Two brothers share a flock of x sheep. They take the sheep to the market and sell each sheep for $x. At the end of the day, they put the money from the sales on the table to divide it equally. All money is in $10 bills, except for a few $1 bills (less than 10 of them). One at a time they take out $10 bills. The brother who draws first also draws last. The second brother complains about getting one less $10 bill, so the first brother offers him all the $1 bills. The second brother still received a total less than the first brother so he asks the first brother to write him a cheque to balance the things out. How much was the cheque for?
Answer
Since they sell x sheep for $x apiece, the total amount of money they will make is $x^2. In other words, the total is a perfect square. Since the first brother to draw a $10 bill is also the last one to draw, there must be an odd number of $10 bills. In other words, the number in the 10’s place of the total amount earned must be odd.
Any perfect square with an odd number in the 10’s place always has a 6 in the unit’s place (check it out – it’s true). Therefore, the brothers must have $6 in one-dollar bills. If the first brother has an extra $10 and the second brother has the $6, then the difference is $4, and the first brother has to write a cheque for $2 to even things out.
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