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#1 2006-02-26 16:54:49
- ganesh
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Black and Red
A deck of 52 playing cards is cut into three separate piles.
In the first pile there are three times as many Blacks as Reds.
In the second pile there are three times as many Reds as Blacks.
In the third pile there are twice as many Blacks as Reds.
How many cards of each colour are there in each of the three piles?
Character is who you are when no one is looking.
#2 2006-02-27 22:01:34
- krassi_holmz
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Re: Black and Red
I get this:
IPBLE: Increasing Performance By Lowering Expectations.
#4 2006-02-27 22:35:39
- MathsIsFun
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Re: Black and Red
B1+R1+B2+R2+B3+R3=52
B1=3R1
3B2=R2
B3=2R3
Then ... ?
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#5 2006-02-28 00:09:51
- krassi_holmz
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Re: Black and Red
Rob, you are forgeting something: this is a diophantine equation. the numbers are integers.
here are all the solutions:
Last edited by krassi_holmz (2006-02-28 00:18:36)
IPBLE: Increasing Performance By Lowering Expectations.
#6 2006-02-28 01:36:39
- ganesh
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Re: Black and Red
krassi_holmz,
Have you taken B1+B2+B3=26 and R1+R2+R3=26 in all your solution sets? I see grom the GIF image that there are many cases where this isn't true (I have not examined all of the solutions yet), which means the solution sets are not acceptable 
Character is who you are when no one is looking.
#7 2006-02-28 04:53:35
- Ricky
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Re: Black and Red
Using the two equations Ganesh posted:
x, y, z are the three piles, xr is the number of red cards in x.
8xr + 5zr = 26
This is only true for integers when xr = 4, so xr = 4 and zr = 4. Thus, yr = 18
xr = 4, so xb = 12
yr = 18, so yr = 6
zr = 4, so zb = 8
x = 4+12 = 16
y = 18 + 6 = 24
z = 8 + 4 = 12
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#8 2006-02-28 15:54:34
- ganesh
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Re: Black and Red
Well done, Ricky 
Character is who you are when no one is looking.