The multiplicative content of a set is the product of its elements. If the set has only one element, its multiplicative content is this single element, and if the set is empty, its multiplicative content is 1. For example, the multiplicative content of $\{1,2,3\}$ is $1 \cdot 2 \cdot 3=6$.

a) Determine the sum of the multiplicative contents of all subsets of $\{1,2,3,4\}$.

b) Determine the sum of the multiplicative contents of all subsets of

$$
\left\{1, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{2016}\right\}
$$
Solution

a) Observe that in the product $(1+a)(1+b)$, when we apply the distributive property of multiplication of real numbers, the products $1 \cdot 1, 1 \cdot b, a \cdot 1$, and $a \cdot b$ appear. These products coincide with the multiplicative contents of the sets $\varnothing, \{b\}, \{a\}$, and $\{a, b\}$, respectively. These are exactly the subsets of $\{a, b\}$. In the multiplication $(1+a)(1+b)(1+c)$, when we again use the distributive property of multiplication, $2^{3}=8$ terms will appear. Each of these terms is associated with elements within each parenthesis that need to be chosen to appear in the corresponding term. For example, the term $1 \cdot b \cdot c = b \cdot c$ corresponds to not choosing $a$ in the first parenthesis and then choosing both $b$ and $c$ in the next two parentheses. The subsets of $\{a, b, c\}$ are associated with all these 8 terms, including the empty set, which has a multiplicative content of the number 1 and is represented in the sum by the term $1 \cdot 1 \cdot 1$. The same applies to any number of terms, and thus the sum of all multiplicative contents of the set $\{1, 2, 3, 4\}$ is the result $(1+1)(1+2)(1+3)(1+4) = 120$.

b) The argument used in the previous item also applies here, and consequently, the sought multiplicative content is

$$
\begin{aligned}
\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{2016}\right) & = \\
\left(\frac{2}{1}\right)\left(\frac{\not{3}}{2}\right)\left(\frac{4}{\not{2}}\right) \ldots\left(\frac{2017}{2016}\right) & = 2017
\end{aligned}
$$