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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
2
Scan the series for triplets: (Perfect Square) (Odd Digit) (Even Digit). Note that 1, 4, 9 are perfect squares in this context.
Scan the series for triplets: (Perfect Square) (Odd Digit) (Even Digit). Note that 1, 4, 9 are perfect squares in this context.
Identify criteria
We are looking for a sequence of 3 digits: (Perfect Square, Odd Digit, Even Digit). Perfect squares in single digits are 1, 4, and 9. 1 is explicitly included.
Scan the sequence
Series: 1 2 8 2 4 1 7 6 1 7 3 2 9 7 5 1 7 8 6 8 2 4 3 9 4 6 6 2. Let's check every triplet: (1,2,8), (2,8,2), (8,2,4), (2,4,1), (4,1,7), (1,7,6) тЖТ [1:Square, 7:Odd, 6:Even - Valid!], (7,6,1), (6,1,7), (1,7,3), (7,3,2), (3,2,9), (2,9,7), (9,7,5), (7,5,1), (5,1,7), (1,7,8), (7,8,6), (8,6,8), (6,8,2), (8,2,4), (2,4,3), (4,3,9), (3,9,4), (9,4,6), (4,6,6), (6,6,2).
Evaluate findings
Reviewing the list: The only segment satisfying (Square, Odd, Even) is '1 7 6' at indices 6, 7, 8. No other consecutive triplets meet all three conditions.
A: 0 is incorrect; B: 2 is incorrect; D: 3 is incorrect. Only one triplet (1, 7, 6) fits the criteria.
C is correct because the only sequence of three numbers that follows the pattern (Perfect Square, Odd, Even) is '1 7 6' in the given series.
When counting sequences like this, write down the series and mark triplets as you go to avoid missing any potential matches.
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