S O L U T I O N . ( 1 S T . P A R T )
Then the first part of the problem can be stated as the following seven equations in eight unknowns:
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The First Part of the Cattle Problem
Archimedes 1 2 1 1/2
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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 |
(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: Spaceward ho 5 0 5 download free.
6 | -5 | -6 | 0 | 0 | 0 | 0 | 0 |
0 | 20 | -20 | -9 | 0 | 0 | 0 | 0 |
-13 | 0 | -42 | 42 | 0 | 0 | 0 | 0 |
0 | -7 | 0 | 0 | 12 | -7 | 0 | 0 |
0 | 0 | 0 | -9 | 0 | 20 | 0 | -9 |
0 | 0 | -11 | 0 | 0 | 0 | -11 | 30 |
-13 | 0 | 0 | 0 | -13 | 0 | 42 | 0 |
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 |