Step 1: Understanding the Question:
We need to find the number of times the digit '8' appears in the given sequence such that 8 is completely divisible (leaving a remainder of 0) by both the number immediately before it (preceding) and the number immediately after it (succeeding).
Step 2: Detailed Explanation:
Let us identify all the occurrences of 8 and check their preceding and succeeding numbers:
1. First 8: \( 2, \mathbf{8}, 3 \)
- Preceding: 2 (8 is divisible by 2)
- Succeeding: 3 (8 is not divisible by 3) \(\rightarrow\)
No
2. Second 8: \( 3, \mathbf{8}, 2 \)
- Preceding: 3 (8 is not divisible by 3) \(\rightarrow\)
No
3. Third 8: \( 4, \mathbf{8}, 2 \)
- Preceding: 4 (8 is divisible by 4)
- Succeeding: 2 (8 is divisible by 2) \(\rightarrow\)
Yes (1)
4. Fourth 8: \( 4, \mathbf{8}, 6 \)
- Preceding: 4 (8 is divisible by 4)
- Succeeding: 6 (8 is not divisible by 6) \(\rightarrow\)
No
5. Fifth 8: \( 6, \mathbf{8}, 2 \)
- Preceding: 6 (8 is not divisible by 6) \(\rightarrow\)
No
6. Sixth 8: \( 2, \mathbf{8}, 2 \)
- Preceding: 2 (8 is divisible by 2)
- Succeeding: 2 (8 is divisible by 2) \(\rightarrow\)
Yes (2)
7. Seventh 8: \( 4, \mathbf{8}, 3 \)
- Preceding: 4 (8 is divisible by 4)
- Succeeding: 3 (8 is not divisible by 3) \(\rightarrow\)
No
8. Eighth 8: \( 3, \mathbf{8}, 2 \)
- Preceding: 3 (8 is not divisible by 3) \(\rightarrow\)
No
9. Ninth 8: \( 2, \mathbf{8}, 6 \)
- Preceding: 2 (8 is divisible by 2)
- Succeeding: 6 (8 is not divisible by 6) \(\rightarrow\)
No
Thus, there are exactly 2 such 8s in the series (in the triplets \( 4, \mathbf{8}, 2 \) and \( 2, \mathbf{8}, 2 \)).
Step 3: Final Answer:
(B) 2