How many even numbers are there in the following sequence which are immediately followed by an odd number and also immediately preceded by an even number? 8 5 8 6 7 6 8 9 3 2 7 5 3 4 2 2 3 5 5 2 2 8 1 1 9 3 1 7 5 1

Introduction / Context:
This problem involves scanning a sequence of numbers and identifying elements that satisfy a specific condition based on their neighbours. In this case we must count all even numbers that are immediately preceded by an even number and immediately followed by an odd number. Such questions test detailed observation and a clear understanding of even and odd parity.

Given Data / Assumptions:

• The sequence is: 8 5 8 6 7 6 8 9 3 2 7 5 3 4 2 2 3 5 5 2 2 8 1 1 9 3 1 7 5 1.
• We are concerned only with positions that have both a previous and a next neighbour (so the first and last terms cannot be counted).
• A number is even if it is divisible by 2, and odd otherwise.
• We must find numbers that are even and have an even neighbour on the left and an odd neighbour on the right.

Concept / Approach:
The strategy is to scan through the sequence from the second element to the second last element and check three conditions for each position i: the element at i is even, the element at i - 1 is even, and the element at i + 1 is odd. Every index that meets all three conditions contributes 1 to the count. It is important to check all three conditions together rather than partially.

Step-by-Step Solution:
Step 1: Mark each term as E (even) or O (odd). Sequence with parity: 8(E), 5(O), 8(E), 6(E), 7(O), 6(E), 8(E), 9(O), 3(O), 2(E), 7(O), 5(O), 3(O), 4(E), 2(E), 2(E), 3(O), 5(O), 5(O), 2(E), 2(E), 8(E), 1(O), 1(O), 9(O), 3(O), 1(O), 7(O), 5(O), 1(O). Step 2: Move through the sequence from the second to the second last element and test each even element. At position 4: the number is 6 (E), preceded by 8 (E) and followed by 7 (O). This satisfies all three conditions. At position 7: the number is 8 (E), preceded by 6 (E) and followed by 9 (O). This also satisfies the conditions. At position 16: the number is 2 (E), preceded by 2 (E) and followed by 3 (O). This qualifies. At position 22: the number is 8 (E), preceded by 2 (E) and followed by 1 (O). This also qualifies. Step 3: No other even number in the sequence has both an even neighbour on the left and an odd neighbour on the right. Step 4: Therefore, exactly 4 numbers satisfy the required condition.

Verification / Alternative check:
We can double check by listing only the triplets (previous, current, next) where the current is even. The four valid triplets are (8, 6, 7), (6, 8, 9), (2, 2, 3) and (2, 8, 1). Each of these clearly has an even previous term, an even current term and an odd next term. Counting these triplets again gives 4, confirming the result.

Why Other Options Are Wrong:
Counts of 2 or 3 arise if some qualifying positions are missed, while a count of 5 would require one more triplet satisfying the condition, which does not exist in the sequence.

Common Pitfalls:
Students often forget to check both neighbours or mistakenly treat the first or last element as valid even though they lack one neighbour. Others may confuse the parity of certain numbers when scanning quickly. Marking E and O clearly and then checking each candidate systematically helps to avoid such mistakes.

Final Answer:
There are 4 such even numbers in the sequence.