先设公的白、黑、花、棕牛的数量分别是x1,x2,x3,x4只,母的白、黑、花、棕牛的数量分别是y1,y2,y3,y4只,依题意可知：
x1=x4+5/6*x2
x2=x4+9/20*x3
x3=x4+13/42*x1
y1=7/12*(x2+y2)
y2=9/20*(x3+y3)
y3=11/30*(x4+y4)
y4=13/42*(x1+y1)
整理,化简后,得到
x1=5936/2376*x4
x2=178/99*x4
x3=1580/891*x4
y1=2402120/1383129*x4
y2=543694/461043*x4
y3=3709101600773436857/4377498837804122112*x4
y4=73640654275250721919/56177901751819567104*x4
因为牛的数量必定是整数,x4=56177901751819567104*K?(K=1,2,3,...),取K=1得到一组
x1=140350178787374137344
x2=101006732442665484288
x3=99619623757435371520
x4=56177901751819567104
y1=97565781178820502702
y2=66248892435312513234
y3=47600137209925772010
y4=73640654275250721919
后面的数字太大了,是用计算机算的,真佩服当年阿基米德用手算出来的结果啊,牛人就是牛人!不过,太阳神的牛好多啊,估计比太阳系的恒星还要多!