1
$\begingroup$

Nine months ago I started my journey on Puzzling in a rough way with this question.

I honesty hope this time I won't generate so extended discussion...!
I love self-referential questions, puzzles or tests. And I still think those make the most efficient way to evaluate the intelligence of a person. Simply because it doesn't require any level of math knowledge or algebraic familiarity. They just rely on pure logics.

Recently I made the next test to give my students when they begin to learn logic. It's a simple one (with just five questions and three options for each one). Can you figure the answers out?

THE TEST:

QUESTION #1: The number of questions of this test with the answer A is
A) zero
B) one
C) two

QUESTION #2: The maximum number of consecutive questions with the same answer is
A) zero
B) one
C) two

QUESTION #3: The answer to the question 4 is
A) B
B) C
C) A

QUESTION #4: In total, the number of questions with the same answer as this one is:
A) zero
B) one
C) two

QUESTION #5: The number of questions between this one and the last one with the same answer as this one is
A) one
B) two
C) three

GOOD LUCK!

PLEASE NOTE that every time a question refers the answer to another question, it refers to the letter option, not to the answer itself.

NOTE TO CLEAR THING UP: Please be literal on your interpretation of the questions. There are some contradictions which rule out some of the answers.

$\endgroup$
4
  • 3
    $\begingroup$ A few clarifications: is Question 2 talking about pairs of answers, or does a single answer automatically set the minimum at 1 (ie is the maximum number of consecutive questions with the same answer for a sequence of ABCAB 1 or 0?). Also does question 4 count itself? $\endgroup$
    – StephenTG
    Mar 3 at 14:36
  • 1
    $\begingroup$ Further to StephenTG's question, with the sequence ABCAB, would the answer to question 5 be two or three? $\endgroup$
    – hexomino
    Mar 3 at 14:43
  • $\begingroup$ @StephenTG, please be literal (you can exclude some options with your reasoning and that's the intended way ;) ) $\endgroup$
    – Pspl
    Mar 3 at 15:12
  • $\begingroup$ @hexomino, in your scenario, the answer to the question 5 would be two. $\endgroup$
    – Pspl
    Mar 3 at 15:13
3
$\begingroup$

Without depending on other questions we can already deduce:

Q1 cannot be A, because then the number of answers that are A would be zero
Q2 must be B or C, because we must have a least one consecutive question with answer X
Q4 cannot be A, because then none of the answer can be A. (This is, if "the number of questions with the same answer as this one" includes this question, which I think OP meant)

Then,

because Q4 cannot be A, Q3 cannot be C
So:
Q1: .,B,C
Q2: .,B,C
Q3: A,B,.
Q4: .,B,C
Q5: A,B,C

If we assume

Q3 to be B, and therefore Q4 is C, then Q1 cannot be C (only 1 answer is A) must be B (A already excluded). Therefore, Q5 must be A. Meaning:
Q1: .,B,.
Q2: .,B,C
Q3: .,B,.
Q4: .,.,C
Q5: A,.,.

However, since Q5 is A, there must be another question that answers A. This is not possible since we showed only one question can be A.

Therefore, our assumption is false

and Q3 is NOT B. Q3 must therefore be A.
If Q3 is A, and therefore Q4 is B, then only Q4 can be B.
Then, Q1 must C (because no more B's allowed)
And Q2 is C (same reason)

Giving:

Q1: .,.,C
Q2: .,.,C
Q3: A,.,.
Q4: .,B,.
Q5: A,.,C

Then Q5 must be A, to leave gap 1 between the last A (Q3) and Q5. This would not hold for C

Giving our final answer:

Q1: .,.,C
Q2: .,.,C
Q3: A,.,.
Q4: .,B,.
Q5: A,.,.

All statements hold (2xA, 2 consecutive same answers, Q4=B, 1xB, gap between A's is 1)

$\endgroup$
1
  • 1
    $\begingroup$ That's it. You read the questions as they are. That's the answer I was looking for :) $\endgroup$
    – Pspl
    Mar 3 at 21:48
1
$\begingroup$

I didn't feel like doing logic today, so as there's no tag, I just got an Excel spreadsheet to tell me the answer:

1=C, 2=C, 3=A, 4=B, 5=A

Checking:

  • 2 questions with answer A

  • 2 consecutive questions with same answer

  • Q4 answer is B

  • this is the only (one) question with answer 'B'

  • just one question between this and the last with same answer.

