This is unsolvable... right?

sonyafterdark · 2006-09-09 10:05:30

Stupid question, yes, but I want to be thorough and know for sure I'm not missing anything. What do you guys think, can this be solved?

K=A-B
L=C-D
M=A+C
N=B+D
Q=A+D
P=B+C
S=A+B+C+D
K, L, M, N, Q, P, S are known. What is A?

Haven't been here in awhile. Nice to be back.

Cheers.:D

sonyafterdark · 2006-09-09 10:44:21

mathsyperson · 2006-09-09 12:03:51

To say I'd be lying if I told you that assuming that it's impossible to solve this is wrong, would be incorrect. Not.

sonyafterdark · 2006-09-09 18:48:22

Someone posted something, which then promptly disappeared. For a moment there...
It was probably incorrect because the person that made took it off.

I just need one more relation. Which I've no idea how to get. Then it will be.
Such a nasty problem this is.

krassi_holmz · 2006-09-09 21:10:15

hA
The existence of a solution depends on the known values.
If P=S-Q,M=P+Q-N,L=P-N, there for every D C=D+L and B=N-D and A=M-C, so in this case we haxe infinitly many solutions.
There are other cases, in which we won't have solutions.
It can be solved fully, but the solution will be very long because you have 7 parameters.

sonyafterdark · 2006-09-10 00:30:53

So it is possible to fully solve it? As in obtaining a relation for A based only on knowns, without B, C, D?

I get the distinct sensation I'm being made fun of.
First, someone posted something then promptly deleted it.
Then Mathsy said it's not possible to solve it, in a rather roundabout way.
Now you say it is possible to solve it, just that it would be a really long solution.

Please, if it's obviously impossible to solve and I'm just too stupid to notice, just tell me. If it is possible to fully solve it, also, please tell me. If you just don't want to help and I'm boring you, again, just tell me.

I've no peace of mind, I keep trying to solve it for A without even knowing if it's pointless effort. I just want to know if I should give up or not. I'm not asking you to solve it for me, although I wouldn't mind, just to know if I should give up or not.

krassi_holmz · 2006-09-10 02:02:18

IT CAN BE SOLVED!
in sertain cases it may not have a solution, but the absence of solution is a solution...
if you understand...

sonyafterdark · 2006-09-10 02:13:52

Can I write A as a relation of K, L, M, N, Q, P, S, without B, C, D or not? That I could understand.
A solution of knowing I cannot would not be much fun though. Or do you mean that I can but it involves division to one of the knowns and as such the possibilty of /0?:o

Oh dear, not getting anywhere.

krassi_holmz · 2006-09-10 02:26:27

Can I write A as a relation of K, L, M, N, Q, P, S, without B, C, D or not?

Yes (When it's possible in terms of the parameters). This is obvios, because B,C,D are unknowns.
But sometimes one of the unknowns may have more of one values, and then the other unknowns will be expressed in terms of it.
For example, if you have the equation:
x=y, then x can be every number, but y MUST BE EQUAL to x (not any)
And in this system you can't have division by 0.
I'm talking about something like:

0=0+const;
0-const=10;

Get it now?

sonyafterdark · 2006-09-10 03:27:29

Nope. It just compounded my confusion. I really can't see how I can solve it, how I could reach such a relation.

krassi_holmz · 2006-09-10 05:09:18

OK I'll try another.
Using:
M=A+C
N=B+D
Q=A+D
P=B+C
S=A+B+C+D

We have:M+N+Q+P=A+C+B+D+A+D+B+C= (A+B+C+D)+(A+B+C+D)=2S.
So M+N+Q+P=2S should be right.
But if you pick some values for M,N,Q,P and S, for which
M+N+Q+P=2S
is false, then the system will have no solutions for the chosen M,N,Q,P and S.
clear now?

Last edited by krassi_holmz (2006-09-10 05:09:38)

sonyafterdark · 2006-09-10 05:35:33

But there are also K and L that are known. You don't use those. I don't understand why not.
Also, I don't pick values for any of K, L, M, N, Q, P or S. They're merely the known result of individually unknown A, B, C, D. I CANNOT have any values for K, L, M, N, Q, P or S that contradict the relations I gave.

K, L, M, N, Q, P and S are known. I know their values. I do not know the value of A. Or B, or C, or D, whatever. You can't find one without finding all the rest. Trouble is I don't think you can find any out.

M+N+Q+P=2S IS RIGHT. I don't understand what you mean. It CANNOT possibly be FALSE. For it to be false would contradict the hypothesis. It would contradict the relations I gave at the start of the thread.

