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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 an odd digit and also immediately followed by an even digit?
6
0
4
3
3
Look for triplets in the format: Odd Number - Even Number - Even Number (O-E-E).
Look for triplets in the format: Odd Number - Even Number - Even Number (O-E-E).
Define the pattern
We need to find an even digit that is preceded by an odd digit (Odd тЖТ Even) and followed by an even digit (Even тЖТ Even). This forms the triplet: Odd, Even, Even.
Scan the series
Series: 6, 8, 9, 4, 4, 8, 6, 5, 3, 5, 1, 7, 5, 8, 6, 5, 8, 9, 7, 3, 4, 7, 0, 1, 6, 8, 7, 4, 5, 4. Scanning: (9, 4, 4), (5, 8, 6), (5, 8, 6), (1, 6, 8).
Count the instances
1. 9-4-4, 2. 5-8-6 (after 6,5,3,5,1,7,5...), 3. 5-8-6 (later in the series), 4. 1-6-8. Total is 4.
A: Incorrect, count is not 3. B: Incorrect, count is not 5. C: Incorrect, count is not 6.
D is correct because there are exactly 4 instances of the O-E-E triplet pattern in the given series.
In such series questions, mark the odd/even status of each digit first to make the scanning process much faster.
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