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Congratulations for achieving this final stage of "Millionaire Super Logic" show!

Once again, my name is Athin, and I'll be your host to help you win the jackpot of one, milLION, DOLLARS! *audience cheering and clapping*

All you have to do is just to pick a single door between these 20 doors, each labeled as an integer 1 to 20. Exactly one of them has the jackpot, and the rests are just having *minor* prizes ranging from \$10,000 to \$190,000.

To help you out, listen carefully to these 8 assistants of mine. Each of them will give you a statement as a hint. But be cautious, because exactly X of those 8 people are lying. What is X? It is written on a piece of paper inside my pocket. I will not share it with you by the way, because you will know for sure where the jackpot is if you know the value of X! Aww, don't be sad, I'm not cheating, it is part of a game, haha!


So now, please welcome my 8 assistants! *audience clapping*

Ok, be ready! You only have 3 minutes after all of my assistants tell their statements. Hopefully, you will win the jackpot of \$1,000,000 tonight!

My assistants! The game... starts... NOW! *more audience cheering and clapping*

Assistant 1: "If the jackpot's door is an even number, then the jackpot's door is a square number."
Assistant 2: "If the jackpot's door is a square number, then the jackpot's door is a number that has less than 6 distinct divisors."
Assistant 3: "If the jackpot's door is a number that has less than 6 distinct divisors, then the jackpot's door is between 1 to 10 inclusive."
Assistant 4: "If the jackpot's door is between 1 to 10 inclusive, then the jackpot's door is an even number."
Assistant 5: "From the previous statements, the statement of exactly 1 person is false."
Assistant 6: "From the previous statements, the statements of exactly 2 people are false."
Assistant 7: "From the previous statements, the statements of exactly 3 people are false."
Assistant 8: "From the previous statements, the statements of exactly 4 people are false."


The time starts ticking now, think like a brilliant millionaire! Which door leads you to the jackpot?

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  • 9
    $\begingroup$ milLION I've found the real answer, there's lions behind every door and you get eaten! $\endgroup$ – Aequitas Feb 12 '18 at 4:10
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    $\begingroup$ So a lazy contestant who picks at random would get $145000? That's more than fair! $\endgroup$ – Eric Duminil Feb 12 '18 at 14:26
  • $\begingroup$ @EricDuminil, well it's already on the final stage tho :p $\endgroup$ – athin Feb 12 '18 at 23:55
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Plotting every answer's truth combinations would take a lot of room, so let's take a couple of shortcuts first:

Assistant 2 is speaking the truth. There are no square numbers with 6 distinct divisors between 1 and 20. (The only numbers with 6 distinct divisors are 12, 18 and 20.)

Assistant 5's statement always adds one liar, except if there was exactly one liar among the first four. Therefore, among the first 5 assistants, there are 1,3,4 or 5 liars. Therefore, assistant 6's statement is always false. Because 6's statement is false, then so is 7's, and because of that, so is 8's. Therefore, 6, 7, and 8 are lying.

For the remaining assistants, let's plot a truth table, and count the number of liars for each possible jackpot door. Since the host said that knowing the X would give away the jackpot, there should be at least one X for which the answer is unique.

 Jackpot     1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
 Ass. 1      -  L  -  -  -  L  -  L  -  L  -  L  -  L  -  -  -  L  -  L
 Ass. 3      -  -  -  -  -  -  -  -  -  -  L  -  L  L  L  L  L  -  L  -
 Ass. 4      L  -  L  -  L  -  L  -  L  -  -  -  -  -  -  -  -  -  -  - 
 Ass. 5      -  -  -  L  -  -  -  -  -  -  -  -  -  L  -  -  -  -  -  -
 Liars (X-3) 1  1  1  1  1  1  1  1  1  1  1  1  1  3  1  1  1  1  1  1
 

Therefore,

X is 6 and the Jackpot is behind door 14.

