Cattle Problem:

[BMAD]

**i didn't see this one posted:

The sun god had a herd of cattle consisting of bulls and cows, one part of which was white, a second black, a third spotted, and a fourth brown. Among the bulls, the number of white ones was one half plus one third the number of the black greater than the brown; the number of the black, one quarter plus one fifth the number of the spotted greater than the brown; the number of the spotted, one sixth and one seventh the number of the white greater than the brown. Among the cows, the number of white ones was one third plus one quarter of the total black cattle; the number of the black, one quarter plus one fifth the total of the spotted cattle; the number of spotted, one fifth plus one sixth the total of the brown cattle; the number of the brown, one sixth plus one seventh the total of the white cattle. What was the composition of the herd?

[bonanova]

. . . Bulls Cows Total
white . a . . b . a+b
black . c . . d . c+d
spotted e . . f . e+f
brown . g . . h . g+h

Equations:

(1) a = g + (5/6)c
(2) c = g + (9/20)e
(3) e = g + (13/42)a

(4) b = 7/12 (c+d)
(5) d = 9/20 (e+f)
(6) f = 11/30 (g+h)
(7) h = 13/42 (a+b)

These equations are invariant under multiplication,
so we can only determine relative numbers.
The first three equations lead to

2376a = 5936g

If we set

a = 5936
g = 2376

we get

c = (a-g)(6/5) = 4272
e = (c-g)(20/9) = 4213.3333

So we multiply all values by 3:

a = 17808
c = 12816
e = 12640
g = 7128

Substituting into (4)-(7) we get

b = bo + (7/21)d where bo = (7/12)c = 7476
d = do + (9/20)f where do = (9/20)e = 5688
f = fo + (11/30)h where fo = (11/30)g = 2613.6
h = ho + (13/42)b where ho = (13/42)a = 5512

Because (11/30)g is not integral, we multiply a c e and g by 10
and also bo do fo and ho

a = 178080
c = 128160
e = 126400
g = 71280

Substituting to eliminate d, f, h gives

b{1-(7/12)(9/20)(11/30)(13/42)} = bo + (7/12)do + (7/12)(9/20)fo + (7/12)(9/20)(11/30)ho
0.9702083333b = 120106

b = 123794.0305

Now to get integer values we multiply by 10000

b = 1237940305

substituting, we get

h = 934371999.2
f = 603963066.3
d = 840583379.7
b = 1237940305

On final multiplication by 10 gives

a = 17808000000
c = 12816000000
e = 12640000000
g = 7128000000

h = 9343719992
f = 6039630663
d = 8405833797
b = 12379403050

A large herd, even by Texas standards.

[Prime]

Seems inconsistent.

[BMAD]

W = (5/6)X + Z

[BMAD]

1. X = (9/20)Y + Z
2. Y = (13/42)W + Z