Two brothers have a flock of sheep. One day a buyer appears at home and makes the following offer:
“I offer you as many euros per sheep as there are sheep in the herd”.
Seeing the amount of money he represented, the brothers accepted the deal without hesitation and the buyer paid them using 10-euro bills and some 1-euro coins.
To distribute the money, the older brother took the wad of bills and kept a ticket, then gave one to the brother, and continued to distribute tickets alternately to one and the other until they were finished so that the last ticket was also for the.
Then, the younger brother said: “You took the first and last ticket, so you have 10 euros more than me”.
To which the major responded: "Well, you keep all the coins".
Not satisfied, the minor said: "But there are less than 10 coins, you will still have more money".
And the elder replied: "Okay, so keep the coins and I'll give you a check so we have the same money.".
What value will the check have?
The value of the check will be 2 euros.
Since the price per sheep is equal to the number of sheep, the total price must be a perfect square of the type:
1×1 = 1
2×2 = 4
3×3 = 9
4×4 = 16
In addition, for the distribution of tickets to be odd, the number of tens of the total price must be odd and this is only true for squares of numbers ending in 4 or 6.
Therefore we have the following possibilities:
4×4 = 16
6×6 = 36
14×14 = 196
16×16 = 256
24×24 = 576
26×26 = 676
34×34 = 1156
36×36 = 1296
If you look, we see that everyone ends in 6, therefore, the first brother always has (10 - 6 =) 4 euros more than the second, so You must return 2 euros to be at peace.