/wsr/ - Worksafe Requests » Thread #1550508

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Anonymous Tue 02 Dec 2025 01:54:34 No.1550508 View View Reply Original Report

Quoted By: >>1550515 >>1550525 >>1550537 >>1550998 >>1551638 >>1551685

I'm playing a game and it asked this question. It says:

D^3-H=26
H^3+D=4
D=?

I'd also like it if you could explain the process to get the answer instead of just giving me the answer, because I still want to learn it. Explain it to me like I'm a retarded monkey please. I spent all my middle and high school years being pretty mentally ill so I couldn't pay much attention, but I have an understanding of basic algebra. Pic unrelated (I was gonna use a picture from the game but the file was too big).

Anonymous

Anonymous Tue 02 Dec 2025 03:05:27 No.1550514 Report

Quoted By: >>1550516 >>1550525

There is no clever algebraic technique I can think of to solve this one. I'm pretty sure the game just wants you to plug some random small numbers in until you get something (there is in fact a small-number solution that you can find this way), without any clever systematic solution.

And no, if you find a solution in this way, that is not a guarantee that that is the *only* solution. There might be multiple solutions, and there is no clever way to check using algebra. For this particular puzzle the computer can crunch the numbers and definitely tell you that only a single (real) solution exists, but there is no easy way to reach that conclusion with algebra.

Anonymous

Anonymous Tue 02 Dec 2025 03:27:12 No.1550515 Report

Quoted By: >>1550591

>>1550508

2 equations, 2 variables
-> reform until one is alone at one side and put right side in. Or set the two right sides equal and solve for the remaining variable. Put result in and solve to get the other.

Only D is asked but to get it you obviously still have to get both. But techniocally you can get around it by using the one you are not interested in as the one you delete.

D^3-26 = H, 3rd root(4-D) = H

--> D^3-26 = (4-D)^1/3
--> D = 3 --> H = 3^3-26 = 1

alternative:

D = 3rd root(26+H), D = 4-H^3

--> (26+H)^1/3 = 4-H^3 --> H = 1 --> D = 4-1^3 = 3

Anonymous

Anonymous Tue 02 Dec 2025 03:33:19 No.1550516 Report

Quoted By: >>1550591

>>1550514
To follow up on this:
The way you would approach this problem algebraically is to rewrite
D^3-H=26
into
D^3 = H + 26
D^3 - 26 = H
H = D^3 - 26
And then substitute H = D^3 - 26 into H^3+D=4, yielding
(D^3 - 26)^3 + D = 4
Expanding and simplifying gives
((D^3)^3 - 3*26*(D^3)^2 + 3*26^2*(D^3) - 26^3) + D = 4
D^9 - 78*D^6 + 2028*D^3 - 17576 + D = 4
D^9 - 78*D^6 + 2028*D^3 + D - 17572 = 0

But now we're stuck. This is a polynomial equation of the ninth order (the highest exponent is 9), which cannot be solved algebraically. Polynomial equations of the second order, like 3x^2 - 5x + 2 = 0, can be solved algebraically using the quadratic formula; third- and fourth-order polynomials can also be solved, by using a formula that is so miserably complicated that noone ever does it in practice. But fifth-order and above cannot be solved algebraically, you can only have a computer crunch the numbers. I don't see a way forward from here.

Anonymous

Anonymous Tue 02 Dec 2025 07:33:01 No.1550525 Report

Quoted By: >>1550526 >>1550591

>>1550508
>>1550514
In general, if you want to solve Diophantine equations, you can reduce modulo a prime number first to eliminate most possibilities. Use Fermat's Little Theorem which says that when p is prime, a^p ≡ a (mod p).

mod 2 this becomes
D+H≡0
H+D≡0
solutions for (D,H) mod 2:
(0,0), (1,1)

mod 3 it becomes
D+2H≡2
H+D≡1
solution for (D,H) mod 3:
(0,1)

Combining these (by the Chinese Remainder Theorem) for possibilities mod 6:
(0,4),(3,1)

So we have eliminated 16/18 ( = 94.4%) of possibilities to brute force. Solving mod an additional prime further reduces the brute force process drastically.
In this case it wasn't necessary because the solution was small anyway.

>Explain it to me like I'm a retarded monkey please.
No, I don't feel like it.
>basic algebra
No, this is number theory. You know they want an integer solution because they didn't give you an option to input complicated expressions.

Anonymous

Anonymous Tue 02 Dec 2025 07:34:23 No.1550526 Report

Quoted By:

>>1550525
>we have eliminated 16/18 of possibilities
(I meant 34/36 = 17/18)

Anonymous

Anonymous Tue 02 Dec 2025 12:27:15 No.1550537 Report

Quoted By: >>1550591

>>1550508
What game?

Anonymous

Anonymous Tue 02 Dec 2025 23:36:19 No.1550591 Report

Quoted By:

>>1550515
>>1550516
>>1550516
>>1550525
Thank you.
>>1550537
World of Solitaire. There are some quests to do, and for some reason some of the quests are math questions. The more you progress the harder they get.

Anonymous

Anonymous Sat 06 Dec 2025 13:06:00 No.1550998 Report

Quoted By: >>1551072

>>1550508
It has been demonstrated that there is no algorithm capable of solving these equations.

Anonymous

Anonymous Sun 07 Dec 2025 04:58:23 No.1551072 Report

Quoted By: >>1551074

>>1550998
Someone got it right though

Anonymous

Anonymous Sun 07 Dec 2025 05:19:20 No.1551074 Report

Quoted By: >>1551219

>>1551072

That was through sharp watching and combination tho.
After noticing that we have to deal with 3rd roots I quickly noticed that 26 is suspiciously close to 27, so 3^3. Checking by putting it in confirmed it.
But there is no all-around way to solve something like D^3-26 = (4-D)^1/3.
You gotta look, assume and check.
Had it not worked immediately I next would have by plugging in other neighbouring values tried to estimate in which interval the solution would need to be.

Anonymous

Anonymous Mon 08 Dec 2025 09:09:20 No.1551219 Report

Quoted By:

>>1551074
Oh okay I see what you're saying. Thanks.

Anonymous

Anonymous Thu 11 Dec 2025 20:39:37 No.1551638 Report

Quoted By:

>>1550508
A pretty easy solution:

The only cube that is less than 4 is 1^3.
26 + 1 = 27. and 3^3 = 27

Anonymous

Anonymous Fri 12 Dec 2025 02:58:44 No.1551685 Report

Quoted By:

>>1550508
This is what I'd do:

OK we have two equations. And they both have to be true. What you usually do there is add the left hand sides of both equations, put that on the left of a new equation, then add the right hand side of both equations, and put that on the right of a new equation. For example:

if these are the equations we're given
2x = y
2x + y = 30

then i'd do like this
2x + 2x + y = y + 30
4x + y = y + 30
4x = 30
x = 7.5

So it should be the same thing here.

(add the left hand sides and right hand sides)
d^3 - h + h^3 + d = 26 + 4

Now here is where I fail because I don't know how to solve such equations and can't think of anything to do.

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