Kill Your Mind with this

I invite you all to have fun with this evil, evil game.

Hope you enjoy it.

http://www.drunkmenworkhere.org/170

You do realize that the only thing that tops this in sheer diabolical frustration is “14 k of g in a f p d,” right? Curse you unto the tenth circle of Hell. Dante may only have described 9, but he never foresaw this.

Good. Lord.

There goes me getting to sleep at a reasonable hour.

You didn’t realize?

This is the 10th circle of hell.

I bet there’s an answer to it though. Fuck if I’m going to spend the ‘number of vowels in this sentence equal to the number of consonants minus 2 multiplied by my favorite ice cream color’ hours to find it.

It wouldn’t be too difficult to set up a spreadsheet in excel to figure it out. Also, you should be able to get 5 of them without killing yourself. The last one, and there are two sets of questions that refer directly to the other one, just go through the options and see which lead to contradictions.

Yeah, I just noticed that for 10 and 16. Also, apparently if you have answers that don’t match, it highlights them in red, which ought to make “guess and check” approaches a bit easier.

EDIT: 500th post! Woohoo! Only took, what, seven and a half years? :smiley:

OK, here’s what I have so far:

[spoiler]20 is E - Standardized tests aren’t measurements of intelligence, just as barometers don’t measure any of the above. Take the 10/16 pair. Experimentation leads to only one answer: A and D, respectively. Take the 6/17 pair. The answer to one must be B, and the other must be D, since any other possible individual answer for one leads to a contradiction with the other. Now consider 8 and 12. 8 basically asks whether the number of answers that are consonants is 16, 15, 14, 13, or 12. Compare that with question 12. Each of the possibilities for 8 give two possible answers, except for C (14) and E (12). Either way, the answer for 12 must be A, which give us the answer for 15, which in turn gives us the answer for 13, meaning no other odd-numbered question’s answer is A (which is already deducible for questions such as 1).

Speaking of question 1, we know the answer to that can’t be B. The answer for 2 can’t be D or E, since we already know that no other odd numbered question besides 14 can be A, which we know 10 is.[/spoiler]

Continuing to look…

How many should I be able to get without killing anybody else?

#7 cannot be A or B since they both refer to E in the next question. Unless you have to switch it around to make something else work I guess. That’d be cheating IMO though

That was beautiful. The best logic problem I’ve ever ‘solved’.

Ah, solved it. Here’s the rest of the reasoning that leads you to the right answer (continued from my first post above):

[spoiler]The answer for Q9 can’t be C, because we already know that the answer to Q12 is A. We know the answer for Q2 can’t be B (or else it’d lead to an answer of B for Q1, which is contradictory). Therefore, Q2 must be A or C.

Speaking of Q2, that tells us what to do with the 6/17 pair. We know that 16 is D, and since there are no consecutive repeated answers beyond 11, the answer to 17 must be B, and 6 is D. It also tells us that Q5 isn’t D. Also, Q5 can’t be A or B (A would lead to Q1 contradicting itself, and B would cause Q2 and Q1 to contradict). Therefore, Q5 must be C or E.

Suppose Q5 were C. That means Q3 would also be C. That means that, since Q8 must be C or E, that Q4 would have to be A or C, leaving no B answers in the first five questions for Q1. Therefore, Q5 is E, and E is eliminated as answers for Q1-4. Now that we have two E answers, we know Q3 can’t be B. Therefore, Q4 must be B, making Q1 D. This means, by Q8, Q3 must be D, and Q8 E.

Now that we have 3 E’s, no other answer is E. Now that we know that there are five A answers, we can tackle Q18. We know there are three E’s, so it’s not D. By Q14, we know it can’t be C, either, so it must be A or D.

Q2 must be A, because we know that Q8 is E, and there can be no more E’s. So Q7 is also D, which fits in with Q8. We now have four out of five A answers, and the only question left that can be A is Q18. So we now know there are five B answers. There are two already present, and four questions that could be B, so only one of those is not B.

Since that’s the case, we can answer Q14 - six present D answers, plus a maximum of only one other D, means the answer must be B (it can’t be A, since we already have all our A’s). So Q 9, 11, and 19 must all be either B or D. Q19 can’t be D (because then Q9 and 11 would be B, which would make Q11 wrong). Q11 can’t be D, because it would make itself wrong. Therefore, Q9 must be D, making the other two B, solving the puzzle.[/spoiler]

Leaper, I hereby dub you

Smartest Person Ever to Live

Novelty certificate is in the mail.

You see therein lies the difference between some of us - I just looked at it for about a minute and thought “screw this for a game of soldiers”.

Can we curse in the game room? It’s not good form? Well, then… I don’t have anything to say.

Finally solved it. Woohoo!!! Now I can finally read Leaper’s solution.

[time passes]

Ok, now that I have… Turns out we did things very differently in some places. For example, I didn’t figure out Q15 and Q12 until very late in the game, when I’d answered at least half of the other questions already. Also, I have a question on this:

Q19 can’t be D (because then Q9 and 11 would be B, which would make Q11 wrong). Q11 can’t be D, because it would make itself wrong. Therefore, Q9 must be D, making the other two B, solving the puzzle.

I don’t follow your reasoning for Q19 there. Specifically, what’s inside the parenthetical statement.

Qustion 11 is B, isn’t it? Or are there multiple solutions?

atomicbadgerrace: Heh, no, just an old hand at logic puzzles. :slight_smile:

Sir Dirx: What I meant is that [spoiler]of Q9, 11, and 19, the answer to one of them must be D, and the other two must be B. What I was doing was trying out cases by assigning one of them the D answer and the others the B answer.

Suppose Q19 is D. That would mean that Q9 and 11 must both be B. However, that would make TWO B answers precede Q11, making Q11’s answer of B (one) contradict itself.[/spoiler]

Hope that’s clearer.

Ah, much clearer, thanks.

You know, after the first 5 questions, I don’t think we did anything in the same order :smiley:

At one point I was convinced that I had screwed up my logic somewhere (again). It looked like some of the “how many answers are X” questions weren’t going to add up right.

I really didn’t like questions 7 and 8 though. That’s where I screwed up the first time.

I realized that all of Q7’s answers would be the same if Q8 was E. I assumed Q8 could NOT be E for that reason, and crossed it off. Luckily, when I started over and got to that point again, I decided not to make that assumption, even though it seemed “right” to me.

Wait a minute, Illuminatiprimus…

You mean it’s actually okay on this board to admit that a game is either way too ridiculously hard for you to even contemplate or not worth a second of your time? Awesome! :smiley: :smiley:

(No, I’m dead serious. Ever try espousing this belief on a typical game board? Several months of flaming is usually the best you can hope for.)

Oh, and of course, major wicked ultimate superb job, Leaper! A brilliant step-by-step solution is always a joy to read.

…yeah, that’s all I got. (Like, I think we all get the concept of “completely insane”. No need for me to stomp it into the ground.)

I think it is, but then I have a flame retardant suit. :stuck_out_tongue:

It was when I looked at the questions and saw the “The answer to Q12 is the answer to Q19 - the answer to Q19 is the answer to Q12” bit that I thought I had better things to do with my time than subject myself to that.