## Archimedes' Cattle Problem

This is Archimedes' cattle problem, also called the bovinum problem:

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?

Step #1: The first task is to convert all of that into equations. Using W, X, Y, Z, w, x, y, z for white bulls, black bulls, spotted bulls, brown bulls, white cows, black cows, spotted cows, and brown cows, respectively, we get these equations:

1. W = (5/6)X + Z
2. X = (9/20)Y + Z
3. Y = (13/42)W + Z
4. w = (7/12)(X+x)
5. x = (9/20)(Y+y)
6. y = (11/30)(Z+z)
7. z = (13/42)(W+w)

Solution:

There are seven equations with eight unknowns. So there are many possible solutions. And we want the smallest positive integer solution (positive integer because we don't want negative numbers or fractions of cattle). I will explain my reasoning below. Here is the solution:

• W = 10,366,482
• X = 7,460,514
• Y = 7,358,060
• Z = 4,149,387
• w = 7,206,360
• x = 4,893,246
• y = 3,515,820
• z = 5,439,213

Sometimes further restrictions are set on the numbers, in order to make the herd much larger. Such numbers were far beyond the limits of the Greek or Roman or Egyptian number systems. And that may have been the entire idea behind the problem. We can precisely describe numbers beyond the limits of our number systems.

Reasoning:

Combining the first three equations, we get W = (5/6)((9/20)Y+Z)+Z = (5/6)((9/20)((13/42)W+Z)+Z)+Z = (13/112)W + (3/8)Z + (5/6)Z + Z = (13/112)W + (53/24)Z, so (99/14)W = (53/3)Z, and 297W = 742Z. This is reduced to its smallest values, so W is divisible by 742 and Z is divisible by 297. By equation #3, W is divisible by 42. 2226 (3x742) is the smallest number divisible by both 742 and 42, so W is divisible by 2226. Let's try some W's:

```    W = 2226 4452 6677 8903 ...
X = 1602 3204 4806 6408 ...
Y = 1580 3160 4740 6320 ...
Z =  891 1782 2673 3564 ...```

The second column is twice the first; the third is three times the first, etc. For each of these columns, we can plug in the values of W, X, Y, and Z into equations #4 through #7. So we have four equations with four unknowns, and we should be able to solve for w, x, y, and z. Choosing the first column, we get these four equations:

• w = (7/12)(1602+x)
• x = (9/20)(1580+y)
• y = (11/30)(891+z)
• z = (13/42)(2226+w)

Solving these four equations for z, we get:

• w = 7206360 / 4657
• x = 4893246 / 4657
• y = 3515820 / 4657
• z = 5439213 / 4657

after reducing z to its lowest terms. So for these four number to be positive integers, W, X, Y, and Z must be 4657 times the first column above. W = 2226 x 4657 = 10366482. And the rest of the number follow.

Problems which call for integer solutions are called Diophantine problems.