Let's break this down step by step to solve for $(A' \cup B') \cap C$.

Definitions:

1. Universe Set (U):
   $U = \{x : x \in \mathbb{N}, x < 25\} = \{1, 2, 3, \dots, 24\}$
2. Set A (Even Numbers < 25):
   $A = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\}$
3. Set B (Perfect Squares < 25):
   $B = \{1, 4, 9, 16\}$
4. Set C (Numbers Divisible by 4 < 25):
   $C = \{4, 8, 12, 16, 20, 24\}$

Step 1: Find $A'$ (Complement of A)

The complement of A is the set of elements in $U$ that are not in A: $A' = U - A = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23\}$

Step 2: Find $B'$ (Complement of B)

The complement of B is the set of elements in $U$ that are not in B: $B' = U - B = \{2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24\}$

Step 3: Find $A' \cup B'$ (Union of A' and B')

The union of A' and B' is the set of elements that are either in A' or in B', or in both: $A' \cup B' = \{1, 2, 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24\}$

Step 4: Find $(A' \cup B') \cap C$ (Intersection with C)

The intersection of $A' \cup B'$ with C is the set of elements that are in both $A' \cup B'$ and C: $(A' \cup B') \cap C = \{4, 8, 12, 16, 20, 24\} \cap \{4, 8, 12, 16, 20, 24\} = \{8, 12, 20, 24\}$

Final Answer:

The elements of $(A' \cup B') \cap C$ are: $\{8, 12, 20, 24\}$

Would you like further details on this, or have any other questions?

Here are some related questions for further exploration:

1. What is the difference between a union and an intersection in set theory?
2. How do you find the complement of a set?
3. Can you list all the prime numbers in $U$?
4. What would $A' \cap B'$ look like instead of $A' \cup B'$?
5. How would the solution change if $C$ contained multiples of 5 instead of multiples of 4?

Tip: Always double-check the complement of a set by comparing the set with the universal set to ensure no elements are missed.