Monday, August 27, 2012


Call me a masochist, but I do love these self-referential or recursive logic puzzles (that shred my brain to itty-bitty, miniscule, pulsating pieces!). The one below I've encountered multiple times over the last year-or-so, and seen several different answers offered as solutions. Most of them don't check out when applied carefully, but one does seem to work well, and because there is at least some wiggle room in interpretation of the statements there may be one or more other successful answers.

So I'm curious if anyone knows the precise origin of this logic puzzle, and definitively what the proper answer(s) are supposed to be?

And, if you've never seen it before, well, knock yourself out…:

Given the following 12 statements which of the statements below are true?

1.  This is a numbered list of twelve statements.
2.  Exactly 3 of the last 6 statements are true.
3.  Exactly 2 of the even-numbered statements are true.
4.  If statement 5 is true, then statements 6 and 7 are both true.
5.  The 3 preceding statements are all false.
6.  Exactly 4 of the odd-numbered statements are true.
7.  Either statement 2 or 3 is true, but not both.
8.  If statement 7 is true, then 5 and 6 are both true.
9.  Exactly 3 of the first 6 statements are true.
10.  The next two statements are both true.
11.  Exactly 1 of statements 7, 8 and 9 are true.
12.  Exactly 4 of the preceding statements are true.


...I've already used several of these type puzzles on the site before, but if you have a favorite you think I may like send it along to me via email and I'll consider posting it and acknowledging the sender.

[...In a few days I'll state one of the answers that works on the above in the comments section, for those interested.]


Anonymous said...

1,5,6,9,11 ??

"Shecky Riemann" said...

I think the simplest, 'cleanest' answer perhaps is having only statements #1 & #5 as true (all the rest false), but Anonymous's answer above works as well.

any others???

MKP said...

If 7 is false then you can't evaluate 8. Ditto if 5 is false you can't evaluate 4. This is only solvable if you use false for those situations I think!

Gabriela Cristian said...

5 and 7 can't both be true. If you regard 8 as vacuously true when 7 is false, then 1, 5, 8, 11, works. 1, 2, 5, 6, 8, 9, 11 also works.

Unknown said...

1,5,6,9,11 is contradicted by 8.
1,5 is contradicted by 8.
1,5,8,11 is contradicted by 12.
1,2,5,6,8,9,11 is contradicted by 5,7,9,11.


Unknown said...

Oh, and I believe the only correct answer turns out to be 1,3,4,6,7,11.


René said...

I tried it by brute force in Java. Must have made an error however, since I get two possible answers.

1 5 6 9 11
1 3 4 6 7 11

If anyone is interested in the code, mail me.

René said...

Now I found a bracket error in my code. (And another error, when I checked the result!) Awful. Here is the only answer:

1 3 4 6 7 11

Gute Mathe Fragen (Mathe-Forum) said...

We tried to solve this logic puzzle as well, our solution is: 1, 5, 6, 9, 11

@Larry "1,5,6,9,11 is contradicted by 8." -> 8 is irrelevant since 7 is not true. Having 8 incorrect does not lead to contradiction.

René said...

I posted the Java code in my blog. See

"Shecky Riemann" said...

The above site gives various computer code for arriving at the 1,3,4,6,7,11 answer (true statements)

Wheat Kracker said...

After analyzing the various groups listed in each statement: Only statement number one is the only true statement. All the others statements are running us around in circles with if, exactly and either statements.