Divide Evenly

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 less than ten excess $1 bills. 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 check to balance the things out. How much was the check?


Answer:


Let the total number of sheep be 10x+y, where y<10. The total money raised is (10x+y)^2 = 100x^2 + 20xy + y^2. Regardless of the values of x and y 100x^2 + 20xy will be divisible by 20. Because the number of $10 bills is odd y^2 mod 20 must be greater than 10 and less than 20. The only values of y where this is true is 4 and 6, where y^2 is either 16 or 36. Either way there will be 6 $1 bills left over. Before the check the first brother will have $4 more than the second brother. A $2 check will balance things out.