Refer to the following series and answer the question. (Counting to be done from left to right only.) (all numbers are single dig…

Question

Refer to the following series and answer the question. (Counting to be done from left to right only.) (all numbers are single digit numbers only.) (Left) 9 2 3 1 4 5 6 8 5 4 3 2 1 9 3 4 5 6 7 8 1 2 9 8 3 4 5 2 6 8 2 (Right) How many such even digits are there, each of which is immediately preceded by a perfect square and immediately followed by an odd digit? (NOTE: 1 is also a perfect square.)

1. A. Three
2. B. More than three (Correct answer)
3. C. Two
4. D. One

Correct Answer

Option B — More than three

Detailed Solution & Explanation

The correct answer is More than three.

Key Points

• Three conditions must hold together for a digit to count: it is even, the digit immediately before it is a perfect square, and the digit immediately after it is odd.
• Single-digit perfect squares are 1, 4 and 9 (the note confirms 1 counts).
• Scanning the series 9 2 3 1 4 5 6 8 5 4 3 2 1 9 3 4 5 6 7 8 1 2 9 8 3 4 5 2 6 8 2 from left to right:
  • 9 → 2 → 3 ✓ (2 is even, preceded by 9, followed by 3)
  • 1 → 4 → 5 ✓
  • 1 → 2 → 9 ✓
  • 9 → 8 → 3 ✓
• That is four such digits, so the answer is more than three.

Additional Information

• Work left to right in a single pass, testing each even digit as you reach it. Trying to hold the whole series in mind at once is where errors creep in.
• The note that 1 is a perfect square is decisive — without it, two of the four cases (1→4→5 and 1→2→9) would be excluded and the answer would drop to two, which is exactly why option C is offered.
• Digits at the very ends cannot qualify, since they lack a preceding or following digit — the final 2 here is automatically excluded.
• Note that "immediately preceded/followed" means directly adjacent; a digit two places away does not count.