2007-01-11 08:32:24 · 29 answers · asked by foster0676@sbcglobal.net 1 in Science & Mathematics ➔ Mathematics
That's pretty darned easy. Give them each 6 and keep the left over for himself.
Of course, there is always more than one way to skin a cat. He could always give them each 6 and have them share the responsibility for the remainder.
You also stated that "you" can not kill any of the cows so...
You can give them each 6 and just let the other one go.
You could also give them each 6 and have the butcher kill the remainder. Technically "you" would not have killed the cow.
2007-01-11 08:38:38 · answer #1 · answered by Anonymous · 2⤊ 0⤋
one cow dies from old age - it happens and that means that i didnt kill the cow - it just died. so as a result each son would then get 6 cows. but now what if all the cows are in calf? that would mean a possible 38 animals. and what if one or more have twins? then you really dont know how many animals there are and the sons still dont have thier cows. and if some of the calves are sent for slaughter (bc they are steers or crapy heifers) then once again ur number changes.
2007-01-11 08:39:13 · answer #2 · answered by bananananana 3 · 1⤊ 0⤋
Easy give each son 6 cows each and let the 19th cow go free
Hope this helps
2007-01-11 08:36:16 · answer #3 · answered by Police Artist 3 · 0⤊ 0⤋
Wait until 1 dies (giving you 18 cows) or two are pregnant (giving you 21 cows)
In both cases you have a number divisible by 3
2007-01-11 08:37:11 · answer #4 · answered by SusanB 5 · 0⤊ 0⤋
Kill two of the sons and give all 19 cows to the survivor.
2007-01-11 08:36:38 · answer #5 · answered by Anonymous · 2⤊ 1⤋
breed 2 cows & when they have their calves each son gets 7 cows...unless the calves are males then one son gets 7 cows & the other 2 sons each get 6 cows & a bull!
2007-01-11 08:40:46 · answer #6 · answered by SmallVoiceInBigWorld 6 · 0⤊ 0⤋
6 for each son and the last one do a drawing to see who gets the 19th cow.
2007-01-11 08:42:30 · answer #7 · answered by Anonymous · 0⤊ 0⤋
if there is a bull among them ... add the bull to one division ... have that the most trustworthy son coordinate pregnacy in 2 of his cows then trade those 2 pregnant cows for non pregnant cows of the other two sons.
2007-01-11 08:42:04 · answer #8 · answered by simply_made 4 · 0⤊ 0⤋
sell one cow and divide the money among the three so all his sons get an equal share of everything
2007-01-11 08:38:05 · answer #9 · answered by Anonymous · 0⤊ 0⤋
Give seven to one and six to the other two. The sons that get six, get a cow each that is pregnant.
2007-01-11 08:37:40 · answer #10 · answered by Anonymous · 0⤊ 0⤋