![]() |
![]() |
#1 (permalink) |
Addict
|
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.. |
![]() |
![]() |
#3 (permalink) |
Devils Cabana Boy
Location: Central Coast CA
|
if you are going to give us a math problem dont use 2 inplace of to, its just silly
__________________
Donate Blood! "Love is not finding the perfect person, but learning to see an imperfect person perfectly." -Sam Keen |
![]() |
![]() |
#4 (permalink) |
Rookie
Location: Oxford, UK
|
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?
__________________
I can't understand why people are frightened of new ideas. I'm frightened of the old ones. -- John Cage (1912 - 1992) |
![]() |
![]() |
#6 (permalink) |
Addict
|
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
|
![]() |
![]() |
#8 (permalink) |
Rookie
Location: Oxford, UK
|
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.
__________________
I can't understand why people are frightened of new ideas. I'm frightened of the old ones. -- John Cage (1912 - 1992) |
![]() |
![]() |
#16 (permalink) |
Rookie
Location: Oxford, UK
|
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...
__________________
I can't understand why people are frightened of new ideas. I'm frightened of the old ones. -- John Cage (1912 - 1992) |
![]() |
Tags |
math, problem |
|
|