Avery , Blake , Clark and Doyle each live in an apartment. Their apartment are arranged like this :
+---+---+---+---+
| A | B | C | D | --> East
+---+---+---+---+
- One of the four is the landlord.
- If Clark's apartment is not next to Blake's apartment, then the landlord is Avery & lives in apartment A.
- If Avery's apartment is east of Clark's apartment, then the landlord is Doyle and lives in apartment D.
- If Blake's apartment is not next to Doyle's apartment, then the landlord is Clark and lives in apartment C.
- If Doyle's apartment is east of Avery's apartment, then the landlord is Blake and lives in apartment B.
Who is the landlord?
I thought I would expand a little on the contrapositive approach.
Statement 2 is equivalent to: "If it is not the case that Landlord Avery lives in apartment A, then Clark's apartment IS next to Blake's."
Statement 3 is equivalent to: "If it is not the case that Landlord Doyle lives in apartment D, then Avery's apartment is WEST OF Blake's."
Statement 4 is equivalent to: "If it is not the case that Landlord Clark lives in apartment C, then Doyle's apartment IS next to Blake's."
Statement 5 is equivalent to: "If it is not the case that Landlord Blake lives in apartment B, then Doyle's apartment is WEST OF Avery's."
Since at most one of them is the landlord, at least three of the following predicates are true.
Predicate 2. C is next to B
Predicate 3. A is west of C
Predicate 4. B is next to D
Predicate 5. D is west of A
Assume predicates 3, 4 and 5 are true.
Then the only valid arrangements are DBCA and BDCA, both of which contradict statement 3.
Assume predicates 2, 4 and 5 are true.
Then the only valid arrangements are CBDA and DBCA, both of which contradict statement 3.
Assume predicates 2, 3 and 4 are true.
Then the only valid arrangements are ADBC and ACBD, both of which contradict statement 5.
So, it must be the case that 2, 3 and 5 are true.
The only valid arrangements are DABC and DACB.
DABC contradicts statement 4, so the only solution is DACB.
All in all, it is easier to just examine all 24 combinations. This approach works hard to reduce the 24 combinations to 8 combinations, which still need to be examined.
Edited on February 7, 2013, 10:59 am