Six people P, Q, R, S, T, V live on different floors (1 lowest to 6 topmost). Who lives on floor 6? I. Exactly one floor lies between R and Q; P lives on an even-numbered floor. II. T does not live on an even-numbered floor. Q lives on an even-numbered floor and not on the top floor. III. S lives on an odd-numbered floor. There are two floors between S and P. P lives immediately above R.

Introduction / Context:
This is a floor-arrangement data-sufficiency problem. We must determine whether the topmost resident (floor 6) is uniquely identified based on the given statements. Because each statement imposes multiple constraints, we check independence and pairwise combinations before concluding necessity of all three.

Given Data / Assumptions:

• Floors are 1 (lowest) to 6 (topmost); each of P, Q, R, S, T, V occupies exactly one floor.
• I: |floor(R) − floor(Q)| = 2; P is on an even floor.
• II: T is on an odd floor; Q is on an even floor and not on floor 6.
• III: S is on an odd floor; there are two floors between S and P; P is immediately above R (P = R + 1).

Concept / Approach:
Translate each statement into arithmetic constraints and test whether they yield a unique topmost floor holder. If multiple consistent placements remain, the set considered is insufficient.

Step-by-Step Solution:
Using I alone, possibilities for (R, P) include (1,2), (3,4), (5,6) (from III’s linkage, but without III we don’t know “P above R”), and Q only constrained to be two floors away from R; multiple models can put different people on 6.I + II restrict Q to {2,4}, T to {1,3,5}, but still permit several placements that change who is on floor 6.I + III impose P even, R immediately below P, and two floors between S and P; still, without II, Q could occupy 2 or 4 in different ways, keeping the topmost unresolved.II + III restrict odd/even assignments further, yet without I’s “gap” between R and Q, multiple consistent top-floor occupants remain possible.

Verification / Alternative check:
Construct counter-examples under each insufficient combination where different persons can occupy floor 6; uniqueness fails until all three are enforced together.

Why Other Options Are Wrong:
Any pair leaves at least two viable layouts; only I+II+III together pin down a unique topmost resident.

Common Pitfalls:
Misreading “two floors between S and P” vs. “immediately above” or ignoring parity constraints for Q and T.

Final Answer:
All three statements together are necessary.