The first and last messages on the answering machine could be the first and second messages left by which one of the ...

Irina August 1, 2019

@davidmaya,

No, the rules only tell us that if F leaves a message, so do H & L. It does not mean that the number of F messages have to equal the number of H & L messages, in fact, we know that only one person left more than one message.

Let's look at the setup of this game more closely:

6 messages on the answering machine were left by some combination of FGHLPT. Notice that there is no requirement that each of the persons left at least one message, meaning it is possible that someone left 0 messages.

Now let's look at the rules.

(1) At most one person left more than one message.
(2) No person left more than three messages.

What does this mean for matching names to messages?

It means at most one person left two or three messages.

If one person left two messages, then four left one message, and one left 0 messages.
If one person left three messages, then three left one message, and 2 left 0 messages.

There is no requirement though that anyone leaves more than one message, so a scenario where everyone leaves one message is also equally plausible.

(3) If the first is H, then last is P

H(1) ->P(6)

H P
__ __ __ __ __ __
1 2 3 4 5 6

(4) if G left any message, so did F & P.

G ->F&P

(5) if F left any message, so did P & T, all of P preceding any of T

F->P&T

we can combine this rule with (4) to conclude that
G->F&P&T

P >> T

Since all of P messages must precede any of T if F left any messages, we can also conclude per rule (3) that if H left the first message, G left none because in that scenario P left message #6, and it is impossible for any T messages to follow.

(6) If P left any message, H & L did also, all of H preceding any of L.

P -> H&L
H >> L

we can combine this rule with (4) & (5):

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

It turns out if G left any messages, then each person left exactly one message and no one left more than one message.

The question asks us who could be the author of the first and last messages on the answering machine?

Since the question requires than one person leaves exactly two messages, we can conclude that G left 0 messages per our conclusion above. If G were selected, then everyone would leave one message, which would violate the conditions of this question. It means that we can only use a combination of F P T H L to fill all six slots to produce a combination that could be true.

We can right away eliminate answer choice (B) per rule (3) because if H leaves message #1, P must leave message #6.

We can also eliminate (C), (D), (E) per rules (5) (6). Since all H must precede all L, L cannot be #1. Since all P must precede all T, T cannot be #1, and P cannot be #6 because if P is #6 - all P messages cannot possibly precede all T messages.

Let's try the remaining answer choice (A):

F F
__ __ __ __ __ __
1 2 3 4 5 6

PTHL must also be selected per rules (5) and (6) but the exact order does not matter as long as P > T and H >L thus a complete order could be as follows:

F P H T L F
__ __ __ __ __ __
1 2 3 4 5 6


Since F could be #1 and #6, (A) is the correct answer choice.

Does this make sense?
Let me know if you have any other questions.

davidamaya August 2, 2019

This explanation makes sense and helped me out. Thank you for your time!

Ravi August 2, 2019

@davidamaya, let us know if you have any other questions!

garrick November 18, 2020

I'm missing something... maybe you can help.

Does TTPHLT break the rules?
1. Only T occurs more than once.
2. T occurs three times.
3. H1 condition does not exist
4. If F, then... well there's no F in TTPHLT so rule 4 does not apply
5. If P then H & L and H>>L... checks for TTPHLT

In this case, T is the 1st & 2nd and the last...

So I'm obviously missing something, 'cause "e" is lookin' pretty good from here.