The First Part of the Cattle Problem

Let

W= number of white bulls
B= number of black bulls
Y= number of yellow bulls
D= number of dappled bulls
w= number of white cows
b= number of black cows
y= number of yellow cows
d= number of dappled cows
Then the first part of the problem can be stated as the following seven equations in eight unknowns:

(1)W= (1/2 + 1/3)B + Y (the white bulls were equal to a half and a third of the black [bulls] together with the whole of the yellow [bulls])
(2)B= (1/4 + 1/5)D + Y (the black [bulls] were equal to the fourth part of the dappled [bulls] and a fifth, together with, once more, the whole of the yellow [bulls])
(3)D= (1/6 + 1/7)W + Y (the remaining bulls, the dappled, were equal to a sixth part of the white [bulls] and a seventh, together with all of the yellow [bulls])
(4)w= (1/3 + 1/4)(B + b ) (The white [cows] were precisely equal to the third part and a fourth of the whole herd of the black)
(5)b= (1/4 + 1/5)(D + d ) (the black [cows] were equal to the fourth part once more of the dappled and with it a fifth part, when all, including the bulls, went to pasture together)
(6)d= (1/5 + 1/6)(Y + y ) (the dappled [cows] in four parts [i.e., in totality] were equal in number to a fifth part and a sixth of the yellow herd)
(7)y= (1/6 + 1/7)(W + w ) (the yellow [cows] were in number equal to a sixth part and a seventh of the white herd)

Equations (1-7) constitute a homogeneous linear system for W, B, Y, D, w, b, y, d with the following 7 x 8 coefficient matrix:

6-5-600000
020-20-90000
-130-42420000
0-70012-700
000-90200-9
00-11000-1130
-13000-130420

Using a symbolic algebra program (Maple©, MatLab©, Mathematica©, etc.) it is easily determined that this matrix has rank seven and a one-dimensional nullspace given by

W= 10,366,482k
B= 7,460,514k
Y= 4,149,387k
D= 7,358,060k
w= 7,206,360k
b= 4,893,246k
y= 5,439,213k
d= 3,515,820k

where k is an arbitrary number and where the integers multiplying k have no common divisor. There are thus infinitely many possible positive integer solutions, corresponding to k = 1, 2, 3, ... . The smallest positive integer solution arises when k = 1, and so is

W= 10,366,482= number of white bulls
B= 7,460,514= number of black bulls
Y= 4,149,387= number of yellow bulls
D= 7,358,060= number of dappled bulls
w= 7,206,360= number of white cows
b= 4,893,246= number of black cows
y= 5,439,213= number of yellow cows
d= 3,515,820= number of dappled cows
and the total number of cattle of the Sun is 50,389,082. If you could have gotten this far, then you "wouldst not be called unskilled or ignorant of numbers, but not yet shalt thou be numbered among the wise". Only by solving the second part of the problem can you be numbered among the wise.

The Second Part of the Cattle Problem

The second part of the cattle problem imposes two additional conditions that restrict the possible values of k beyond k = 1, 2, 3, ... . The first additional condition states

When the while bulls mingled their number with the black, they stood firm, equal in depth and breadth ...
The most direct interpretation of this condition is that
W + B = a square number
or
10,366,482k + 7,460,514k = a square number
or
17,826,996k = a square number
or
(2)(2)(3)(11)(29)(4657)k = a square number
where, in the last equation, the number 17,826,996 has been expressed as a product of prime numbers. For the left-hand side of this equation to be a square number, it follows that k must be of the form
k = (3)(11)(29)(4657)r ^2
or
k = 4,456,749r ^2
where r is any positive integer and r ^2 denotes "r-squared".

The second additional condition, which further restricts the allowable value of k, states

... when the yellow and the dappled bulls were gathered into one herd they stood in such a manner that their number, beginning from one, grew slowly greater till it completed a triangular figure ...
This means that
Y + D = a triangular number
where triangular numbers are numbers of the form
1 + 2 + 3 + 4 + 5 + . . . + m
where m is some positive integer. By using the formula for the sum of the first m integers, we can also characterize triangular numbers as those numbers of the form
m(m + 1) /2
where m is some positive integer. At this point we have
4,149,387k + 7,358,060k = m(m + 1) /2
or
11,507,447k = m(m + 1) /2
Using our previous condition for the allowable values of k, this becomes
(11,507,447)(4,456,749)r ^2 = m(m + 1) /2
or
102,571,605,819,606r ^2 = m(m + 1)
The problem now is to find positive integers r and m that satisfy this last equation. A partial solution to this problem was given in
A. Amthor
"Das Problema bovinum des Archimedes"
Zeitschrift fur Math. u. Physik. (Hist.-litt.Abtheilung)
Volume XXV (1880), pages 153-171
Amthor found that the smallest integers r and m that satisfy this equation result in a total number of cattle expressed by an integer with 206,545 digits that begins with the digits 776.

Further progress on the solution had to await the computer age. In 1965 three researchers at the University of Waterloo (Waterloo, Ontario, Canada) announced a complete solution to the cattle problem in

H. C. Williams, R. A. German, and C. R. Zarnke
"Solution of the cattle problem of Archimedes"
Mathematics of Computation
Volume XIX (1965) pages 671-687
However, they described their computer results rather than reproducing them in full. In 1981 Harry L. Nelson of the Lawrence Livermore National Laboratory (Livermore, California, USA) published the 47-page printout from a CRAY-1 computer containing the 206,545 digits of the total number of cattle in
Harry L. Nelson
"A solution to Archimedes' cattle problem"
Journal of Recreational Mathematics
Volume 13 (1980-81) pages 162-176
The first and last fifty digits of the total number of cattle found by these researchers are

77602714064868182695302328332138866642323224059233
. . .

05994630144292500354883118973723406626719455081800

There are few numerical problems in mathematics that have taken 22 centuries to solve.