Alice picks four numbers from the set $\{1,2,3,4,5,6\}$, tells Bob their product and asks him to guess their sum. Bob realizes he cannot even determine for sure whether the sum is odd or even. What is the product of the numbers Alice chose?
Let $P$ be said product. Evidently there are two distinct sets of numbers $\left\{x_{1}, x_{2}, x_{3}, x_{4}\right\}$ and $\left\{y_{1}, y_{2}, y_{3}, y_{4}\right\}$ such that $x_{1} x_{2} x_{3} x_{4}=y_{1} y_{2} y_{3} y_{4}=P$, but $x_{1}+x_{2}+x_{3}+x_{4}$ and $y_{1}+y_{2}+y_{3}+y_{4}$ have different parity.

Instead of considering the four numbers Alice picks, we consider instead the two numbers Alice does not pick; $\left\{x_{5}, x_{6}\right\}$ and $\left\{y_{5}, y_{6}\right\}$. They have the same properties we described above, since

$$
x_{5} x_{6}=\frac{6 !}{x_{1} x_{2} x_{3} x_{4}}=\frac{720}{P}=\frac{6 !}{y_{1} y_{2} y_{3} y_{4}}=y_{5} y_{6}
$$

and analogously the sums also have different parities.

Thus we are looking for pairs of distinct numbers in $\{1,2,3,4,5,6\}$ which have the same product but different sums. We can record the entire multiplication table, as below.

| $\times$ | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 |  | 2 | 3 | 4 | 5 | 6 |
| 2 |  |  | 6 | 8 | 10 | 12 |
| 3 |  |  |  | 12 | 15 | 18 |
| 4 |  |  |  |  | 20 | 24 |
| 5 |  |  |  |  |  | 30 |
| 6 |  |  |  |  |  |  |

The numbers with duplicate entries are

$$
\begin{aligned}
6 & =6 \cdot 1=2 \cdot 3 \\
12 & =2 \cdot 6=3 \cdot 4
\end{aligned}
$$

Of these, 12 has the desired property but 6 does not. Hence, Alice chose one of the quadruples $\{1,3,4,5\}$ or $\{1,2,5,6\}$; the product is then $P=1 \cdot 3 \cdot 4 \cdot 5=1 \cdot 2 \cdot 5 \cdot 6=60$.