ryan333

January 25th 2009, 11:51 PM

Question:

Mr. Brown raises chickens. Each can be described as thin or fat, brown

or red, hen or rooster. Four are thin brown hens, 17 are hens, 14 are thin

chickens, 4 are thin hens, 11 are thin brown chickens, 5 are brown hens, 3

are fat red roosters, 17 are thin or brown chickens. How many chickens does

Mr. Brown have?

Any help is appreciated. Just looking for the most efficient method. Thanks in advance!

ctperng

January 26th 2009, 12:56 PM

Question:

Mr. Brown raises chickens. Each can be described as thin or fat, brown

or red, hen or rooster. Four are thin brown hens, 17 are hens, 14 are thin

chickens, 4 are thin hens, 11 are thin brown chickens, 5 are brown hens, 3

are fat red roosters, 17 are thin or brown chickens. How many chickens does

Mr. Brown have?

Any help is appreciated. Just looking for the most efficient method. Thanks in advance!

Have you played Rubik's cube? You could use 3d ar-ray to solve this.

Let T= thin = top face, F = fat = bottom, H = hen = front, R0 = rooster = rear, B = brown = left, and R = red = right.

Then you get 8 blocks, each of which is the intersection of 3 conditions. The intersection of two conditions gives the sum of two blocks, etc. It helps to break into top and bottom faces, as follows:

TR0B TR0R

THB THR

and

FR0B FR0R

FHB FHR

Then you start with the intersection of 3 conditions such as TBH = 4, etc, and successively derive all the rest. (It is just straitforward using addition or subtraction.)

You get for top face

7 3

4 0 and for bottom face

2 3

1 12

Therefore the total is 7+3+4+0+2+3+1+12 = 32. (If my arithmetic is correct.)

ctperng

SinJei

January 26th 2009, 07:52 PM

totally amazing ...

use rubik's cube to solve problem...

smart!

ryan333

January 26th 2009, 11:03 PM

''' that is by far the most interesting way to answer that question.

thanks for the reply :)

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