An officer has to solve a case with 20 suspects, 10 from colony A, and 10 from colony B. He can solve the case once at least 19 of them answer truthfully during an investigation.

The officer has two identical boxes labeled P and Q, which each have 20 cards, one for each suspect. Before each investigation, he takes one card from each box. He interrogates these two people during the investigation; the suspect from box P will tell the truth, and the suspect from box Q will tell the truth if and only if the suspect from box P is from colony A. (The officer can tell who's telling the truth.)

After each investigation, the officer will discard cards from truthful suspects (from both boxes) and return cards from lying suspects to the original box.

Find the number of possibilities that he can solve the case in 10 investigations.

(In reply to A question of procedure by Dej Mar)

I think the fact that he could interrogate the same person twice in the same investigation does seem a little weird, but it was probably a choice the author made to keep it simpler. My guess is he would want that card discarded, since that suspect was truthful during an investigation (even though he lied later in that investigation)

Your initial assumption makes sense -- but really only a "tricks" problem would try to confuse you using plurals. As you had thought, there are only two interrogations in an investigation, but he could have "liars" in an investigation if zero liars were present.

Posted by Gamer on 2008-08-30 23:20:58