Q9

[Ofideas]

where is the diagram for this one?

[noah]

The diagram for each game is in its own thread - take a look here: diagram-t221.html

If you'd like help with a particular question, please tell us.

For this question, there's not much to say since 4 answer choices must be true, and the correct one doesn't have to be (or, in other words, it could be false).

I'll tell you that just looking at the answers, before even digging into the game, I could predict that the answer was probably (not definitely) going to be (D) or (E) since those are so much more involved in the game's numerical distribution issues. Here, it's totally possible that one person leaves 3 messages. You just have to be careful not to "trigger" most of the chain by using G or F.

[hilarykustoff]

Can someone explain this one? I don't know where to start.

[noah]

Tackling this question requires understanding the chain in the initial set-up.

G --> P + F --> P - T --> H - L

So, for example, G triggers P, F, T, H and L. Thus, G triggers everyone being in.

You must also remember that we can either have everyone leave 1 message, or 5 people leave 1 message and 1 person leave 2 messages, or 4 people leave 1 message and 1 person leave 3 messages. Thus, we definitely need 4 people.

With that in mind, let's dive in, remembering that 4 answers must be true. Thus, we're trying to "break" these answers:

(A) Can we avoid L leaving a message? If so, we'd have to avoid triggering G, P and T. If we knock out 3, how do we get to 5 people leaving messages? We have to have an L in.

(B) This suffers from the same problem. To avoid triggering T, we can't have G or F in. That leaves us with 4 people. Not enough. We have to have a T in.

(C) This is exactly the same as (A). We'd have to leave out G, P and T. There's got to be an H.

(D) We could have G trigger P, F, T, H and L. No doubling up allowed in this scenario! Thus this doesn't have to be true.

(E) This is something we already figured out.

That clear it up?

[hilarykustoff]

Thank you!