/sci/ - Science & Math

[Anonymous No.3721996]

(a + b + c + d + e)^3 = abcde

98.9% of the population is unable to solve this problem. Can /sci/ find the value of every variable?

[Anonymous No.3722002]

a=b=c=d=e=0 works. Or do you want the complete set of valid solutions>

[Anonymous No.3722010]

>>3721996

There isnt any.
All i can see are constants.

[Anonymous No.3722015]

>>3722010
How the fuck do you know?

[EK !EKFQOBUFnQ No.3722024]

<span class="math"> (1+7+5+7+6)^3 = 17576 [/spoiler]

[Anonymous No.3722031]

The simpest non-trivial solution would be for a=b=c=d=e=k!=0 :

(5k)^3 = k^5
125 = k^2
k = +/- sqrt(125)

I imagine there's a way to find the whole set of solutions, but I don't know it.

[Anonymous No.3722033]

>>3722024
0/10

[Anonymous No.3722045]

>>3721996

OP, is "abcde" supposed to be 5 numbers multiplied together, or a series of digits?

[EK !EKFQOBUFnQ No.3722046]

>>3722033
are you implying that i was trolling?
http://www.wolframalpha.com/input/?i=%281%2B7%2B5%2B7%2B6%29^3%3D17576

[Anonymous No.3722049]

>>3722015

Because i am the Emperor of (1.1% of Population) Master Race.
Now kneel before me and weep in face of my mathematical might and magic!

[Anonymous No.3722051]

>>3722045
multiplied

[EK !EKFQOBUFnQ No.3722052]

>>3722045
i presumed series of digits...fuck
which is it OP?

[Anonymous No.3722056]

assume:
a=1 b=1 c=1 d=1

then

(4+e)^3=e
This equation can be solved
>just an example

[EK !EKFQOBUFnQ No.3722059]

>>3722051
...fine
give me a minute.

[EK !EKFQOBUFnQ No.3722065]

>>3722059
actually fuck it, i have no idea to approach the problem

[Anonymous No.3722076]

>>3722065
EK, did you, even momentarily, really think you could solve this?

[EK !EKFQOBUFnQ No.3722080]

>>3722076
yeh...
i did it in like 2 minutes when i thought abcde meant a string of digits.

[Anonymous No.3722085]

>>3722076
>EK
>think you could solve

[Anonymous No.3722088]

>>3722085

>EK
>think

[Anonymous No.3722090]

Sci don't know how to binomial theorem

[Anonymous No.3722092]

OP are you looking for an example of a solution, which are very easy to find
eg >>3722002
>>3722031
>>3722056

Or the general set of solutions?

[Anonymous No.3722097]

>>3722090

>(a + b + c + d + e)
>binomial

[Anonymous No.3722098]

>everybody laughing at people failing to awnser correctly can't awnser it themselves

[Anonymous No.3722099]

>>3722092
This is up to you. A general set of solutions would obviously be the most viable answer.

[Anonymous No.3722105]

>>3722080

Congratulations.
Did that make you feel like you had something worthwhile to post in the science and mathematics forum?
What was it like to feel that way for the first time?

[Anonymous No.3722108]

>>3722080
A correct solution (of the problem) would include a proof of uniqueness of your solution (of the equation), which you did not provide. Your solution seems like a "brute force" one. There could be other solutions, and if that is the case, then you certainly did not solve the problem.

[Anonymous No.3722115]

>>3721996
another solution is when two of the abcde's are equal to -1 and 3 of them equal to one , or three equal to -1 and two equal to 1

[Anonymous No.3722113]

redirecting
>>3722105
to
>>3722098

[EK !EKFQOBUFnQ No.3722119]

>>3722105
well i thought it was worthwhile if abcde is a string of digits. this is maths related.
and i've made worthwhile posts before.

