A dilemma in dividing up chocolates

The day after Valentine’s Day, a father ran into a math dilemma. He had bought a heart-shaped box of chocolates on clearance to be divided amongst his three daughters.

The issue came when he tried to divide the chocolates. The box contained 17 pieces of chocolates. He had promised 1/2 of the chocolates to the eldest daughter, 1/3 of the chocolates to the middle daughter, and 1/9 of the chocolates to the youngest daughter.

The problem: how was he supposed to divide the 17 pieces of candy? He wondered since the numbers did not divide evenly, and the chocolates were too special to be cut into fractions.

He asked his wife for help, and she came upon a solution.

From the box of chocolates she received (she got her own on Valentine’s Day), she added one chocolate piece to the group to make 18 in total. The wife then portioned out exactly 9 (= 1/2 x 18) chocolate pieces to the oldest daughter; 6 (= 1/3 x 18) chocolate pieces to the middle daughter; and 2 (= 1/9 x 18) chocolate pieces to the youngest daughter.

Having solved the problem of dividing the 17 chocolate pieces (17 = 9 + 6 + 2), the wife then took her piece of chocolate back!

It’s a fun story, and it is often told in terms of three sons dividing camels.

If you’re wondering why it worked out, I will refer to an explanation given on this page:

How did it work out? Add the fractions, 1/2 + 1/3 + 1/9. To add them, you must convert each to a fraction with their LEAST COMMON DENOMINATOR (LCD). Well, LCD = 2 · 9 = 18. So, 1/2 = 9/18; 1/3 = 6/18; 1/9 = 2/18. Hence, 1/2 + 1/3 + 1/9 = 9/18 + 6/18 + 2/18 = (9 + 6 + 2)/18 = 17/18. Then, the Algebraic Problem reads: (17/18)x = 17, or (dividing both sides by 17) x/18 = 1, or (multiplying both sides by 18) x = 18.

That is, x must be made 18 …. So, the stranger ADJOINS HIS [own] TO THE 17 MAKING THE TOTAL 18, which is a MULTIPLE of 2, 3, 9, hence, easily APPORTIONED.