All the relations in the hypothesis and their consequences are TRUE in any and all cases:

K=A-B ALWAYS TRUE.
L=C-D ALWAYS TRUE.
M=A+C ALWAYS TRUE.
N=B+D ALWAYS TRUE.
Q=A+D ALWAYS TRUE.
P=B+C ALWAYS TRUE.
S=A+B+C+D ALWAYS TRUE.
M+N+Q+P=2S ALWAYS TRUE.
S=M+N ALWAYS TRUE.
S=Q+P ALWAYS TRUE.
M+N=Q+P ALWAYS TRUE.
S=K+2N+L ALWAYS TRUE.
S=K+P+N ALWAYS TRUE.
P=N+L ALWAYS TRUE.
etc... ALWAYS TRUE.

K, L, M, N, Q, P, S KNOWN.
A, B, C, D UNKNOWN.
What is A?

Please, please help. I'm losing my sanity over this problem. At least tell me it's insane and unsolvable so I can sleep easily.

sonyafterdark · 2006-09-10 05:55:47

OMG! Can anyone help me?! Will anyone bother?
Not.

krassi_holmz · 2006-09-10 07:02:25

But there are also K and L that are known. You don't use those. I don't understand why not.

Because in my example it doesn't matter what K and L are.
They can be any.
(I'm tlking about a particular case, when M+N+Q+P≠ 2S. I'm not talking about the whole problem.)

krassi_holmz · 2006-09-10 07:03:56

But sorry. If you don't like my comments, I'll stop posting.

mathsyperson · 2006-09-10 07:54:51

sonyafterdark wrote:

I get the distinct sensation I'm being made fun of.
First, someone posted something then promptly deleted it.
Then Mathsy said it's not possible to solve it, in a rather roundabout way.
Now you say it is possible to solve it, just that it would be a really long solution.

I don't think people are making fun of you. At least not intentionally. I don't know what happened about the phantom post, but I presume that someone posted an incorrect solution, realised it was wrong and quickly took it down again.

Then I came and looked at it for a while, wrote down a few equations and came to the conclusion that whenever you tried to rearrange equations and substitute them into other equations to make A the subject of an equation involving only known values, you always ended up with
A = A + {various known values}

And so the A's cancel out and all you get is a combination of known values that adds up to 0, which is useless.

But I was in a weird mood and so I decided to give a cryptic answer. Which, in retrospect, was a mistake and I'm sorry. But yeah, as far as I can tell, it's not possible. I can't give you a formal proof though.

numen · 2006-09-10 08:03:19

This is what I get using Gauss elimination (Sorry, I couldn't figure how to do proper LaTeX code here):

K =A-B
N = B +D
L = C-D
M = K+N+L
Q= K+N
P= N+L
S= K+2N+L

So basically, we have 3 equations and 4 unknowns, since the equations for M,Q,P and S are useless here.

numen · 2006-09-10 08:05:53

http://en.wikipedia.org/wiki/Gaussian_elimination

sonyafterdark · 2006-09-10 08:13:07

Super. 10x a lot. At least I know I shouldn't be wasting anymore time on this. Alas, now I've got to find another approach to the problem. :|

Math Is Fun Forum

#1 2006-09-09 10:05:30

This is unsolvable... right?

#2 2006-09-09 10:44:21

Re: This is unsolvable... right?

#3 2006-09-09 12:03:51

Re: This is unsolvable... right?

#4 2006-09-09 18:48:22

Re: This is unsolvable... right?

#5 2006-09-09 21:10:15

Re: This is unsolvable... right?

#6 2006-09-10 00:30:53

Re: This is unsolvable... right?

#7 2006-09-10 02:02:18

Re: This is unsolvable... right?

#8 2006-09-10 02:13:52

Re: This is unsolvable... right?

#9 2006-09-10 02:26:27

Re: This is unsolvable... right?

#10 2006-09-10 03:27:29

Re: This is unsolvable... right?

#11 2006-09-10 05:09:18

Re: This is unsolvable... right?

#12 2006-09-10 05:35:33

Re: This is unsolvable... right?

#13 2006-09-10 05:55:47

Re: This is unsolvable... right?

#14 2006-09-10 07:02:25

Re: This is unsolvable... right?

#15 2006-09-10 07:03:56

Re: This is unsolvable... right?

#16 2006-09-10 07:54:51

Re: This is unsolvable... right?

#17 2006-09-10 08:03:19

Re: This is unsolvable... right?

#18 2006-09-10 08:05:53

Re: This is unsolvable... right?

#19 2006-09-10 08:13:07

Re: This is unsolvable... right?

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