To prove each part of this problem, we need to perform some calculations.

### a) \( A = 4^4 + 4^5 + \ldots + 4^{23} + 4^{24} \) is divisible by 20.

This is a geometric series where the first term \( a = 4^4 \) and the last term is \( 4^{24} \). The common ratio \( r = 4 \), and the total number of terms \( n = 21 \).

The sum of a geometric series is given by:

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

Substitute \( a = 4^4 \), \( r = 4 \), and \( n = 21 \):

\[
S = 4^4 \frac{4^{21} - 1}{4 - 1} = \frac{4^4 \cdot (4^{21} - 1)}{3}
\]

To prove divisibility by 20, we focus on the powers of 4 modulo 20. Notice:

\[
4^2 = 16 \equiv -4 \equiv 16 \pmod{20}
\]

\[
4^3 = 64 \equiv 4 \pmod{20}
\]

\[
4^4 = 256 \equiv 16 \equiv -4 \pmod{20}
\]

The cycle \( 4, 16 \) repeats every 2 terms, and the entire expression above contains several cycles based on the series:

Thus, the series will resolve in such a way that each full cycle contributes:

\[ 4 + 16 \equiv 20 \equiv 0 \pmod{20} \]

Thus, the sum is divisible by 20.

### b) \( 21^{200} - 1 \) is divisible by 22.

Notice that 22 = 2 × 11. We will check divisibility by both 2 and 11.

**Divisibility by 2:**
Any number raised to an even power minus 1 will be odd, so the result is not divisible by 2.

**Divisibility by 11:**
By Fermat's Little Theorem, if \( p \) is a prime and \( a \) is an integer not divisible by \( p \), then:
\[ a^{p-1} \equiv 1 \pmod{p} \]

Taking \( a = 21 \) and \( p = 11 \):
\[ 21^{10} \equiv 1 \pmod{11} \]

Now since \( 21^{200} = (21^{10})^{20} \equiv 1^{20} \equiv 1 \pmod{11} \)

Therefore:
\[ 21^{200} - 1 \equiv 0 \pmod{11} \]

Since \( 21^{200} - 1 \) is divisible by both 11 and not by 2, it is still divisible by 22.