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teliot
teliot
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March 17th, 2021 at 7:00:33 PM permalink
Quote: Koi

First time poster to the forum.
Here is a nice algebra problem to chew on.
Solve the following system of equations for a, b, c, d, e, and f in real numbers other than the trivial solution of a = b = c = d = e = f = 0.
a + b + c + d + e = f^2
a + b + c + d + f = e^2
a + b + c + e + f = d^2
a + b + d + e + f = c^2
a + c + d + e + f = b^2
b + c + d + e + f = a^2

The trivial solution a=b=c=d=e=f=5 occurred to me in about 3 seconds. The spoiler is too small for the details, but that's the only non zero solution.
Last edited by: teliot on Mar 17, 2021
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Koi
Koi
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March 17th, 2021 at 7:23:05 PM permalink
Quote: teliot

The trivial solution a=b=c=d=e=f=5 occurred to me in about 3 seconds. I took about 2 minutes to figure out that's the only non trivial solution. The spoiler is too small for the details.


Your (correct) trivial solution is one I hadn't realized. I came up with more than one non-trivial solution for this problem.
charliepatrick
charliepatrick
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March 18th, 2021 at 12:59:38 AM permalink
One easy way to look at it is assume you add f, e, ... a to each side to get
a+b+c+d+e + f = f2 + f.
a + b+c+d+e+f = a2 + a.
So f2 + f = a2 + a etc.
giving for all a,b,c,d,e,f
(a2-b2) + (a-b) = 0;
(a-b)(a+b)+(a-b)=0
(a-b)(a+b+1)=0
So a=b or (a+b+1)=0
If a=b then a+a+a+a+a=a2 giving 5a=a2 giving a=0 or 5.

Not sure about (a+b+1)=0 as I think this gives a=c=e and b=d=f and a=-1-b and b2+b+3=0.
ThatDonGuy
ThatDonGuy
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March 18th, 2021 at 7:31:54 AM permalink
Quote: Koi

First time poster to the forum.
Here is a nice algebra problem to chew on.
Solve the following system of equations for a, b, c, d, e, and f in real numbers other than the trivial solution of a = b = c = d = e = f = 0.
a + b + c + d + e = f^2
a + b + c + d + f = e^2
a + b + c + e + f = d^2
a + b + d + e + f = c^2
a + c + d + e + f = b^2
b + c + d + e + f = a^2




The only solution I have found so far is a = b = c = d = e = f = 5.

unJon
unJon 
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March 18th, 2021 at 8:07:23 AM permalink
Quote: ThatDonGuy


The only solution I have found so far is a = b = c = d = e = f = 5.



I agree. Found two complex solutions by setting a=b=c and d=e=f and deriving a^4 - 4a^3 - 2a^2 - 15a = 0.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
ThatDonGuy
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Thanks for this post from:
unJonteliot
March 18th, 2021 at 8:16:12 AM permalink

a + b + c + d + e = f^2
a + b + c + d + f = e^2
a + b + c + e + f = d^2
a + b + d + e + f = c^2
a + c + d + e + f = b^2
b + c + d + e + f = a^2

Add f to the first one, e to the second one, and so on:
a (a + 1) = b (b + 1) = c (c + 1) = d (d + 1) = e (e + 1) = f (f + 1)

a (a + 1) = b (b + 1)
a^2 + a + 1/4 = b^2 + b + 1/4
(a + 1/2)^2 = (b + 1/2)^2
a + 1/2 = +/- (b + 1/2)
a = (b + 1/2) - 1/2 or a = (-b - 1/2) - 1/2
a = b or a = -(b + 1)
However, b + 1/2 = +/- (a + 1/2)
b = (a + 1/2) - 1/2 or b = (-a - 1/2) - 1/2
b = a or b = -(a + 1)
If a <> b, then a = -(b + 1) and b = -(a + 1)

Do the same with c, d, e, and f replacing a
Each term b, c, d, e, f equals a or -(a + 1)

All five equal a:
a^2 = 5a
a = 5 or 0

Four equal a:
a^2 = 4a - (a + 1) = 3a - 1
a^2 - 3a + 1 = 0
a = 3/2 +/- sqrt(5)/2
a = (3 + sqrt(5))/2: {(3 + sqrt(5))/2, (3 + sqrt(5))/2, (3 + sqrt(5))/2, (3 + sqrt(5))/2, (3 + sqrt(5))/2, -(5 + sqrt(5))/2)

a + b + c + d + e = 5 (3 + sqrt(5))/2 = (15 + 5 sqrt(5))/2
f^2 = (-(5 + sqrt(5)))^2 / 4 = (30 + 10 sqrt(5))/4 = (15 + 5 sqrt(5))/2
This is a solution

a = (3 - sqrt(5))/2: {(3 - sqrt(5))/2, (3 - sqrt(5))/2, (3 - sqrt(5))/2, (3 - sqrt(5))/2, (3 - sqrt(5))/2, -(5 - sqrt(5))/2)
a + b + c + d + e = 5 (3 - sqrt(5))/2 = (15 - 5 sqrt(5))/2
f^2 = (-(5 - sqrt(5)))^2 / 4 = (30 - 10 sqrt(5))/4 = (15 - 5 sqrt(5))/2

