To solve the problem, we need to analyze the sum \( A = 5 + 5^2 + 5^3 + 5^4 + \cdots + 5^{18} \).

This is a geometric series with the first term \( a = 5 \) and the common ratio \( r = 5 \). The sum of the first \( n \) terms of a geometric series is given by:

\[ S_n = a \frac{r^n - 1}{r - 1} \]

Here, \( n = 18 \), \( a = 5 \), and \( r = 5 \). So, the sum \( A \) can be written as:

\[ A = 5 \left( \frac{5^{18} - 1}{5 - 1} \right) = 5 \left( \frac{5^{18} - 1}{4} \right) \]

Now, let's analyze the divisibility of \( A \) by 5, 6, and 31.

### a. \( A \) divisible by 5

Since \( A \) is a sum of powers of 5, each term in the series is divisible by 5. Therefore, \( A \) is clearly divisible by 5.

### b. \( A \) divisible by 6

For \( A \) to be divisible by 6, it must be divisible by both 2 and 3.

- **Divisibility by 2**: Since \( A \) is a sum of powers of 5, and 5 is odd, each term in the series is odd. The sum of odd numbers is also odd, so \( A \) is not divisible by 2.
- **Divisibility by 3**: We need to check if \( A \) is divisible by 3. Notice that \( 5 \equiv 2 \pmod{3} \). Therefore, each power of 5 can be written as \( 5^k \equiv 2^k \pmod{3} \).

The series \( 5 + 5^2 + 5^3 + \cdots + 5^{18} \) modulo 3 becomes \( 2 + 2^2 + 2^3 + \cdots + 2^{18} \).

The powers of 2 modulo 3 cycle every 2 terms: \( 2 \equiv 2 \pmod{3} \) and \( 2^2 \equiv 1 \pmod{3} \).

So, the series modulo 3 is: \( 2 + 1 + 2 + 1 + \cdots + 2 + 1 \) (9 pairs of \( 2 + 1 \)).

Each pair sums to 3, which is divisible by 3. Therefore, the entire sum is divisible by 3.

Since \( A \) is not divisible by 2 but is divisible by 3, \( A \) is not divisible by 6.

### c. \( A \) divisible by 31

We need to check if \( A \) is divisible by 31. Notice that \( 5 \equiv 5 \pmod{31} \). Therefore, each power of 5 can be written as \( 5^k \pmod{31} \).

The series \( 5 + 5^2 + 5^3 + \cdots + 5^{18} \) modulo 31 becomes \( 5 + 5^2 + 5^3 + \cdots + 5^{18} \).

Using Fermat's Little Theorem, \( 5^{30} \equiv 1 \pmod{31} \). Therefore, \( 5^{18} \equiv 5^{18 \mod 30} \equiv 5^{18} \pmod{31} \).

We need to check if the sum \( 5 + 5^2 + 5^3 + \cdots + 5^{18} \) modulo 31 is 0.

The sum of a geometric series modulo 31 is:

\[ S = 5 \left( \frac{5^{18} - 1}{5 - 1} \right) \pmod{31} = 5 \left( \frac{5^{18} - 1}{4} \right) \pmod{31} \]

Since \( 5^{18} \equiv 1 \pmod{31} \) (by Fermat's Little Theorem), we have:

\[ 5^{18} - 1 \equiv 0 \pmod{31} \]

Thus, the numerator \( 5^{18} - 1 \) is divisible by 31, and hence \( A \) is divisible by 31.

### Summary

- \( A \) is divisible by 5.
- \( A \) is not divisible by 6.
- \( A \) is divisible by 31.