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Refer to the given series and answer the question that follows (all numbers are single-digit numbers only, and all counting is to be done from left to 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 will be considered as a perfect square.)
2
3
1
0
2
Look for a 3-digit sequence: [Perfect Square] тЖТ [Even Digit] тЖТ [Odd Digit]. Remember 1 is a perfect square. Scan the series quickly for 1, 4, 9 as starts.
Look for a 3-digit sequence: [Perfect Square] тЖТ [Even Digit] тЖТ [Odd Digit]. Remember 1 is a perfect square. Scan the series quickly for 1, 4, 9 as starts.
Define Criteria
We need to find even digits that are (i) preceded by a perfect square (1, 4, 9) and (ii) followed by an odd digit.
Scan the sequence
Sequence: 3, 4, 8, 5, 6, 9, 2, 2, 1, 6, 6, 9, 3, 2, 4, 8, 9, 1, 6, 8, 3, 2, 2, 3, 8, 3, 2, 1. Scanning for [Square][Even][Odd]: 4-8-5 (Yes), 1-6-6 (No), 9-2-2 (No), 4-8-9 (Yes), 1-6-8 (No).
Final Count
The valid occurrences are: (4,8,5) and (4,8,9). Total = 2.
B: 1 is incorrect as we found 2 matches. C: 3 is incorrect. D: 4 is incorrect.
A is correct because there are exactly two instances of an even digit preceded by a perfect square and followed by an odd digit: 4-8-5 and 4-8-9.
Always list the digits carefully and mark them as S(Square), E(Even), or O(Odd) to avoid missing overlapping patterns.