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 fewer 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?

The check was for 2 dollars.

Since the x sheep were sold for x dollars, they received x^2 dollars. Since the brother who drew first also drew last, we know that the 10’s place of their sheep-money is odd. Let’s first see what would happen if they sold 1-10 sheep. The possible sheep-money they received is 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. Now, I know that the first three items aren’t really possible for this problem statement (since the first brother is supposed to make out with more money than the second), but bear with me here =)

Technically, the 10’s place of all of those numbers is even, except for 16 and 36. This looks promising, since we have more than one possibility for how much sheep-money they received, but the 1's place is the same for each. However, we should prove that there aren’t any more numbers that could fit the bill.

Let’s say x = (n*10 + m) where n is an element of [0, 1, 2, 3, 4, 5…} and m is an element of [1, 2, 3, … 8, 9, 10]. Then x^2 = (n*10 + m)*(n*10 + m) = 100*n^2 + 20*n*m + m^2.

Remember, we only care about if the 10’s place is odd, and then what the corresponding 1’s place is. Let’s look at each term. No matter what value n is, 100*n^2 will always have zero in its 10’s place. And if we consider 20*n*m, this term will always have an even value for it’s 10’s place. So the only term that can have an odd 10’s place (and thus make x^2 have an odd 10’s place) is m^2.

And we have already shown that only happens when m is a 4 or a 6. So, x must end in a 4 or a 6, and x^2 will always end in a 6. So the difference between the extra 10 that the first brother gets, and all 6 ones that the second brother gets is 4. So if the first brother writes a check for $2, that means he loses $2 while his brother gains $2, thus closing the $4 gap that was between them.

Later!

Posted by nikki on 2003-10-06 10:07:28