Let $A$ be an even number but not divisible by $10$. The last two digits of $A^{20}$ are:

(A): $46$, (B): $56$, (C): $66$, (D): $76$, (E): None of the above.
1. **Understanding the problem**:
   We need to find the last two digits of \(A^{20}\) where \(A\) is an even number not divisible by 10. This means \(A\) is not divisible by 2 and 5 simultaneously.

2. **Using Euler's Totient Theorem**:
   Euler's Totient Theorem states that \(A^{\phi(n)} \equiv 1 \pmod{n}\) where \(\gcd(A, n) = 1\). Here, we choose \(n = 25\) because we are interested in the last two digits, which can be found using modulo 100. Since \(A\) is even and not divisible by 10, \(\gcd(A, 25) = 1\).

3. **Calculating \(\phi(25)\)**:
   \[
   \phi(25) = 25 \left(1 - \frac{1}{5}\right) = 25 \cdot \frac{4}{5} = 20
   \]
   Therefore, \(A^{20} \equiv 1 \pmod{25}\).

4. **Considering modulo 4**:
   Since \(A\) is even, \(A \equiv 0 \pmod{4}\). Therefore, \(A^{20} \equiv 0^{20} \equiv 0 \pmod{4}\).

5. **Using the Chinese Remainder Theorem**:
   We now have the system of congruences:
   \[
   \begin{cases}
   A^{20} \equiv 1 \pmod{25} \\
   A^{20} \equiv 0 \pmod{4}
   \end{cases}
   \]
   We need to find a number \(x\) such that:
   \[
   \begin{cases}
   x \equiv 1 \pmod{25} \\
   x \equiv 0 \pmod{4}
   \end{cases}
   \]

6. **Solving the system**:
   We can solve this system by checking numbers that satisfy \(x \equiv 1 \pmod{25}\) and also \(x \equiv 0 \pmod{4}\). We start with \(x = 1\) and keep adding 25 until we find a number that is divisible by 4:
   \[
   \begin{aligned}
   1 &\equiv 1 \pmod{25} \quad \text{but} \quad 1 \not\equiv 0 \pmod{4}, \\
   26 &\equiv 1 \pmod{25} \quad \text{but} \quad 26 \not\equiv 0 \pmod{4}, \\
   51 &\equiv 1 \pmod{25} \quad \text{but} \quad 51 \not\equiv 0 \pmod{4}, \\
   76 &\equiv 1 \pmod{25} \quad \text{and} \quad 76 \equiv 0 \pmod{4}.
   \end{aligned}
   \]
   Therefore, \(x = 76\) satisfies both congruences.

Conclusion:
\[
\boxed{76}
\]