Cheap Sheep...Bleat!
A brother and a sister inherited a herd of sheep. They sold all of them, receiving for each sheep the same number of dollars as there were sheep in the herd.
The money was given to them in $10 bills except for an excess, less than $10, that was in silver dollars. They divided the bills between them by placing them on a table and alternately taking a bill until there was none left.
"It isn't fair," complained the sister. "You drew first and you also took the last bill, so you got $10 more than I did."
To even things up partially, the brother gave his sister all of the silver dollars, but she was still not satisfied. "You gave me less than $10," she argued. "You still owe me some money."
"True," said the brother. "Suppose I write you a check that will make the total amounts we each end up with exactly the same."
This he did. What was the value of the check?
Source: A. N. Myous Student
Hint: Suppose there were n sheep...then the total amount received is....
And be careful in the very last step.
Solution Commentary: The submitter's commentary: If the number of sheep is n, the total number of dollars received is n^{2}. This was paid in $10 bills, plus an excess, less than $10, in silver dollars.
By alternately taking bills, the brother drew both first and last, and so the total amount must contain an odd number of $10's. Since the square of any ultiple of 10 contains an even number of 10's, we conclude that n must end in a digit the square of which contains an odd number of 10's. Only two digits, 4 and 6, have such squares: 16 and 36, Both squares end in 6. The excess amount consisted of 6 silver dollars.
After the sister took the $6 she still had $4 less than her brother, so to even things up the brother wrote a check for $2.
It is surprising how many good mathematicians will work the problem correctly up to this last step, then forget that the check must be $2 instead of $4.
