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Eureka Airlines operates 5 daylong flight sequences serving City A, City B, City C, City D, and City E. The following list shows the order of the cities served by each flight sequence.
Due to the number of available flight crews, Eureka can operate only 3 flight sequences in a given day. Exactly 10 flights are flown each day. No sequence is used for more than 2 consecutive days, and no sequence is idle for 2 consecutive days.
For City A, select the number of Eureka flights that will arrive in City A over the span of any 3-day period. For City D, select the number of Eureka flights that will arrive in City D over the span of any 3-day period. Make only two selections, one in each column.
3
6
7
8
9
Let's map out each sequence and count the flights:
| Sequence | Route | Total Flights | Arrivals at A | Arrivals at D |
|---|---|---|---|---|
| 1 | A→B→C→A | 3 | 1 (from C) | 0 |
| 2 | A→D→B→A | 3 | 1 (from B) | 1 (from A) |
| 3 | A→E→D→A | 3 | 1 (from D) | 1 (from E) |
| 4 | A→E→B→D→A | 4 | 1 (from D) | 1 (from B) |
| 5 | A→C→B→D→A | 4 | 1 (from D) | 1 (from B) |
Key insight: Each sequence has exactly 1 arrival at City A (the final flight returns to A).
We need exactly 10 flights per day with 3 sequences. Let's check combinations:
So each day MUST have: Two 3-flight sequences + One 4-flight sequence
This means daily we need:
The constraints state:
This means over any 3-day period:
With 5 sequences and 3 sequences/day × 3 days = 9 total uses:
Given our daily requirements:
With the at-most-twice constraint:
Each sequence brings exactly 1 arrival to A.
Over 3 days: 9 sequence uses = 9 arrivals at City A
Let's calculate based on our forced distribution:
Total arrivals at D: \(0 + 2 + 2 + 3 = 7\) arrivals at City D
Based on our analysis, over any 3-day period:
These numbers are fixed due to the constraints forcing a specific distribution of sequence usage.