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  • $\begingroup$ This is nice, but I think my answer is more systematic and easier to follow :-) $\endgroup$ – Rand al'Thor Feb 11 '18 at 14:49
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    $\begingroup$ Yes, but did you get it in under three minutes? $\endgroup$ – HaveSpacesuit Feb 12 '18 at 22:07
  • $\begingroup$ Is this site a speed contest? That would detract massively from the idea of sitting down and thinking in peace $\endgroup$ – Timwi Feb 13 '18 at 13:39
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    $\begingroup$ @Timwi, the asker can give the ”correct answer” tick to whomever, but usually it goes to the first answer without errors and reasong explained. The reputation point system is from Stack Exchange, and it’s somewhat bass-ackwards for puzzling. So by all means, sit back, enjoy the puzzles, and try not to let the points distract you. That’s what I try to do, at least. $\endgroup$ – Bass Feb 13 '18 at 14:36
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Bass's answer is correct, but I wanted to give what I think is a slicker proof.

  1. If Assistant 5 is telling the truth, then Assistant 6 is lying, Assistant 7 is lying, and Assistant 8 is lying. If Assistant 5 is lying, then Assistant 6 is lying, Assistant 7 is lying, and Assistant 8 is lying.

    Therefore we know that Assistants 6, 7, and 8 are lying.

  2. The first four assistants are saying respectively $A\Rightarrow B$, $B\Rightarrow C$, $C\Rightarrow D$, $D\Rightarrow A$, where $A,B,C,D$ are the statements that the jackpot's door is (respectively) an even number, a square number, a number with less than 6 distinct divisors, a number $\leq10$.
  3. The negations of the first four assistants' statements are respectively $A\land¬B$, $B\land¬C$, $C\land¬D$, $D\land¬A$ (i.e. "A and not B", "B and not C", "C and not D", "D and not A"). Clearly no adjacent pair of these negated statements can be true simultaneously. So the false statements among the first four assistants must be either {1,3}, {2,4}, any one of them, or none.
  4. The second assistant's statement is true, because all the possible square numbers (1, 4, 9, 16) have fewer than 6 distinct divisors. So the false statements among the first four assistants must be {1,3}, {1}, {3}, {4}, or none.
  5. So the false statements among all assistants are {1,3,5,6,7,8}, {1,6,7,8}, {3,6,7,8}, {4,6,7,8}, or {5,6,7,8}. Given the host's claim that knowing X would be enough to solve the puzzle, we know

    $X=6$ and the false statements are {1,3,5,6,7,8}.

  6. So the jackpot's door is

    even, not square, having less than 6 distinct divisors, and greater than 10. A quick check shows that 12, 18, and 20 each have six distinct divisors, so the answer is 14.

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  • 25
    $\begingroup$ Very nice! I’m afraid the ”easier to follow” bit only applies to those who speak PSEs official language ”formal logic” as their mother tongue :-) $\endgroup$ – Bass Feb 11 '18 at 15:51
  • $\begingroup$ Can you explain what "So the false statements among the first four assistants must be either {1,3}, {2,4}" means? I read it as meaning either 1 and 3 are both false, or 2 and 4 are both false or they are all false or none are false but then you go on to say something different. I can't actually work out what restriction you are putting on them here.... $\endgroup$ – Chris Feb 12 '18 at 14:30
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    $\begingroup$ @Chris At a glance, that's set notation. The last sentence of that statement basically reads "either: 1) both statements 1 and 3 are false [and the other two statements are true], 2) both statements 2 and 4 are false [and the other two statements are true], 3) any one statement is false [and the other three true], or 4) no statement is false [all four are true]." $\endgroup$ – KitsuneZeta Feb 12 '18 at 15:47
  • $\begingroup$ @KitsuneZeta: Ah! That makes sense. I got the set part but managed to completely misparse it. :) $\endgroup$ – Chris Feb 12 '18 at 16:12
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Partial working out:

