Page:Popular Science Monthly Volume 67.djvu/667

The Wolfenbüttel manuscript has an appendix, also in Greek, giving numbers in answer to the problem, the total number of cattle being stated as 4,031,126,560, but the results satisfy only the first seven conditions. Lessing also gives results computed by Leiste, a clergyman of Wolfenbüttel, whose solution satisfies these seven conditions and likewise the eighth one. In 1821 J. and K. L. Struve published a critical and mathematical discussion, which was followed in 1828 by another from G. Hermann. The latter makes the interesting remark that Gauss had arrived at a complete solution; but, if so, no further information regarding it has been obtained. Many other attempts at solution were made, but the large numbers required to satisfy the nine conditions discouraged many investigators. Some critics thought that the original problem of Archimedes included only the first seven conditions and that the two others had been added by a later writer. It is, of course, clearly seen that the exercise as stated includes two problems, the first to find integral numbers that satisfy the first seven conditions, and the second to find integral numbers that satisfy all the nine conditions. As the poem says, the first problem may be solved by those of moderate proficiency in numbers, while the second can only be done by those of the highest skill.

The first problem is an easy one for a boy in the high school. Let W, B, D, Y represent the number of white, black, dappled and yellow bulls and let w, b, d, y represent the number of white, black, dappled and yellow cows. The seven conditions then give the seven equations:

${\begin{aligned}W&={\tfrac {5}{6}}\;B+Y,&(1)&\qquad \qquad \qquad &w&={\tfrac {7}{12}}(B+b),&(4)\\B&={\tfrac {9}{26}}D+Y,&(2)&&b&={\tfrac {9}{26}}(D+d),&(5)\\D&={\tfrac {12}{42}}W+Y,&(3)&&d&={\tfrac {11}{36}}(Y+y),&(6)\\&&&&y&={\tfrac {13}{42}}(W+w),&(7)\\\end{aligned}}$

and these contain eight unknown quantities. The problem, therefore, is of the kind called indeterminate, for many sets of numbers may be found to satisfy the seven equations. That set having the smallest numbers is the one required, for any other set may be found by multiplying these numbers by the same integer. If B and W are eliminated from equations (1), (2), (3), there will be found the single equation $891\,D=1{,}580\,Y,$ and hence $T=891$ and $D=1{,}580$ are the smallest integral numbers satisfying it; from these are found $B=1{,}602$ and $W=2{,}226.$ These values, before insertion in equations (4) to (7), are to be multiplied by a factor m, the value of which is later to be determined, so that the number of cows in each herd shall be an integer. Proceeding with the elimination, the values of w, b, d, y are deduced in terms of m, and it is then seen that 4,657 is the least value of m which will make the results integers. It is thus easily found that