Let's analyze and solve the problem step-by-step.

Problem Interpretation

We have three subsets $A_1$, $A_2$, and $A_3$ of the set $S = \{1, 2, 3, 4, 5\}$. The subsets must satisfy: $(A_1 \cup A_2) \cap (A_2 \cup A_3) = \{1, 2, 3, 4, 5\}.$ This implies that every element in $S$ must be in either $A_1 \cup A_2$ or $A_2 \cup A_3$, meaning every element of $S$ must appear in at least one of the subsets $A_1$, $A_2$, or $A_3$.

Our goal is to find $N$, the number of ways to construct the list of sets $(A_1, A_2, A_3)$ that satisfies this condition, and then find the last two digits of $N$.

Solution Outline

To ensure each element $x \in S$ is in at least one of $A_1$, $A_2$, or $A_3$, we have several choices for each element:

1. Include $x$ in $A_1$.
2. Include $x$ in $A_2$.
3. Include $x$ in $A_3$.
4. Include $x$ in both $A_1$ and $A_2$.
5. Include $x$ in both $A_2$ and $A_3$.
6. Include $x$ in both $A_1$ and $A_3$.
7. Include $x$ in all three subsets $A_1$, $A_2$, and $A_3$.

This provides 7 choices per element in $S$.

Total Number of Ways

Since there are 5 elements in $S$ and each element has 7 choices, the total number of ways to construct the list of sets $(A_1, A_2, A_3)$ is: $N = 7^5.$

Calculating $7^5$ Modulo 100

We need to find the last two digits of $N$, which is $7^5 \mod 100$.

1. Compute $7^5$: $7^5 = 16807.$

2. Find the last two digits of 16807: $16807 \mod 100 = 07.$

Answer

The last two digits of $N$ are 07.

Would you like more details on this approach or any other aspect? Here are some related questions:

1. What happens if we change the set $S$ to have more elements?
2. How would the solution change if we had different subset requirements?
3. Can we use combinatorial arguments to simplify the solution for other similar problems?
4. How can we compute powers modulo 100 more efficiently?
5. What if each element had a constraint to be in exactly two sets?

Tip: When calculating powers modulo a number, breaking down the power into steps or using modular exponentiation can help simplify large computations.