To prove that \( A = 5^1 + 5^2 + 5^3 + \ldots + 5^{20} \) is divisible by 5, 6, and 13, we can use the properties of geometric series and modular arithmetic.

### Step 1: Calculate \( A \)

The sum \( A \) is a geometric series, which can be calculated using the formula for the sum of a geometric progression:

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

where
- \( a = 5^1 = 5 \),
- \( r = 5 \),
- \( n = 20 \).

Now, applying the formula:

\[
A = 5 \frac{5^{20} - 1}{5 - 1} = 5 \frac{5^{20} - 1}{4}
\]

### Step 2: Check divisibility by 5

Clearly, since \( A = 5 \cdot \frac{5^{20} - 1}{4} \), \( A \) is divisible by 5.

### Step 3: Check divisibility by 6

To check divisibility by 6, we need to show that \( A \) is divisible by both 2 and 3.

1. **Divisibility by 2**:
- Since \( 5^{20} \) is odd, \( 5^{20} - 1 \) is even. Therefore, \( \frac{5^{20} - 1}{4} \) is an integer, and hence \( A \) is even. So, \( A \) is divisible by 2.

2. **Divisibility by 3**:
- We can calculate \( A \mod 3 \):
- \( 5 \equiv 2 \mod 3 \), so \( 5^k \equiv 2^k \mod 3 \).
- Therefore:
\[
A \mod 3 = 2^1 + 2^2 + 2^3 + \ldots + 2^{20}
\]
- The sum of a geometric series gives:
\[
= 2 \frac{2^{20} - 1}{2 - 1} = 2(2^{20} - 1)
\]
- Since \( 2^{20} \equiv 1 \mod 3 \), we have:
\[
2(1 - 1) \equiv 0 \mod 3
\]
- Thus, \( A \) is divisible by 3.

Since \( A \) is divisible by both 2 and 3, it is also divisible by 6.

### Step 4: Check divisibility by 13

Now we check \( A \mod 13 \):

1. We find that \( 5^1, 5^2, ..., 5^{12} \) can help us with finding a repeating cycle:
- Calculate powers of 5 mod 13:
- \( 5^1 \equiv 5 \)
- \( 5^2 \equiv 12 \)
- \( 5^3 \equiv 8 \)
- \( 5^4 \equiv 1 \) (this means the cycle repeats every 4)

Since the cycle is \( 5, 12, 8, 1 \):

To find \( A \):
- Collect terms from cycles:
- For 17 terms, we can find complete cycles:

\[
5 + 12 + 8 + 1 = 26 \equiv 0 \mod 13
\]
Repeating this for the 20 terms results in:

\[
5^1 + 5^2 + 5^3 + 5^4 + 5^1 + 5^2 + 5^3 + 5^4 + 5^1 + 5^2 + 5^3 + 5^4 + 5^1 + 5^2 + 5^3 \equiv 0 \mod 13
\]

Thus, \( A \) is divisible by 13.

### Conclusion

In conclusion, we have shown that \( A \) is divisible by 5, 6, and 13. Therefore, \( A \) satisfies the required conditions.