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-   -   A harder problem...one for GoT (http://archives2.twoplustwo.com/showthread.php?t=108981)

GuyOnTilt 08-02-2004 08:40 PM

Re: A harder problem...one for GoT
 
i UNDERSTAND that you cannot just pick ANY number, but what i'm saying is you're just PICKING a number for a question so that it works.

I understand what you're saying too. But instead of thinking of it as picking an answer for Q1 just so the rest of the questions work, think of it as finding THE ONLY solutino for Q1 that will make the rest of the puzzle make sense.

GoT

RocketManJames 08-02-2004 08:54 PM

I think I got it... FINALLY.
 
Answers in white:

<font color="white">

Q1 : 144
Q2 : 2
Q3 : T
Q4 : 2
Q5 : F
Q6 : 24
Q7 : F
Q8 : -12
Q9 : T
Q10: -16

</font>

Solved using a few simultaneous equations... thanks for the hints.

-RMJ

aloiz 08-02-2004 08:55 PM

Re: I think I got it... FINALLY.
 
congrats, took me much longer to get it.

aloiz

RocketManJames 08-02-2004 09:11 PM

SPOILER: Equations Used
 
Now that Aloiz verified the solution, I'll post my steps.

In white... (in case the message subject wasn't enough).

<font color="white">
As most have deduced, and as GoT pointed out, these answers must be correct:

Q2: 2
Q3: T
Q4: 2
Q5: F
Q7: F
Q9: T

Leaving us with these unknowns:

Q1 : Z^2
Q6 : A
Q8 : Z
Q10: X

1) Z^2 + 2 + 2 + A + Z + X = Z^2 (Sum) =&gt;
A + Z + X = -4

2) (Z^2 + 4 + A + Z + X) / 6 = A (Average) =&gt;
Z^2 / 6 = A (by Eq 1)

3) A = -2Z (by Q9's True Condition)

4) Z^2 / 6 = -2Z (by Eq 2 and Eq 3) =&gt;
Z^2 = -12Z =&gt;
Z = -12, Z^2 = 144, A = 24

5) X = -16 (by Eq 1 and Eq 4)

</font>

-RMJ

jwvdcw 08-03-2004 02:26 PM

Re: A harder problem...one for GoT
 
[ QUOTE ]
That was supposed to make it clearer. If it had just been the sum off all numerical answers someone would have inevitably asked if that included the answer to question 1. It does.

aloiz

[/ QUOTE ]

I'm so confused...does it include #1 or not?

Basically, this is what I think it means, tell me if I'm wrong:

(2+3+4+5+6+7+8+9+10)x2

right? Since the part in the parentheses is #1 and then you add #1 in its just doubled.

jwvdcw 08-03-2004 02:29 PM

Re: A harder problem...one for GoT
 
[ QUOTE ]
i don't think you understand what i'm saying.


i understand you can't just "pick numbers from the sky" and have everything still work.

i'm just saying, the questions just start to lose all meaning.

for example, if one of the questions was:

What color is a tree?


and you answered


15.


because you "had to" .. it would be a dumb puzzle.

[/ QUOTE ]

I agree...it seems arbitrary to me as well.

pudley4 08-03-2004 03:09 PM

Re: A harder problem...one for GoT
 
[ QUOTE ]
[ QUOTE ]
i don't think you understand what i'm saying.


i understand you can't just "pick numbers from the sky" and have everything still work.

i'm just saying, the questions just start to lose all meaning.

for example, if one of the questions was:

What color is a tree?


and you answered


15.


because you "had to" .. it would be a dumb puzzle.

[/ QUOTE ]

I agree...it seems arbitrary to me as well.

[/ QUOTE ]

It is arbitrary - if you look only at that single question/equation. Once you look at all the other questions simultaneously, then you can find the answer.

It's the same idea as this quiz:

1: A + 0 = A
2: B + C + D = 0
3: A/C = B
4: A - B = C
5: D

Obviously A can be anything and it will satisfy the first equation. Once you start plugging it into the other equations, you realize that A can't be just any old number.

(Answer: A=4, B=2, C=2, D=-4)

aloiz 08-03-2004 03:21 PM

Re: A harder problem...one for GoT
 
[ QUOTE ]

Basically, this is what I think it means, tell me if I'm wrong:

(2+3+4+5+6+7+8+9+10)x2

right? Since the part in the parentheses is #1 and then you add #1 in its just doubled.


[/ QUOTE ]

If I understand this correctly not quite.
Let x = SUM(all other answers in quize not including #1)
Let y = the answer for number 1.

In order for the question to be answered, namely that the answer to #1 equals the sum of all numerical answers in the quiz including #1 we must satisfy the equation y = x + y. Now if x is not equal to 0 then y must be equal to infinity, however if x does equal 0 then y could be any integer. I did say at some point that all numerical answers were integers, which seemed like a logical inference. Therfore, x must be 0. Hope that helps.

aloiz


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