Three equal a:
a^2 = 3a - 2 (a + 1) = a - 2
a^2 - a + 2 = 0
a = 2 or -1
a = 2: {2, 2, 2, 2, -1, -1} does not fit into the first equation (7 = 1)
a = -1: {-1, -1, -1, -1, 2, 2} does not fit into the first equation (-2 = 4)

Two equal a:
a^2 = 2a - 3 (a + 1) = -a - 3
a^2 + a + 3 = 0 has no real solutions

One equals a:
a^2 = a - 4 (a + 1) = -3a - 4
a^2 + 3a + 4 = 0 has no real solutions

None equal a:
a^2 = -5 (a + 1) = -5a - 5
a^2 + 5a + 5 = 0 has no real solutions

Thus, the only real nonzero solutions are:
(a) All six variables = 5;
(b) Five of the variables = (3 + sqrt(5)) / 2, and the sixth = -(5 + sqrt(5)) / 2


Koi
Koi
Joined: Feb 24, 2021
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March 18th, 2021 at 8:14:44 PM permalink
Quote: ThatDonGuy


a + b + c + d + e = f^2
a + b + c + d + f = e^2
a + b + c + e + f = d^2
a + b + d + e + f = c^2
a + c + d + e + f = b^2
b + c + d + e + f = a^2

Add f to the first one, e to the second one, and so on:
a (a + 1) = b (b + 1) = c (c + 1) = d (d + 1) = e (e + 1) = f (f + 1)

a (a + 1) = b (b + 1)
a^2 + a + 1/4 = b^2 + b + 1/4
(a + 1/2)^2 = (b + 1/2)^2
a + 1/2 = +/- (b + 1/2)
a = (b + 1/2) - 1/2 or a = (-b - 1/2) - 1/2
a = b or a = -(b + 1)
However, b + 1/2 = +/- (a + 1/2)
b = (a + 1/2) - 1/2 or b = (-a - 1/2) - 1/2
b = a or b = -(a + 1)
If a <> b, then a = -(b + 1) and b = -(a + 1)

Do the same with c, d, e, and f replacing a
Each term b, c, d, e, f equals a or -(a + 1)

All five equal a:
a^2 = 5a
a = 5 or 0

Four equal a:
a^2 = 4a - (a + 1) = 3a - 1
a^2 - 3a + 1 = 0
a = 3/2 +/- sqrt(5)/2
a = (3 + sqrt(5))/2: {(3 + sqrt(5))/2, (3 + sqrt(5))/2, (3 + sqrt(5))/2, (3 + sqrt(5))/2, (3 + sqrt(5))/2, -(5 + sqrt(5))/2)

a + b + c + d + e = 5 (3 + sqrt(5))/2 = (15 + 5 sqrt(5))/2
f^2 = (-(5 + sqrt(5)))^2 / 4 = (30 + 10 sqrt(5))/4 = (15 + 5 sqrt(5))/2
This is a solution

a = (3 - sqrt(5))/2: {(3 - sqrt(5))/2, (3 - sqrt(5))/2, (3 - sqrt(5))/2, (3 - sqrt(5))/2, (3 - sqrt(5))/2, -(5 - sqrt(5))/2)
a + b + c + d + e = 5 (3 - sqrt(5))/2 = (15 - 5 sqrt(5))/2
f^2 = (-(5 - sqrt(5)))^2 / 4 = (30 - 10 sqrt(5))/4 = (15 - 5 sqrt(5))/2

Three equal a:
a^2 = 3a - 2 (a + 1) = a - 2
a^2 - a + 2 = 0
a = 2 or -1
a = 2: {2, 2, 2, 2, -1, -1} does not fit into the first equation (7 = 1)
a = -1: {-1, -1, -1, -1, 2, 2} does not fit into the first equation (-2 = 4)

Two equal a:
a^2 = 2a - 3 (a + 1) = -a - 3
a^2 + a + 3 = 0 has no real solutions

One equals a:
a^2 = a - 4 (a + 1) = -3a - 4
a^2 + 3a + 4 = 0 has no real solutions

None equal a:
a^2 = -5 (a + 1) = -5a - 5
a^2 + 5a + 5 = 0 has no real solutions

Thus, the only real nonzero solutions are:
(a) All six variables = 5;
(b) Five of the variables = (3 + sqrt(5)) / 2, and the sixth = -(5 + sqrt(5)) / 2



You mentioned all of them except for one. The case where a = (3 - sqrt(5))/2 is also a valid solution. You did correctly solve for that value of a, but didn't mention that as a solution at the bottom of your post.

