There are eight floors in the building. 1 is the lowermost floor, then 2 is just above 1, then 3 is just above 2, and so on till 8.
Eight boys B1, B2, B3, B4, B5, B6, B7, and B8 live on these eight floors.
i) B8 lives on the lowermost floor.
ii) B3 lives on the fifth floor
iii) B2, B3, and B6 don’t live below B7.
To satisfy this condition B2, B3, and B6 cannot live on floors 2, 3, and 4 because they don't live below B7.
iv) Only B4 lives between B1 and B5.
v) B1 lives above B5.
vi) B4 lives on an odd-numbered floor. (Thus it can be placed on the floor 7th only)
vii) B2 lives on an even-numbered floor. (Thus only one place left i.e floor 4th).
viii) B2 lives just above B6. ( This means B6 is on the 3rd floor.
Thus the final arrangement is:
Option 1: B5, B6
We can see there is a two-floor gap between them.
Op[tion 2: B1, B3
We can see there is a two-floor gap between them.
Option 3: B3, B7
We can see there is a two-floor gap between them.
Option 4: B8, B4
We can see there is a five-floor gap between them.
All the options except option (4) following the common logic of a 2-floor gap between them while option 4 does not follow the common pattern.
Hence, the correct answer is "Option 4".