Assistant 2 is telling the truth. 12, 18 and 20 are the only numbers which have 6 or more factors

scenario if: everyone else is lying: we would know that the jackpot is an even number, but not square, because if it were not an even number, assistant 1 would be telling a vacuous truth, and should it be square, a regular truth. we would know it has <6 factors (not 12, 18, or 20), and it is greater than 10 however, this conflicts with assistant 4 being a liar, so we know that there are at least two truth. we know at least one of assistant 3 and 4 tells us the truth

scenario: everyone but assistant 4 and assistant 2 lies we would know that the jackpot is an even number, but not square. we would know it has <6 factors (not 12, 18, or 20), and it is greater than 10 this gives us, as the only possibility, 14, because 12, 18, 16, and 20, even numbers greater than 10, are ineligible I could stop here with the meta solution of, we know there is no ambiguity. but that's cheating also 5 to 8 are lying because there is the wrong number of people/ 2+ however many of 5-8 are before them lying, which is never correct for them.

it seems like 5 to 8 are irrelevant or red herrings to me at this point. however, in my later state of solving, it seems like they're actually the key to this somehow, if i could just find out how

scenario, 2 and 3 only tell the truth. even non-square, however assistant 4 lying means number from 1-10 which isn't even. impossible scenario

scenario. 2, 3 and 4 tell the truth, as does assistant 5 even non-square. 2,6,8,10, 12, 18 or 20 (assistant 3 helps here) it doesn't seem like i can divine the truth in this scenario. i can't find a contradiction

scenario 5 through 8 lie only: odd number or square number. 12, 18, 20 or between 1-10. greater than 10 or even. this leaves only 4. i can't see a contradiction

perhaps 5-8 are key... is one of the liars the presenter?

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  • $\begingroup$ Asistant 6 through 8 are all liars. If asistant 5 is telling the truth then there is exactly one liar before assistant 6, two before assistant 7 and three before assistant 8. If assistant 5 is lying then the number of liars among 1 thorugh 4 is some number other than 1. If follows that the numbers of liars preceding assistant 6 is some number other than 2. Etc. $\endgroup$ – Taemyr Feb 11 '18 at 21:17
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Of the eight statements given by the assistants, four of them are completely useless and can be ignored due to either always being false or always being true.

Statement 2 would be false if and only if the jackpot door's number was a square that had 6 or more distinct divisors. All four squares in the range of 1 to 20 (specifically, the numbers 1, 4, 9, 16) have fewer than 6 distinct divisors, therefore Statement 2 must be true.

Statement 5 is true if and only if there is exactly one false statement among the first four statements. The possible number of false statements, regardless of what the statements are, among the first five statements is therefore either 1, 3, 4, or 5. Because of this, it is impossible to have exactly 2 false statements among the first 5 statements, therefore Statement 6 cannot be true.
Because statement 6 cannot be true, the possible number of false statements among the first 6 statements is 2, 4, 5, or 6. Therefore, it is impossible for there to be exactly 3 false statements among them, and Statement 7 cannot be true.
Because statement 7 cannot be true, the possible number of false statements among the first 7 statements is 3, 5, 6, or 7. Therefore, it is impossible for there to be exactly 4 false statements among them, and Statement 8 cannot be true.


After reducing the number of statements to check the validity of to 4, we must now figure out when a given statement would be false. Statement 5's truthiness has been elaborated above (since it is required in the proof of the truthiness of other statement(s)), and the remaining three statements are of the form "If A, then B".

For any statement of the form "If A, then B" to be false, A must be true and B must be false.

Statement 1 has the condition of "If the door is an even number", so it only applies if the door is an even number (2, 4, 6, 8, 10, 12, 14, 16, 18, 20).
Statement 1 then states that the only "correct" even numbers are squares (4, 16). Note that while 1 and 9 are squares, they aren't even, and don't matter for this statement.
Statement 1 is therefore false on any non-square even number. (These are the numbers 2, 6, 8, 10, 12, 14, 18, and 20)

Statement 3 has the condition of "If the door is a number with less than 6 distinct divisors".
There are three numbers in the range of 1 to 20 that have 6 or more divisors, and therefore don't match the condition: 12 (1, 2, 3, 4, 6, 12), 18 (1, 2, 3, 6, 9, 18), and 20 (1, 2, 4, 5, 10, 20).
all other numbers match this condition.
Statement 3 states that the door must be in the range of 1 to 10 (1, 2, 3, 4, ..., 10)
Therefore, Statement 3 is false on the following numbers: 11, 13, 14, 15, 16, 17, 19

