A humble farmer goes to the market to exchange all his sheep for goats and horses. Upon arriving at the market, he finds that a vendor offers 56 horses for all his sheep and another one agrees to deliver 63 cows for the 72 sheep he owns.
The farmer wants to get rid of all the sheep and wants to buy both horses and cows.
How do you have to sell the 72 sheep to get some cows and some horses without getting over any sheep?
It is not possible to buy cows and horses without sheep.
The reasoning is as follows:
The first seller offers 7 cows for every 8 sheep while the second offers to exchange 9 sheep for 7 horses. This means that in order to get rid of all the animals you have to find a divisible number between 8 and another divisible number between 9 that add up to 72 sheep, but this is not possible since it is not possible to divide 72 by 8 and the rest is divisible between nine or the other way around.