 23 or 22.. I guess
The minimum of b is I guess -36 maximum is 84
Don't guess, try to form a analytic approach :)
Oh wait, it goes further.
Your problem is hard.
No spoiler please :)
How do we do it by the way?
It'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.
is the answer 5? @FoolForMath ?
No Arnab.
okay, 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
not 5^^
sorry, 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
8?
is it right?
No, not right
the equation is equivalent to (x-8)*(y+12)=-96 so the answer is 11
Na na
is all the way wrong or just the 11?
22
I m pretty sur it's 22
24
is 24 correct answer.?
No, you are close by the way
I 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.
HINT: There are 23 of them.
Real challenge is to find a quick analytic approach :)
yeah i just found the other 12, making 23.
Most people probably got that:\[8y-12x-xy = 0\]There is a way you can factor this to get an idea of how many solutions there are...
That's right joe! Congrats man!
ty :)
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