The Unapologetic Mathematician

Mathematics for the interested outsider

On the road again

Today is my last long drive for a while. I’ll try to post a Sunday Sample tonight, or maybe tomorrow. But for now, here’s a problem to chew on.

In a multiple choice test, one question was illegible, but the following choice of answers was clearly printed. Which must be the right answer?

  1. All of the below
  2. None of the below
  3. All of the above
  4. One of the above
  5. None of the above
  6. None of the above

August 24, 2008 - Posted by | Uncategorized


  1. Option 5: None of the above

    Comment by ashwin | August 24, 2008 | Reply

  2. I got the same answer, but I’m willing to show my work…

    Option 6 (none of the above) being true would mean options 1-5 are false, but since then options 1-4 are false, option 5 (none of the above) would be true, forming a contradiction. Therefore, Option 6 is false.

    Since option 6 is false, option 1 (all of the below) is false, meaning option 3 (all of the above) is false.

    That leaves options 2, 4, and 5. If option 2 is true (none of the below), that would imply that 4 (one of the above) is false (by 2) and 4 is true (since only 2 above 4 is true), a contradiction. Therefore, 2 is false. Since 1, 2, and 3 are all false, 4 (one of the above) must therefore be false.

    What’s left is 5 (none of the above). Since 1, 2, 3, and 4 are known to be false, that means 5 is true. 6 is also known to be false.

    Therefore, the only consistent answer is Option 5: None of the above.

    Comment by blaisepascal | August 24, 2008 | Reply

  3. Option 5, because most multiple choice tests use two similar answers, where one is a “distractor”. The two most similar answer options are 5 and 6, but as stated above, 6 contradicts itself. No need to even consider options 1-4 in that context (though blaisepascal’s reasoning is sound)

    Comment by Bobby Isosceles | August 24, 2008 | Reply

  4. Also option 5, by a much easier argument:

    We assume that there is a unique true option. It cannot be option 4, because then one of options 1, 2, and 3 would be true, contradicting uniqueness. So 4 is false; so 1, 2, and 3 are all false; so 5 is true.

    Comment by Chad Groft | August 25, 2008 | Reply

  5. Why would you ever make such an assumption? It’s not even valid for real multiple-choice tests, as is shown by the ubiquitous “all of the above” response.

    Comment by John Armstrong | August 25, 2008 | Reply

  6. Chad wrote:

    We assume that there is a unique true option.

    John wrote:

    Why would you ever make such an assumption?

    Because in posing your puzzle, you said: “Which must be the right answer?”

    Comment by John Baez | August 25, 2008 | Reply

  7. Similar language (“fill in the bubble corresponding to the correct answer”) is in the instructions to the SAT, and it also includes “all of the above” answers. I doubt I’d ever have met Chad if he took the definite article that literally.

    Comment by John Armstrong | August 25, 2008 | Reply

  8. With a test-designer this devious, we can’t be sure the illegible question wasn’t “Which of the following is false?”…

    Comment by Sridhar | August 25, 2008 | Reply

  9. That’s a good point, Sridhar. What does the problem look like in that case?

    Comment by John Armstrong | August 25, 2008 | Reply

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    Pingback by Blog aggregators? « Shores of the Dirac Sea | December 21, 2008 | Reply

  11. Apologies in advance for the proof…

    If we take this all as Boolean logic the rules boil down to:
    (. == AND, + == OR, ! == NOT, 1,2,3,4,5,6 => A,B,C,D,E,F)

    A: B.C.D.E.F
    B: !C.!D.!E.!F
    C: A.B
    D: A!B!C + B!A!C + C!A!B
    E: !A.!B.!C.!D
    F: !A.!B.!C.!D.!E

    So if we start with rule C (#3)
    C = BCDEF . !C!D!E!F –Substitution
    = 0 –Complementation
    Thus, C must be false.
    And by subsitution A(#1) must also be false (A = B.0.D.E.F = 0)
    Furthermore, D(#4) is as follows:
    D = 0.!B.!0 + B.!0.!0 + 0.!0.!B
    = 0 + B + 0
    D = B

    Next E(#5):
    E = !A.!B.!C.!D
    = !0.!B.!0.!B –Substitution
    = 1.!B.1.!B –Idempotence
    = !B

    Then F(#6):
    F = !A.!B.!C.!D.!E
    = 1.!B.1.!B.B
    = B.!B –Complementation, Idempotence
    = 0
    Thus F must be false.

    Finally, B(#2):
    B = !C.!D.!E.!F
    = 1 .!B.!!B.1 –Complementation, Double Negation
    = 0
    Thus B must be false.

    And now we have all the pieces to put the rest together:
    A = 0
    B = 0
    C = 0
    D = B = 0
    E = !B = 1
    F = 0

    And thus in the negative (by duality):
    A = 1
    B = 1
    C = 1
    D = 1
    E = 0 –E Must be false.
    F = 1

    Therefore, E is the only correct answer.

    Comment by Nik Hodgkinson | December 21, 2008 | Reply

  12. @Sridhar, @John Armstrong: The question cannot be “Which of the following is false?”:

    1. Only one is false and all the others are true. (Given.)
    2. Either 5 or 6 is true. (From 1.)
    3. All “of the above” (1-4) are false. (From 2.)
    4. 1 contradicts 3: 1-4 are false, but only one can be false.

    Therefore, that question is impossible.

    Comment by Tim McCormack | December 21, 2008 | Reply

  13. […] Wow, people are loving my zero-knowledge test. It got 1,743 views yesterday, thanks to someone redditing it. […]

    Pingback by Symmetric Tensors « The Unapologetic Mathematician | December 22, 2008 | Reply

  14. Try this one then:

    Comment by creating2009 | December 23, 2008 | Reply

  15. I chose option 5, and knew I was right, without having to proof anything for anyone.

    Ahh, the “high ground” I keep hearing about. Yep, the view is nice.

    Comment by dziban | January 9, 2009 | Reply

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