Given the universal set `={x : x in N " and " x lt 20}`, find:
(i) `A={x : x=3p, p ne N}`
(ii) `B={y:y=2n+3, n in N}`
(iii) `C={x : x " x is divisible by 4"}`

Answer

To solve the given problem, we need to find the sets A, B, and C based on the universal set defined as \( U = \{ x : x \in \mathbb{N} \text{ and } x < 20 \} \).

### Step 1: Define the Universal Set
The universal set \( U \) consists of all natural numbers less than 20. Therefore, we can list the elements of \( U \):
\[ U = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19 \} \]

### Step 2: Find Set A
Set \( A \) is defined as:
\[ A = \{ x : x = 3p, p \notin \mathbb{N} \} \]
Since \( p \) is not a natural number, it can be 0 (the only whole number that is not a natural number). Thus, we calculate:
\[ x = 3 \times 0 = 0 \]
So, the set \( A \) is:
\[ A = \{ 0 \} \]

### Step 3: Find Set B
Set \( B \) is defined as:
\[ B = \{ y : y = 2n + 3, n \in \mathbb{N} \} \]
We will substitute natural numbers for \( n \) starting from 1:
- For \( n = 1 \): \( y = 2 \times 1 + 3 = 5 \)
- For \( n = 2 \): \( y = 2 \times 2 + 3 = 7 \)
- For \( n = 3 \): \( y = 2 \times 3 + 3 = 9 \)
- For \( n = 4 \): \( y = 2 \times 4 + 3 = 11 \)
- For \( n = 5 \): \( y = 2 \times 5 + 3 = 13 \)
- For \( n = 6 \): \( y = 2 \times 6 + 3 = 15 \)
- For \( n = 7 \): \( y = 2 \times 7 + 3 = 17 \)
- For \( n = 8 \): \( y = 2 \times 8 + 3 = 19 \)

Continuing this process, we find that the values of \( y \) that are less than 20 are:
\[ B = \{ 5, 7, 9, 11, 13, 15, 17, 19 \} \]

### Step 4: Find Set C
Set \( C \) is defined as:
\[ C = \{ x : x \text{ is divisible by } 4 \} \]
We will check the elements of the universal set \( U \) to see which are divisible by 4:
- 4 is divisible by 4
- 8 is divisible by 4
- 12 is divisible by 4
- 16 is divisible by 4

Thus, the set \( C \) is:
\[ C = \{ 4, 8, 12, 16 \} \]

### Summary of the Sets
- \( A = \{ 0 \} \)
- \( B = \{ 5, 7, 9, 11, 13, 15, 17, 19 \} \)
- \( C = \{ 4, 8, 12, 16 \} \)