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Author Topic: [SOLVED] $20 in BTC for an algebra problem  (Read 2386 times)
TriggerX
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April 23, 2015, 06:26:55 AM
Last edit: April 23, 2015, 06:37:58 AM by TriggerX
 #21

I think I have the answer, can you escrow the funds? Also do you think you can extend the date a few more days?

Edit: Nevermind, I'll just post the answers. Only for part a)

alpha = (mu)/[(mu+L)(mu+3L)]
xi = 3L/u
c = (s_k)(u+L)(u+L)(u+3L)(u+3L)/3L
k = (1/[(s_k)/m-(1+1/xi)*c*alpha*alpha*2*L)] + (mu+L)/[2*alpha*mu*L]

If it is correct, please send the BTC to this address: 13MLwx49jT4a2SSwD17qu1uaG1sbGMdrzi

Hi!
oxiyusuf
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April 23, 2015, 07:30:13 AM
 #22

Im still confuse with math question, for this one, very difficult for me! Im usually can answer for aljabar question!
erikalui
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April 23, 2015, 10:31:52 AM
 #23

Checking the solution!

Am I correct that you are suggesting the following?

k=1

alpha=(u + L)/2uL

e =4uL - 1

c=(sk/m - 1)/(1 + 1/(4uL - 1))*2u²L/(u + L)²

EDIT: I still need to check it, but I don't think it will work, since your first three lines contain a mistake...
1/k - 2alpha*u*L/(u + L) <=0;
1/k <= 2alpha*uL/(u + L);
k<=(u + L)/2alpha*uL

EDIT2: This solution does not work, since for u = 1, L =2, the second (original) equation does not hold.
(1+4*1*2-1)*(3)/(2*1*2) - 2/(1+2) = 5.3333... which is not <=0

I just checked and noticed I substituted a wrong value for alpha and so the result is incorrect. I can see that another memebr has tried solving the equation and if by chance his solution too is incorrect, I can try again.  Smiley

Agestorzrxx
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April 23, 2015, 12:30:17 PM
 #24

you can use Mathematica to solve this problem, it's not hard.
arrowguys
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April 23, 2015, 03:03:44 PM
 #25

the answer is n ^ n + b ^ = c n ^
Baldassare (OP)
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April 23, 2015, 07:40:18 PM
 #26

TriggerX,

Let u=1/100, L=2/100, s=1, m=2

Then xi = 6, c = 147/2000000, a = 100/21.

In that case the first three inequalities hold, but the last one doesn't.

BOUNTY STILL OPEN!!!!!

EDIT: I will pay $20 for EITHER a) or b) !!!! Will extend the deadline 24 hours from now.
TriggerX
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April 24, 2015, 01:23:46 AM
Last edit: April 24, 2015, 02:42:57 AM by TriggerX
 #27

xi = 2L/mu
alpha = 1/(1+xi)(mu+L)
c = (s_k)/(2*xi*mu*m*alpha*alpha)
k=(1/[(s_k)/m-(1+1/xi)*c*alpha*alpha*2*L)] + (mu+L)/[2*alpha*mu*L]

Is this correct?

Hi!
Baldassare (OP)
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April 24, 2015, 04:24:39 AM
 #28

TriggerX, this looks extremely close! Still need to carefully check
ShetKid
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April 24, 2015, 05:58:26 AM
 #29

I am not Math genius, but in high school we were told that if we had 5 variables and 4 equations for it, then it wouldn't be unsolvable . Wouldn't that be true for this case ?   
monsterdoge
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April 24, 2015, 06:22:31 AM
 #30

math make me crazy insane lol , goodluck to everyone .
bitspill
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April 24, 2015, 07:57:27 AM
 #31

wouldn't be unsolvable

May want to rephrase that a bit

{ BitSpill }
volatilebtc
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April 24, 2015, 03:40:10 PM
 #32

where is the fund being escrowed ? Show us the proof ?

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Baldassare (OP)
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April 24, 2015, 04:01:53 PM
 #33

TriggerX did not request an escrow. I'm checking his solution at this moment. If it's good, I'll send the money.
monsterdoge
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April 24, 2015, 04:21:07 PM
 #34

so triggerx already won this contest?
Baldassare (OP)
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April 24, 2015, 04:55:31 PM
 #35

Yes, TriggerX got an answer. Thanks everyone!
Contest closed.

TriggerX, money sent.
Would you mind posting the steps to the solution?
TriggerX
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April 25, 2015, 01:17:48 AM
 #36

Yes, TriggerX got an answer. Thanks everyone!
Contest closed.

TriggerX, money sent.
Would you mind posting the steps to the solution?

It took 30 minutes to figure it out and I don't want to type it all out again haha. If I have time I may post it here again.

As for the BTC, I have received it, thanks.

Hi!
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