Problem 602

Solution 1

Each of the $n$ friends is to
receive $\dfrac{1}{n}$ of the
pizza.

Since there are two pieces that are each $\dfrac{1}{6}$ of the pizza and these
pieces cannot be cut, then each friend receives at least $\dfrac{1}{6}$ of the pizza. This means
that there cannot be more than 6 friends; that is, $n \leq 6$.

Therefore, $n =7, 8, 9, 10$ are not
possible. The sum of these is 34.

The value $n=2$ is possible. We
show this by showing that the pieces can be divided into two groups,
each of which totals $\dfrac{1}{2}$
of the pizza.

Note that $\$\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{12} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{3}{6} = \dfrac{1}{2}\$.$

This also means that the other 6 pieces must also add to $\dfrac{1}{2}$.

We show that the value of $n=3$ is
possible by finding 3 groups of pieces, with each group totalling $\dfrac{1}{3}$ of the pizza.

Since $2 $\times \dfrac{1}{6} = \dfrac{1}{3}$$and$4 $\times \dfrac{1}{12} = \dfrac{1}{3}\$,$ then the other 4 pieces must also
add to $\dfrac{1}{3}$ (the rest of
the pizza) and so $n=3$ is
possible.

The value $n = 4$ is possible
since $2 $\times \dfrac{1}{8} = \dfrac{1}{4}$$and$$\dfrac{1}{6} + \dfrac{1}{12} = \dfrac{2}{12} + \dfrac{1}{12} = \dfrac{3}{12} = \dfrac{1}{4}\$$ (which can be done twice). The other 4 pieces must
also add to $\dfrac{1}{4}$.

The value $n = 6$ is possible
since two pieces are $\dfrac{1}{6}$
on their own, two groups of size $\dfrac{1}{6}$ can be made from the four
pieces of size $\dfrac{1}{12}$, and
$\$\dfrac{1}{8} + \dfrac{1}{24} = \dfrac{3}{24} + \dfrac{1}{24} = \dfrac{4}{24} = \dfrac{1}{6}\$$ (which can be
done twice), which makes 6 groups of size $\dfrac{1}{6}$.

The sum of the values of $n$ that
are not possible is either 34 (if $n=5$ is possible) or 39 (if $n = 5$ is not possible). Since 34 is not
one of the choices, the answer must be 39.

(We can see that $n=5$ is not
possible since to make a portion of size $\dfrac{1}{5}$ that includes a piece of
size $\dfrac{1}{6}$, the remaining
pieces must total $\$\dfrac{1}{5} - \dfrac{1}{6} = \dfrac{6}{30} - \dfrac{5}{30} = \dfrac{1}{30}\$.$
Since every piece is larger than $\dfrac{1}{30}$, this is not possible.)

Solution 2

The pizza is cut into 2 pieces of size $\frac{1}{24}$, 4 of $\frac{1}{12}$, 2 of $\frac{1}{8}$, and 2 of $\frac{1}{6}$.

Each of these fractions can be written with a denominator of 24, so we
can think of having 2 pieces of size $\frac{1}{24}$, 4 of $\frac{2}{24}$, 2 of $\frac{3}{24}$, and 2 of $\frac{4}{24}$.

To create groups of pieces of equal total size, we can now consider
combining the integers 1, 1, 2, 2, 2, 2, 3, 3, 4, and 4 into groups of
equal size. (These integers represent the size of each piece 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 $n = 7, 8, 9, 10$ are not
possible.

Here is a way of breaking the slices into $n=6$ equal groups, each with total size
$24 \div 6 = 4$: $$4 \qquad 4 \qquad 3+1 \qquad 3+1 \qquad 2+2 \qquad 2+2$$ Here is a way of breaking the slices into $n=4$ equal groups, each with total size
$24 \div 4 = 6$: $$4 + 2 \qquad 4 + 2 \qquad 3+3 \qquad 2+2+1+1$$ Here is a way of breaking the slices into $n=3$ equal groups, each with total size
$24 \div 3 = 8$: $$4 + 4 \qquad 2+ 2+ 2+2 \qquad 3+3+1+1$$
Here is a way of breaking the slices into $n=2$ equal groups, each with total size
$24 \div 2 = 12$: $$4 + 4 +2+2 \qquad 3+3+2+2+1+1$$ Since
$24$ is not a multiple of 5, the
pieces cannot be broken into 5 groups of equal size.

Therefore, the sum of the values of $n$ that are not possible is $5+7+8+9+10 =39$.