To solve for the last two digits of $(56789)^{41}$, we can use modular arithmetic. Specifically, we are interested in finding $(56789)^{41} \mod 100$.

Steps:

1. Reduce the base modulo 100:
   We start by reducing $56789 \mod 100$, since we only need the last two digits.

   $56789 \mod 100 = 89$

   Thus, we need to find $89^{41} \mod 100$.

2. Use Euler’s Theorem:
   Euler's theorem can simplify this problem, but first we need to compute $\phi(100)$, where $\phi(n)$ is Euler's totient function, which counts the number of integers from 1 to $n$ that are coprime to $n$.

   For $n = 100$, we know: $\phi(100) = 100 \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{5} \right) = 40$ Euler’s theorem tells us that if $a$ is coprime to 100, then $a^{\phi(100)} \equiv 1 \mod 100$. Since 89 is coprime to 100, we can use the property that $89^{40} \equiv 1 \mod 100$.

3. Simplify the exponent: We now need to compute $89^{41} \mod 100$, and we can break it down using the fact that: $89^{41} = 89^{40} \times 89$ From Euler’s theorem, we know $89^{40} \equiv 1 \mod 100$, so: $89^{41} \equiv 1 \times 89 \mod 100$ Therefore, we get: $89^{41} \equiv 89 \mod 100$

Conclusion:

The last two digits of $(56789)^{41}$ are 89.

Sub-topic:

This problem falls under the sub-topic of modular arithmetic and Euler's Theorem. Specifically, it involves calculating large powers modulo a number and applying properties of Euler's totient function.

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Would you like further details on any of the steps or concepts used in this solution? Here are some related questions you might find interesting:

1. What is Euler's Theorem and how does it apply to modular arithmetic?
2. How do you calculate Euler’s totient function for other numbers?
3. How can we use modular exponentiation techniques like "Exponentiation by Squaring"?
4. What happens if the base number is not coprime to 100?
5. How can we compute the last three digits of $(56789)^{41}$?

Tip: Modular exponentiation can be much more efficient for large numbers, especially when the exponent is large, because it avoids computing huge intermediate values.