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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 odd digit?
5
4
2
3
4
Scan the sequence for even numbers (0, 2, 4, 6, 8) and check if the digits immediately to their left and right are both odd (1, 3, 5, 7, 9).
Scan the sequence for even numbers (0, 2, 4, 6, 8) and check if the digits immediately to their left and right are both odd (1, 3, 5, 7, 9).
Analyze the series for the required condition
The series is: 2, 1, 6, 8, 3, 9, 5, 2, 7, 6, 3, 2, 9, 9, 7, 3, 9, 6, 4, 6, 3, 2, 6, 2, 2, 0, 3, 4, 7, 2. We look for (Odd, Even, Odd) triplets.
Identify the triplets
Checking the series: (1, 6, 8)-No, (8, 3, 9)-No, (5, 2, 7)-Yes (1st), (7, 6, 3)-Yes (2nd), (3, 2, 9)-Yes (3rd), (9, 6, 4)-No, (4, 6, 3)-Yes (4th), (6, 3, 2)-No, (3, 2, 6)-No, (2, 6, 2)-No, (6, 2, 2)-No, (2, 2, 0)-No, (0, 3, 4)-No, (3, 4, 7)-Yes (5th).
Final count verification
The valid occurrences are: 5-2-7, 7-6-3, 3-2-9, 3-4-7. Wait, checking again: ...5, 2, 7 (Valid), 7, 6, 3 (Valid), 3, 2, 9 (Valid), 3, 4, 7 (Valid). The total count is 4.
A: 3 is incorrect, B: 4 is correct based on the detected triplets, C: 5 is incorrect, D: 6 is incorrect.
B is correct because there are exactly 4 even digits (2, 6, 2, 4) that are surrounded by odd digits on both sides: 5-2-7, 7-6-3, 3-2-9, and 3-4-7.
When counting elements in a long string, mark them off as you find them to avoid double-counting or missing elements due to fatigue.
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