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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 odd digits are there, each of which is immediately preceded by a perfect square and immediately followed by an even digit? (Note: 1 will be considered as a perfect square.)
4
1
2
3
1
Look for a 3-digit sequence: [Perfect Square] [Odd Digit] [Even Digit]. Remember that 1 is a perfect square.
Look for a 3-digit sequence: [Perfect Square] [Odd Digit] [Even Digit]. Remember that 1 is a perfect square.
Define Criteria
We need to find an odd digit (1, 3, 5, 7, or 9) that is immediately preceded by a perfect square (1, 4, 9) and immediately followed by an even digit (0, 2, 4, 6, 8).
Scan the Series
Series: 8, 7, 7, 8, 5, 2, 7, 1, 4, [3], 2 (Match 1: 4, 3, 2); 4, [5], 2 (Match 2: 4, 5, 2); 3, 1, [3], 2 (Match 3: 1, 3, 2); 6, 7, 9, [7], 8 (Match 4: 9, 7, 8). Other triplets do not satisfy the condition.
A: 3 is incorrect, C: 5 is incorrect, D: 6 is incorrect. The series contains exactly 4 such patterns.
B is correct because there are exactly 4 instances where an odd digit is flanked by a perfect square on the left and an even digit on the right: (4,3,2), (4,5,2), (1,3,2), and (9,7,8).
When counting specific patterns in a long series, always mark the valid triplets as you find them to avoid double-counting or missing values.
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