Statement 4 has the condition of the number being in the range of 1 to 10.
Statement 4 states that the door must then be an even number.
Therefore, Statement 4 is false for any single-digit odd number. (1, 3, 5, 7, 9)


There is one final statement that must be considered to determine which door has the prize, and it's given not by an assistant but by the host himself. Specifically, knowledge of exactly how many liars there are reveals the door.

Factoring in statement 2's guaranteed truthfulness AND the false values for statements 3 and 4 having no overlap, the possible number of false statements among the first five statements are 1 and 3. To find the correct door, we must figure out which door has a unique number of false statements among the first five statements.

Given that it is impossible for there to be 0 false statements among the first five (due to statement 5 being false if the first four statements are all true), any doors where exactly one of the first four statements are false cannot be the door, nor can any door where all four of the first four statements are true be the correct door. This eliminates:
1. all odd numbers below 10 (only statement 4 is false),
2. all even numbers up to and including 10 (With the exception of number 4, these make statement 1 the only false statement. In 4's case, the only false statement is statement 5, so it is wrong as well),
3. all odd numbers above 10 (only statement 3 is false)
4. The numbers 12, 18, and 20 (only statement 1 is false), and
5. The number 16 (only statement 3 is false)

Removing those 19 numbers leaves only the number 14. In the number 14's case, both statement 1 is false (it's an even number that's not a square) and statement 3 is false (it's a number greater than 10 with fewer than 6 distinct divisors). Because this is the only number that results in more than 1 false statement among the first five statements (and therefore more than 4 false statements among the 8 assistant statements), the prize must be behind door 14.

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Out of 5, 6, 7 and 8, if one is telling the truth, the preceding and following assistants must be lying. Either only one of them is telling the truth, or none of them is. If one of them is telling the truth, in any case that would suggest only one of the first half is a liar, making 5 the truth teller.

Also, 1 implies that it can be 1, 3, 4, 5, 7, 9, 11, 13, 15, 16, 17 or 19.

2's statement is a tautology, because the square numbers from 1 to 20 always have fewer distinct divisors than 6.

3 implies that it can be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18 or 20.

4 implies that it can be 2, 4, 6, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19 or 20.

At least one of 1, 3 or 4 must be telling the truth as they cover all the numbers together. If 5 is telling the truth, there are 1+3=4 liars. If he's lying, there are 0v2 + 4 = 4 or 6 liars. Since we're supposed to be able to tell the result if we knew X, X must be unique, meaning X=6 and there are 2 liars out of the first 4, making the answer 14.

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    $\begingroup$ Your initial paragraph misses the possibility that there is one liar amongst the first four assistants. In that case assistant 5 is telling the truth and 6 through 8 is lying. This is important because the anouncer gives one piece of information that you are ignoring. $\endgroup$ – Taemyr Feb 11 '18 at 21:23
  • $\begingroup$ @Taemyr Fixed my answer. $\endgroup$ – Nautilus Feb 12 '18 at 10:05
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Answer is

4th door. If assistants 5 through 8 are all liars then first four statements should be true. Number (4) is an even number, square number, has less than 6 distinct divisors, between 1 to 10 inclusive.

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  • $\begingroup$ How do you figure assistants 5 through 8 are all liars? $\endgroup$ – Eric Dobbs Feb 12 '18 at 14:59
  • $\begingroup$ If we assume assistants 5 through 8 are all liars then we can derive 4th door. Just random assumption but 4th door fits the problem $\endgroup$ – Venkatesh Shanmugam Feb 13 '18 at 8:25
  • $\begingroup$ Could you edit your answer to include spoiler tags, so as not to spoil the solution for anyone who wants to try the puzzle themselves? Thanks! $\endgroup$ – F1Krazy Feb 13 '18 at 10:05

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