Problem D and Solution

Ten Slice Pizza

The DECI-Pizza Company has a special pizza that has \(10\) slices. Two of the slices are each \(\frac{1}{6}\) of the whole pizza, two are each \(\frac{1}{8}\), four are eachÂ \(\frac{1}{12}\), and two are each \(\frac{1}{24}\). A group of \(n\) friends share the pizza by distributing all of these slices. They do not cut any of the slices. Each of the \(n\) friends receives, in total, an equal fraction of the whole pizza. For what values of \(n > 1\) is this possible?

**Solution 1**

Each of the \(n\) friends is to receive \(\frac{1}{n}\) of the pizza.

Since there are two slices that are each \(\frac{1}{6}\) of the pizza and these slices cannot be cut, then each friend receives at least \(\frac{1}{6}\) of the pizza. This means that there cannot be more than \(6\) friends. That is, \(n \leq 6\).

The value \(n = 2\) is possible. We show this by dividing the slices into two groups, each of which totals \(\frac{1}{2}\) of the pizza. Note that \(\frac{1}{6} + \frac{1}{6} + \frac{1}{12} +\frac{1}{12} = \frac{1}{2}\). This means the other six slices must also add to \(\frac{1}{2}\).

The value \(n=3\) is possible. We show this by finding three groups of slices, with each group totaling \(\frac{1}{3}\) of the pizza. Since \(2 \times \frac{1}{6} = \frac{1}{3}\) and \(4\times \frac{1}{12} = \frac{1}{3}\), then the other four slices must also add to \(\frac{1}{3}\) (the rest of the pizza), and so \(n = 3\) is possible.

The value \(n = 4\) is possible since \(2 \times \frac{1}{8} = \frac{1}{4}\) and \(\frac{1}{6} + \frac{1}{12} = \frac{1}{4}\) (which can be done twice). The other four slices must also add to \(\frac{1}{4}\).

The value \(n = 6\) is possible since two slices are \(\frac{1}{6}\) on their own, two groups of size \(\frac{1}{6}\) can be made from the four slices of size \(\frac{1}{12}\), and \(\frac{1}{8} + \frac{1}{24} = \frac{1}{6}\) (which can be done twice), which makes six groups of size \(\frac{1}{6}\).

The value \(n = 5\) is not possible, since to make a portion of size \(\frac{1}{5}\) that includes a slice of size \(\frac{1}{6}\), the remaining slices must total \(\frac{1}{5} - \frac{1}{6} = \frac{1}{30}\). Since every slice is larger than \(\frac{1}{30}\), this is not possible.

Therefore, the possible values of \(n\) are \(2\), \(3\), \(4\), and \(6\).

**Solution 2**

The pizza is cut into two slices of size \(\frac{1}{24}\), four slices of size \(\frac{1}{12}\), four slices of size \(\frac{1}{8}\), and two slices of size \(\frac{1}{6}\).

Each of these fractions can be written with a denominator of \(24\). Thus, this is equivalent to saying there are two slices of size \(\frac{1}{24}\), four slices of size \(\frac{2}{24}\), two slices of size \(\frac{3}{24}\), and two slices of size \(\frac{4}{24}\).

To create groups of slices of equal total size, we can now consider combining the integers \(1\), \(1\), \(2\), \(2\), \(2\), \(2\), \(3\), \(3\), \(4\), and \(4\) into groups with equal sum. (These integers represent the size of each slice measured in units of \(\frac{1}{24}\) of the pizza.)

Since the largest integer in the list is \(4\), then each group has to have size at least \(4\). Since \(4 = 24 \div 6\), then the slices cannot be broken into more than \(6\) groups of equal size, which means that \(n\) cannot be greater than \(6\).

Here is a way of breaking the slices into \(n = 6\) equal groups, each with total size \(24 \div 6 = 4\): \[4, \quad 4, \quad 3+1, \quad 3+1, \quad 2+2, \quad 2+2\]

Here is a way of breaking the slices into \(n = 4\) equal groups, each with total size \(24 \div 4 = 6\): \[4+2, \quad 4+2, \quad 3+3, \quad 2+2+1+1\]

Here is a way of breaking the slices into \(n = 3\) equal groups, each with total size \(24 \div 3 = 8\): \[4+4, \quad 2+2+2+2, \quad 3+3+1+1\]

Here is a way of breaking the slices into \(n = 2\) equal groups, each with total size \(24 \div 2 = 12\): \[4+4+2+2, \quad 3+3+2+2+1+1\] Since \(24\) is not a multiple of \(5\), the slices cannot be broken into \(n=5\) groups of equal size.

Therefore, the possible values of \(n\) are \(2\), \(3\), \(4\), and \(6\).