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K–P fill floors 1(bottom)–6. K is on 2 or 5; L on 1 or 4; M on 3 or 6. N is two floors above O. P is below K, O above L, and M above N. O’s floor?

A1
B3
C2
D5
Answer & Solution
Correct answer: C. 2
If L were on4, O would be above4 and N two floors higher, impossible; so L is1. K cannot2 because P belowK would need occupied1; so K is5. N is at least3, so M cannot3 and must6. N/O can then only use4/2; O is2.
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