Avery , Blake , Clark and Doyle each live in an apartment. Their apartment are arranged like this :

          +---+---+---+---+
          | A | B | C | D |  --> East
          +---+---+---+---+

1. One of the four is the landlord.
2. If Clark's apartment is not next to Blake's apartment, then the landlord is Avery & lives in apartment A.
3. If Avery's apartment is east of Clark's apartment, then the landlord is Doyle and lives in apartment D.
4. If Blake's apartment is not next to Doyle's apartment, then the landlord is Clark and lives in apartment C.
5. If Doyle's apartment is east of Avery's apartment, then the landlord is Blake and lives in apartment B.

Who is the landlord?

(In reply to re: explaining my approach by Steve Herman)

..... the probability is only 25% that the landlord lives in his corresponding  applies to each candidate landlord

the probability is 15/24 that the a  landlord lives in his corresponding  apt and some of the tenants might or  might not
as well  e.g.. AdcB

Only in 9 out of 24 cases none has name-flat matching:

badc,bcda,bdac,cadb,cdab,cdba,dabc,dcab,dcba.

I understood that either only one or none of the 2nd part of the 2-5 statements will have a "true" value and this is the reason to start with the IG of arrangment without capital letters., like abcd  or badc and modify "as you go". the modification and back-stepping will define the owner, and if one of the "then " statements is true you get the landlord, living in nmatching apt.

Since by now you are an expirenced swimmer in the ocean of 4 apartments, try  my method, starting with any guess (small letters, no landlord yet) ,check this IG against the original statements, not the inversions- and see how fast it converges.

You will find that it is faster than writing down by hand the 96 letters of the 24 words.

We both agree, that it was a neat (even if conroversial)

problem.

Best regards and thanks for your remarks.

 

Posted by Ady TZIDON on 2013-02-08 17:56:35