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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.)
1
0
4
2
4
Scan the series for all odd digits, then check if the digit immediately to its left is in {1, 4, 9} and the digit immediately to its right is {0, 2, 4, 6, 8}.
Scan the series for all odd digits, then check if the digit immediately to its left is in {1, 4, 9} and the digit immediately to its right is {0, 2, 4, 6, 8}.
Identify criteria
We need to find an odd digit (1, 3, 5, 7, 9) preceded by a perfect square (1, 4, 9) and followed by an even digit (0, 2, 4, 6, 8).
Analyze the series
Series: 2, 2, 4, 8, 2, 4, 6, 4, 2, 6, 8, 8, 6, 7, 4, 5, 3, 6, 2, 8, 7, 1, 3, 3, 6, 8, 6, 4. Triplets (Square-Odd-Even): (4, 5, 3) - No (3 is odd), (4, 7, 4) - Yes (4 is square, 7 is odd, 4 is even), (8, 7, 1) - No, (1, 3, 3) - No, (6, 7, 1) - No. Wait, re-evaluating: 4,7,4 (Yes), 8,7,1 (No), 3,6,2 (No). Let's check systematically: 4-7-4 works. No other triplet fits the criteria.
Verification
Checking sequence:...6, 7, 4, 5, 3... тЖТ (4, 5, 3 is even), (4, 7, 4 is correct: 4 is square, 7 is odd, 4 is even). Searching remaining:...2, 8, 7, 1, 3... тЖТ (8, 7, 1 is odd), (1, 3, 3 is odd). Only one such occurrence found.
A: Represents 2, which is incorrect. C: Represents 3, which is incorrect. D: Represents 4, which is incorrect.
B is correct because only one triplet (4, 7, 4) satisfies the condition of having an odd digit preceded by a perfect square and followed by an even digit.
When searching for patterns in strings, highlight the target positions (predecessor and successor) to avoid skipping digits accidentally.
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