Nice explanation on why the other values of a don't give valid solutions.
gordonm888
gordonm888
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March 29th, 2021 at 9:55:08 AM permalink
This is an original problem that I have created, and I hope you find it interesting.
*****************
Statement: The game is Texas Hold'em with a 55 card deck - there are 3 jokers in the deck and they all play as being completely wild. (This is an actual game that is offered on-line on a "play money" site.) Hand rankings are as customary in a wild card game: FIve of a KInd is the highest hand category, followed by Royal Flush, and so on.

You have Qs-Js (Queen and Jack of spades) as your two hole cards. The flop has come as Joker-Joker-Joker. Congratulations. you have flopped a Royal Flush! However, there are two more common cards to be dealt and . . .

a) You have one opponent with two random cards. What is your probability of winning, tieing and losing?

b) Same question, but two opponents, each with two random cards.

c) Same question, but three opponents, each with two random cards.
****************
Comments: I have started to analyze this but do not yet have the answers. This problem is easily stated but has "subtleties", so I decided to share it. Results of looping codes or simulations are welcome and would be much appreciated, but I believe a combinatronics solution for the 1 and 2 opponent cases should be obtainable. A combinatronics analysis of the 3 opponent case, however, might turn your brain to jelly and therefore should only be attempted by forum members whose brains are already jelly.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
charliepatrick
charliepatrick
Joined: Jun 17, 2011
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March 29th, 2021 at 10:55:48 AM permalink
I can see a method of approaching this which look at possible board types and then you can determine what other players need to tie or beat you.
Pairs
(i) Pair on the board (AA KK 10s-2s) so you are playing the board and can only be beaten by a higher hidden pair.
(ii) QQ,JJ similar to (i) except everyone has the same as you so you can only be beaten by AA or KK (or QQ).
Suited
(iii) Qx Jx - similar to (ii) except you have the Quins and AA KK (Qx) beats you and (Qx) Jx ties with you.
(iv) A/K with Q/J - similar to (iii) except you lose to AA/KK/(QQ) Ax/Kx/(Qx), as appropriate, beats you and Qx/(Jx) ties you.
(v) AK/AT/KT - here the board has RF but you can be beaten by any hidden pair or matching the board to make Quins.
(vi) Ax/Kx/Tx - here a player only needs a matching high suited card to tie your RF as well as hidden pair or matching a board card.
(vii) 9x or worse - player needs a hidden pair or to match a board card to beat you, or two suited cards that make RF to tie you.
Un-suited
Most of these are as above except sometimes the board won't have the RF.
gordonm888
gordonm888
Joined: Feb 18, 2015
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Thanks for this post from:
charliepatrick
March 29th, 2021 at 1:34:17 PM permalink
Quote: charliepatrick

I can see a method of approaching this which look at possible board types and then you can determine what other players need to tie or beat you.
Pairs
(i) Pair on the board (AA KK 10s-2s) so you are playing the board and can only be beaten by a higher hidden pair.
(ii) QQ,JJ similar to (i) except everyone has the same as you so you can only be beaten by AA or KK (or QQ).
Suited
(iii) Qx Jx - similar to (ii) except you have the Quins and AA KK (Qx) beats you and (Qx) Jx ties with you.
(iv) A/K with Q/J - similar to (iii) except you lose to AA/KK/(QQ) Ax/Kx/(Qx), as appropriate, beats you and Qx/(Jx) ties you.
(v) AK/AT/KT - here the board has RF but you can be beaten by any hidden pair or matching the board to make Quins.
(vi) Ax/Kx/Tx - here a player only needs a matching high suited card to tie your RF as well as hidden pair or matching a board card.
(vii) 9x or worse - player needs a hidden pair or to match a board card to beat you, or two suited cards that make RF to tie you.
Un-suited
Most of these are as above except sometimes the board won't have the RF.



Yes, that is exactly what I came up with as an analytical approach. However, as you implement it you need to differentiate between As,Ks,Ts on the board versus A,K,T of other suits, and you need to be careful to avoid doublecounting opponent hands that can beat you two ways, or both beat you and tie you.
Last edited by: gordonm888 on Mar 29, 2021
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

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