$\endgroup$
1
$\begingroup$

Answer:

CCABA

Reason, considering the questions in order:

Q1 cannot be A. If it was, there would be at least 1 'A'.
Q2 cannot be B. If it was, there would be zero consecutive answers.
Q3 is a different answer (label) from Q4.
Q4 cannot be A because Q4 has the same answer as itself.
Q5 ties it all together:
- if Q5=A then Q3=A, Q4=B, so Q2=C, and Q1=C.
- if Q5=B then Q2=B (contradiction with Q2$\neq$B).
- if Q5=C then Q1=C and Q2,3,4$\neq$C, so Q2=A, Q4=B, Q3=A (but Q2=Q3 contradicts Q2=A).

$\endgroup$
0
$\begingroup$

This may make me look extremely stupid (iq-wise) but here's my guess:

Answers: 1c. 2b. 3a. 4b. 5a.

Reasoning:

If #3 and #5 are both A, then the answer to #1 must be C.

If the above order is correct, then no 2 of the same letter are consecutive, so it must be B for #2.

For 3 and 4, its only what makes any sense at all: #3 could not be B, because it would contradict the answer for #2. #3 could also not be C because by having A as the answer to number 4 would contradict my answer for #1 which states that there are 2 letter As as answers, and my answer to #2 which states that no two consecutive letters can be in a row. So thus, the answers for #3 is A, and #4 is B, which would check out, as #2 would also be B, fulfilling the requirement for #4.

And for #5, A or C could both work in this case, as #4 is the one answer between #s 3 and 5, and there are 3 between #s 1 and 5. But for the sake of the other answers, and I know this is a bit of a stretch, my final answer for #5 is A because if the correct answer was C, it would contradict #1, making it B, and thus making the last answer of C false as well.

$\endgroup$
4
  • $\begingroup$ Probably best to add a caveat about how you are interpreting questions 2,4 and 5 (see Gareth's answer, for example). $\endgroup$
    – hexomino
    Mar 3 at 15:12
  • $\begingroup$ Please, take a look at my second note. Be literal... ;) $\endgroup$
    – Pspl
    Mar 3 at 15:23
  • $\begingroup$ @Pspl Oops. Just realized: 3-1=1 is not correct...): (I forgot that there were 2 between 3 and 5 and not 1) $\endgroup$
    – Joe Kerr
    Mar 3 at 15:31
  • $\begingroup$ @JoeKerr, you got that part right. THERE is one question between questions 3 and 5 (the question number 4). However, your answer to question 4 is not correct: the question starts with the sentence "In total...". $\endgroup$
    – Pspl
    Mar 3 at 15:35
0
$\begingroup$

Q4 has two plausible interpretations. If it is asking about other questions with the same answer (recent comments from OP indicate, IIUC, that this is not the intention):

2A is impossible.
If 4A then no other A, so not 1A, so more than one A, contradiction. So not 4A. (I am taking Q4 to ask for the number of other questions with the same answer.)
If 5C then 1C, so two As among 2,3,4 but 2A and 4A are impossible. So not 5C.
If 5A then 3A so 4B. Also 5A,3A => 1C so we have C-ABA. 4B means two B answers, so must be CBABA.
Otherwise, 5B so 2B. So no two consecutive equal answers, which means no other B. We already know not 4A, so 4C, so we have -B-CB which means two other Cs, which is impossible because no two consecutive equal answers.
So the only possibility is CBABA.

On the other hand, if Q4 is asking for the number of such questions including Q4 (which I think is what's intended), the reasoning is very similar but not quite the same:

2A and 4A are obviously impossible.
If 5C then 1C, so two As among 2,3,4 but 2A and 4A are impossible. So not 5C.
If 5A then 3A so 4B so no other B, so 2C. We now have -CABA and then we must have CCABA. Otherwise 5B so 2B. So no two consecutive equal answers, which means no other B. We already know not 4A, so 4C and therefore 3B, so we have -BBCB which is impossible because no two consecutive equal answers.
So the only possibility is CCABA.

To summarize these jointly:

1,3,4,5 are C,A,B,A. 2 is B if Q4 excludes itself and C if Q4 includes itself.

$\endgroup$
4
  • $\begingroup$ I would say Q2 and Q5 also have more than one interpretation (see comments). $\endgroup$
    – hexomino
    Mar 3 at 15:10
  • $\begingroup$ You might be right. I don't fancy doing the analysis eight times :-). $\endgroup$
    – Gareth McCaughan
    Mar 3 at 15:12
  • $\begingroup$ Okay, Pspl has at least clarified that your interpretation of Q5 is consistent. $\endgroup$
    – hexomino
    Mar 3 at 15:14
  • $\begingroup$ Question 4 starts with the words... $\endgroup$
    – Pspl
    Mar 3 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.