In the series 2 8 3 8 2 4 8 2 4 8 6 8 2 8 2 4 8 3 8 2 8 6, how many times does the number 8 appear such that it is exactly divisible by both its immediate preceding number and its immediate succeeding number?

Introduction / Context:
This question combines sequence scanning with number divisibility. You are asked to examine every occurrence of a particular number (8) in a series and test a specific condition involving the numbers immediately before and after it. Such problems are common in reasoning tests, where careful reading and systematic checking are more important than complex calculations.

Given Data / Assumptions:
- The series is: 2 8 3 8 2 4 8 2 4 8 6 8 2 8 2 4 8 3 8 2 8 6. - We focus only on positions where the middle number is 8. - For each such 8, the number before it and the number after it must both divide 8 exactly (8 % neighbour = 0). - Edge terms that do not have both a predecessor and a successor cannot be counted.

Concept / Approach:
The method is straightforward: scan through the series, locate each 8 that has both a previous and a next term, and then check divisibility. A number a is “exactly divisible” by another number b if the remainder of a / b is zero. Therefore, we test 8 % previous == 0 and 8 % next == 0. Whenever both conditions hold, we count that occurrence of 8. Finally, we add up how many such valid occurrences exist in the entire series.

Step-by-Step Solution:
Step 1: Write down the series with indices for clarity: Index: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 Value: 2 8 3 8 2 4 8 2 4 8 6 8 2 8 2 4 8 3 8 2 8 6 Step 2: Identify positions where the value is 8 and has both neighbours: indices 2, 4, 7, 10, 12, 14, 17, 19, 21. Step 3: At index 2: neighbours are 2 and 3. 8 is divisible by 2 (8 % 2 = 0) but not by 3 (8 % 3 = 2), so do not count. Step 4: At index 4: neighbours are 3 and 2. 8 is not divisible by 3, so do not count. Step 5: At index 7: neighbours are 4 and 2. 8 % 4 = 0 and 8 % 2 = 0, so this 8 is valid and we count one. Step 6: At index 10: neighbours are 4 and 6. 8 is divisible by 4 but not by 6, so not counted. Step 7: At index 12: neighbours are 6 and 2. 8 is not divisible by 6, so not counted. Step 8: At index 14: neighbours are 2 and 2. 8 % 2 = 0 and 8 % 2 = 0, so this 8 is valid; count a second occurrence. Step 9: At indices 17, 19, and 21, at least one neighbour does not divide 8 exactly, so these are discarded. Step 10: The total count of valid 8s is 2.

Verification / Alternative check:
You can quickly re scan only the local triples where 8 is in the middle: (2, 8, 3), (3, 8, 2), (2, 4, 8), (4, 8, 6), (6, 8, 2), (2, 8, 2), (4, 8, 3), (3, 8, 2), (2, 8, 6). Among these, only (4, 8, 2) and (2, 8, 2) have both neighbours as exact divisors of 8. This confirms the answer without re-doing every index step.

Why Other Options Are Wrong:
five: Greatly overcounts because most occurrences of 8 fail the divisibility test with at least one neighbour. four: Also overestimates; only two triples satisfy the condition completely. one: Underestimates by acknowledging only one valid triple instead of two.

Common Pitfalls:
A common mistake is to count every 8 whose neighbours are smaller numbers, assuming they must be divisors, without actually checking the remainder. Another error is to ignore the requirement that both sides must divide 8, not just one side. Keeping the definition of exact divisibility in mind and checking each triple carefully prevents these errors.

Final Answer:
The number of times 8 appears with both neighbours exactly dividing it is two.