[Anonymous No.3722120]

(-1 + 1 + -1 + 1 + -1)^3 = (-1)(1)(-1)(1)(-1)
(-1)^3 = -1
-1 = -1

a = -1, b = +1, c = -1, d = +1, e = -1

[Anonymous No.3722126]

>>3722115
(-1+1+-1+1+1)^3 = -1*1*-1*1*1

[EK !EKFQOBUFnQ No.3722133]

>>3722108
there are other solutions, mine isnt the only one, it is just the first one i found.
<span class="math"> (1+9+6+8+3)^3 = 19683 [/spoiler]
^there's another^
proof of not-unique with 1 example.

[Anonymous No.3722155]

>>3722133
Very nice, assuming the calculations are correct. Now you have to provide ALL the solutions in order for you to say that you "did it".

[Anonymous No.3722163]

(a+b+c+d+e)^3=abcde
(a+b+c+d+e)^3-abcde=0
The above's values range from -inf to +inf, so there MUST be solution to this for every considered four of variables, so this has infinite number of solutions.
I don't know what you wnat more from us

[Anonymous No.3722175]

e = -(a+b+c+d)
Either a, b, c or d = 0.

[EK !EKFQOBUFnQ No.3722177]

>>3722155
how about no?
i have better things to do than find every single solution to this fucking problem.
(actually i just realised it might not actually take that long, only values between 22^3 and 45^3 would need to be considered)
(..i'm still too lazy to do it tho)

[Anonymous No.3722183]

>>3722177
All I'm just sayan is that "i did it in like 2 minutes when i thought abcde meant a string of digits." is a false statement.

[EK !EKFQOBUFnQ No.3722185]

>>3722177
actually i'm pretty sure the only solutions that work are the 2 that i posted anyway.

[EK !EKFQOBUFnQ No.3722195]

>>3722183
define 'i did it'
i found a solution.
when doing problems usually people just find one solution and thats it, if a problem is solved you dont need to go ahead and solve it in a different way just because you can. one answer is usually enough.

[Anonymous No.3722196]

>>3722185
Prove it.

[EK !EKFQOBUFnQ No.3722206]

>>3722196
erm, i dont know how.
other than trial and error and you just presuming that i did it right.

[Anonymous No.3722250]

>>3722206
e=/=4
e=/=9

[Anonymous No.3722310]

One equation, 5 variables
Can't be "solved", can it? It's like the equation of a plane but for 5 dimensions
You have to give a value to 4 then you can probably always solve the 5th
If a,b,c,d = 0, e^3 = 0, e = 0
If a,b,c,d = 1, (4+e)^3 = e, e = ~-5.80

If a = 543, b = 2323, c = -8888888, d = 123456789
(114570767 + e)^3 = (123456789*-8888888*2323*543) * e
e = -1086

[Anonymous No.3722326]

One solution per variable when the other 4 are fixed (to anything) in <span class="math">\mathbb{R}[/spoiler]. Up to three solutions when working in <span class="math">\mathbb{C}[/spoiler].

http://www.wolframalpha.com/input/?i=solve%28%28a%2Bb%2Bc%2Bd%2Be%29%5E3%3Da*b*c*d*e%29

[Anonymous No.3722352]

>>3721996
>one equation
>5 variables
>find the value of every variable
>/sci/ not pointing this out immediately

[OP No.3722369]

a = 1
b = 2
c = 3
d = -6
e = 0

Though, as long as a+b+c+d = 0 and e = 0, any value can be used for a, b, c, and d...
Am I missing something? That solves the equation, doesn't it?

[Anonymous No.3722371]

>>3722352
In some cases, 1 equation with 5 variables can have a unique solution. Very simple example:
(a,b,c,d,e)=(1,1,1,1,1) in <span class="math">R^5[/spoiler].

If you want equations in dimension 1, then there are examples for diophantine equations. This is also true when there is a restriction on the possible solutions (like when some thought OP meant a, b, c, d and e were digits and abcde was the string of digits).

[OP No.3722375]

>>3722369
Sorry, not OP...
Forgot to change name.

[Anonymous No.3722383]

>>3722371
>(a,b,c,d,e)=(1,1,1,1,1)
>mfw (1 + 1 + 1 + 1 + 1)^3 = 5^3 = 125 =/= 1^5