In the following series of numbers, how many times does the pattern 1, 7, 3 or 3, 7, 1 occur, with 7 in the middle and 1 and 3 on either side: 2 9 7 3 1 7 3 7 7 1 3 3 1 7 3 8 5 7 1 3 7 7 1 7 3 9 0 6?

Introduction / Context:
Pattern recognition in number sequences is a common type of reasoning question. Here, you are asked to identify how many times a specific three number pattern appears with a fixed middle number. The middle number is always 7, and the numbers on either side must be 1 and 3 in any order (either 1, 7, 3 or 3, 7, 1). The goal is to scan the entire sequence and count all such occurrences accurately.

Given Data / Assumptions:
- The series is: 2 9 7 3 1 7 3 7 7 1 3 3 1 7 3 8 5 7 1 3 7 7 1 7 3 9 0 6. - We seek triples of consecutive numbers where the middle number is 7. - The two neighbours must be 1 and 3, in any order. - Overlapping patterns are allowed as long as each triple satisfies the condition.

Concept / Approach:
The approach is to slide a window of three consecutive numbers across the sequence and inspect the middle element. Whenever the middle element is 7, check if the left and right elements are exactly 1 and 3 in any order. If they are, we count one valid occurrence. Proceeding this way from the start to the end ensures that no potential triple is missed and that the count is accurate.

Step-by-Step Solution:
Step 1: Rewrite the sequence with indices to avoid confusion: Index: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 Value: 2 9 7 3 1 7 3 7 7 1 3 3 1 7 3 8 5 7 1 3 7 7 1 7 3 9 0 6 Step 2: Consider triples (i - 1, i, i + 1) where the middle index i runs from 2 to 27. Step 3: For each triple, check if the middle value is 7. Step 4: When the middle is 7, check whether the other two numbers are 1 and 3 in any order. Step 5: At indices (4, 5, 6): 3, 1, 7 is not valid because 7 is not in the middle. Step 6: Look at indices (5, 6, 7): 1, 7, 3. Here the middle is 7, and neighbours are 1 and 3. Count 1. Step 7: At indices (13, 14, 15): 1, 7, 3. Again the middle is 7 with neighbours 1 and 3. Count 2. Step 8: At indices (21, 22, 23): 7, 7, 1 has middle 7 but neighbours are 7 and 1, not 1 and 3, so do not count. Step 9: At indices (22, 23, 24): 7, 1, 7 has neighbours 7 and 7, so not valid. Step 10: At indices (23, 24, 25): 1, 7, 3 again has middle 7 with neighbours 1 and 3. Count 3. Step 11: No other triple satisfies the pattern, so the total count is 3.

Verification / Alternative check:
To quickly verify, list all triples where the middle value is 7: (1, 7, 3) at indices 5–7, (1, 7, 3) at indices 13–15, and (1, 7, 3) at indices 23–25 are the only ones with neighbours 1 and 3. Triples like (3, 7, 7), (7, 7, 1), or (7, 1, 7) fail the neighbour condition. This re scan confirms that there are exactly three valid patterns.

Why Other Options Are Wrong:
4: Overcounts by including at least one triple where the neighbours are not 1 and 3. 5: Greatly overestimates and assumes many more matching triples than actually appear. More than 5: Clearly impossible once the sequence is examined carefully.

Common Pitfalls:
A typical error is to count any occurrence of 1, 3, and 7 in any order without insisting that 7 must be in the middle. Another is to double count overlapping triples or to misread the sequence and skip a term. Working with indices and checking each triple systematically avoids these mistakes.

Final Answer:
The pattern with 7 in the middle and 1 and 3 on either side occurs 3 times in the series.