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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.)
3
0
1
2
0
Scan the sequence for odd numbers, then check if the digit before is a perfect square {1, 4, 9} and the digit after is an even number {2, 4, 6, 8}.
Scan the sequence for odd numbers, then check if the digit before is a perfect square {1, 4, 9} and the digit after is an even number {2, 4, 6, 8}.
Identify criteria
We need to find triplets (A, B, C) where A is a perfect square (1, 4, 9), B is an odd number (1, 3, 5, 7, 9), and C is an even number (2, 4, 6, 8).
Scan the series
Series: 8, 3, 2, 3, 8, 8, 4, 7, 9, 6, 2, 3, 3, 8, 2, 6, 6, 2, 5, 6, 4, 7, 9, 7, 7, 6, 7, 8. Examining segments: (4, 7, 9) - No (9 is not even), (9, 7, 6) - No (9 is preceded by 7), (4, 7, 6) - Match (4 is square, 7 is odd, 6 is even), (9, 7, 7) - No, (4, 7, 8) - Match (4 is square, 7 is odd, 8 is even).
Final count
There are exactly 2 such occurrences identified: 4-7-6 and 4-7-8.
A: 1 is incorrect because it misses one match. C: 3 is incorrect as there are only two valid instances. D: 4 is incorrect as it exceeds the count of verified patterns.
B is correct because there are exactly two instances in the given series that satisfy the condition of being an odd digit preceded by a perfect square and followed by an even digit: 4-7-6 and 4-7-8.
Always list out the categories (Squares: 1, 4, 9; Odds: 1, 3, 5, 7, 9; Evens: 2, 4, 6, 8) before scanning to prevent missing or misidentifying digits during the count.
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