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Old 08-19-2003, 10:11 AM   #1 (permalink)
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math problem

250.....b......d
---- = --- x ---
60.......a......c
A has 2 be 30 or 32
bcd can be 34,38,40,41,42,43,44,46,49,50,52,60,66,70,80,90,100
it needs 2 come out even 2 each other. and if anyone can help me and tell me how ecactly u do it?
had 2 put the dots in between the numbers cuz if not they get close 2 geter and they dont look like fractions

Last edited by Smackre; 08-19-2003 at 10:13 AM..
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Old 08-19-2003, 10:14 AM   #2 (permalink)
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I'm confused. Is it (250/60) = (b/a) x (d/c) ??
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Old 08-19-2003, 10:43 AM   #3 (permalink)
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if you are going to give us a math problem dont use 2 inplace of to, its just silly
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Old 08-19-2003, 11:06 AM   #4 (permalink)
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U hav2 cansL T fracshnz

250/60 = 25/6
so 25/6 = bd/ac

a is 30 or 32 I understand; but by 'bcd can be....(list)' do you mean b x c x d must be one of the list, or that each of b, c, and d must be picked from the list. Can we repeat numbers?

If so... I like:

a=30 b=100 d=100 c=80

Can you fill us in on exactly what it's all for?
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Old 08-19-2003, 11:20 AM   #5 (permalink)
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If you can't repeat, these work:

a = 30
b = 50
c = 40
d = 100


250/60 = (50/30)x(100/40) = (50x100)/(30x40) = 5000/1200 = 2500/600 = 250/60
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Old 08-19-2003, 11:59 AM   #6 (permalink)
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dexlarge understands me right. yes those numbers are for each one not added to geter. sorry about the 2's it comes from 3 years of eq heh. how exactly do you do this easy takes me like 4 hours to do one of these problems and i need it for a grinder at work. I need to be able 2 place 250 with other numbers and do the same problem
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Old 08-19-2003, 12:08 PM   #7 (permalink)
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Dang... no kidding about the youth of today not using english correctly. I can see where this become a problem.
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Old 08-19-2003, 12:21 PM   #8 (permalink)
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Just pick one of the numbers for a, then you only have to solve b,c,d. Either b or d has to be quite large in order to get the 125 for bd/c (assuming a=30). So pick high-ish numbers for them Seeing as the smallest c can get is 34, they have to multiply to at least 4250 - restricting your choice even further. So select b and d, divide the product by 125 and see if it's on your list for c.

Probably not the quickest way, but simple and works... perhaps someone kind with a bit of 1337 5k!11z will knock up a program/script for you if you're likely to be doing it regularly.
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Old 08-22-2003, 11:50 AM   #9 (permalink)
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There is no one correct answer. You need 4 different equations to solve for 4 unique variables.
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Old 08-26-2003, 12:21 PM   #10 (permalink)
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Suddenly I remember why I sucked so much ass in Math class.

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Old 08-27-2003, 12:58 AM   #11 (permalink)
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I'm crying this is so hard! *sheesh!* This is math!?!
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Old 08-27-2003, 06:27 AM   #12 (permalink)
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Fremen: You should do a PhD in it!!

Last edited by mathmo; 08-27-2003 at 06:40 AM..
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Old 08-27-2003, 09:25 PM   #13 (permalink)
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Quote:
Originally posted by mathmo
Fremen: You should do a PhD in it!!
If you're doing that, then my hat goes off to you, sir.
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Old 08-28-2003, 12:09 AM   #14 (permalink)
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Well, it's dirty work, but someone has to do it. Might as well be me.
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Old 08-28-2003, 08:02 AM   #15 (permalink)
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a =1
b =1
c = 60
d = 250

As has been said, there are innumerable answers to this problem. To have a finite subset of answers, there would need to be more equations.
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Old 08-28-2003, 09:13 AM   #16 (permalink)
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ermm.. cowlick - there were some more equations

Namely: a list of possible values for the four variables.

If you still want a script knocking up, tell me a bit more about the grinder...
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Old 08-28-2003, 09:35 AM   #17 (permalink)
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