The keys of a safe with five locks are cloned and distributed among eight people such that any of five of eight people can open the safe. What is the least total number of keys?

$ 
\textbf{a)}\ 18
\qquad\textbf{b)}\ 20
\qquad\textbf{c)}\ 22
\qquad\textbf{d)}\ 24
\qquad\textbf{e)}\ 25
$
1. **Understanding the problem**: We need to distribute the keys of a safe with five locks among eight people such that any group of five people can open the safe. We need to find the minimum total number of keys required.

2. **Key distribution requirement**: Each key must be distributed among at least 4 people. If a key is given to fewer than 4 people, then there will be at least 5 people who do not have that key, and thus, they will not be able to open the safe. Therefore, each key must be given to at least 4 people.

3. **Calculating the minimum number of keys**: Since there are 5 keys and each key must be given to at least 4 people, the minimum number of keys required is:
   \[
   5 \times 4 = 20
   \]

4. **Verification with an example**: We can verify this by constructing a distribution table where each key is given to exactly 4 people. Here is an example distribution:

\[
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
A_1 & \times & \times & \times & \times & & & & \\
\hline
A_2 & \times & \times & & & \times & \times & & \\
\hline
A_3 & \times & & \times & & \times & & \times & \\
\hline
A_4 & & \times & & \times & & \times & \times & \\
\hline
A_5 & & & \times & & \times & \times & & \times \\
\hline
\end{array}
\]

In this table:
- Key \(A_1\) is given to people 1, 2, 3, and 4.
- Key \(A_2\) is given to people 1, 2, 5, and 6.
- Key \(A_3\) is given to people 1, 3, 5, and 7.
- Key \(A_4\) is given to people 2, 4, 6, and 7.
- Key \(A_5\) is given to people 3, 5, 6, and 8.

This distribution ensures that any group of 5 people will have at least one person with each key, thus they can open the safe.

Conclusion:
The minimum total number of keys required is \( \boxed{20} \).