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.
Perhaps I am just thinking of the long weekend ahead, but I find this puzzle incoherent. I do not understand either the conditions or the task.
There is the "officer" and there are two groups of 10 each (colony A and colony B) called "suspects." It is not clear whether or not the officer initially knows the "colony" of any or all of the suspects.
There are two "identical boxes" P and Q, but no specification of in what respect they are "identical" (the sizes, shapes, etc. of the boxes do not seem relevant), except that each initially has 20 cards, "one for each suspect" (presumably not necessarily in any given ordering). Does the last clause mean that each card has only the name on one of the 20 suspects? Or are there questions to be asked the suspect(s)? The officer after drawing a pair of cards "interrogates these two people" (could the same person's name be drawn from both boxes in one interrogation?). Is the interrogation related to something written on these cards (the name only -- so question is "are you X"? -- or what -- e.g. "did you do it?" or "do you know who did it? or whatever)?
We are then told that the suspect from box P "will tell the truth" (since all 20 have their names in box P, then they will all do something truthfully: what?). Then we are also told that the suspect from box Q "will tell the truth if and only if the suspect from box P is from colony A" . From the last we must assume that all suspects know the "colony" of each of the others, else the equivalence is meaningless -- since they must know when they should tell the truth and when they should lie. We are told the officer "can tell who is telling the truth" -- is that from knowing the "colony" of each, or what? Truth about what?
What is "an investigation" - each interrogation session with a pair of suspects, or each pass through the entire "boxes"? In discarding cards "from both boxes" does this mean discarding cards drawn in pairs, or finding matching names, or what?? Some "cards" (note plural) are "returned" to "the original box".
We are not told what would constitute "solving the case", but are supposed to test that by a limit of 10 "investigations". We are asked to "find the number of possibilities" of a solution: does this mean the probability of some outcome? The "number of possibilities" is TWO: he solves it, or he doesn't solve it.
This problem is classified as "Just Math" but seems to be anything but. Somewhere between logic and plain obfuscation. Kafka's "Trial" comes to mind. Perhaps some will find their way through this morass. I suspect that a problem has been borrowed from another source, and feeling the need for paraphrase and disguise, become mangled. I'll keep my dunce cap in the corner until further developments.