Do it!!!
: is division?
This is literally a system of equations.
Aka babby shit.
I already solved this for my 8 year old sister. You got this at one of those dumb escape rooms.
>>7863515
>nonlinear system of equations
>babby
It's overdetermined but still.
Go solve:
x^3-xyz+z^2=1
yx^2+x = 2
z^2+2xyz = -1
>>7863388
MATLAB will do it for me.
>>7863388
The equals signs for the vertical rows aren't vertical
>>7863660
No shit, sherlock
>>7863388
This can't be solved.
>>7863663
Also the upmost, left most "x" is smaller than the other Xs. Also in not sure what : means outside of set builder notation.
>>7863677
Let me correct myself. I am assuming that only positive whole numbers can be used. In that case, it cannot be solved. If you change the top right box from 1 to 2, then the solution is easy, but it won't work as posted.
>>7863692
It can't be solved with integers period.
>>7863388
Too many degrees of freedom.
>>7863388
>6 equations
>8 unknowns
not possible
>>7863772
that's not how this works, newfriend. I'm sorry you're taking linear algebra but this isn't linear and we're not on a field
>>7863772
Not all unknowns are contained in all the equations...
>>7863796
how is it nonlinear?
>>7863802
abc = 16
and again, most importantly, we're not in a field. in Z we have prime decomposition
Add two diagonal equations and it would be dank.
>>7863808
It would still be unsolvable, no matter how dank.
>>7863806
There are no solutions in Z
>>7863462
is : division? *
>>7863870
Rephrasing that was pointless, you know.
>>7863877
no i wasnt
>>7863388
I tried a few different ways, the middle column always seemed to stuff up while the rest worked.
Assuming integer division
Was close :(
>>7864099
2+8, equals 10. 10 divided by 8 isn't 1..
>>7864174
That's why I mentioned integer division. It's a shitty result anyway, I know.
>>7864209
Nah, like 5 mins
>>7864174
and theyre wrong anyway... lol
dont waste your time
>>7864174
In which order do we perform operations, ask any of the local high schoolers.
>>7863772
Let's set the following unknowns, p,y,f,g,c,l,a,o.
Let's set 6 equations
p=y=f=g=c=r=l=0
a and o can be chosen freely to satisfy this system of equations.
Top this
>>7864434
>2 + 6 : 8 != 1
dumbass
>>7864439
This
>>7863388
some gay shit
6 equations with 8 unknowns can't resolve ...
Juste take an example with one equation with two unknows you can result with propertie which is if you have more unknowns that equation you can't solve (can't determinate a precise value of each unknown)
here is a part of the solution to make it easier for you fags
it's not that hard.
Finally got it!
I also proved this is the only integer solution. >>7864706 is wrong.
>>7865123
Good job. I failed to solve it because I thought the order of operations was from left to right (because this newspaper puzzles usually say that)
How did you prove it's the only integer solution?
>>7865147
I partially parameterized the equations based on the top left box so that there were only a few hundred options and then used excel to solve by exhaustion.
>>7865153
What do you mean with solve by exhaustion? Brute force?
Because with simple brute force you wouldn't be able to prove it's the only solution for any value Z (only for the finite subset you brute force).
>>7864251
In which way do we read?
Oh yeah, left right and up down
>>7865166
>What do you mean with solve by exhaustion? Brute force?
Yes.
>Because with simple brute force you wouldn't be able to prove it's the only solution for any value Z (only for the finite subset you brute force).
It's not simple brute force because of the way I parameterized the equations. I'll try to explain.
First we notice that the lower boxes G,H,I can only be +/- 1,2,4,8,16
But then 6C-I = 40 so 6C=40+I. This only works if C = 4,6,7,8 and I = -16,-4,2,8
And similarly AD-G = 21 so AD = 21-G = 13,17,19,20,22,23,25,29
Now for the parameterization:
B+D = 14
B-(B-E)+AD/A = 14
1. B must be a multiple of C so B = Ck
2. A must be 1-k
3. B-E = 2-H
Ck - (2-H) + AD/(1-k) = 14
k = ( +/-sqrt( (Ck)^2 + (2-H)^2 - 2(Ck)(2-H - 2AD + 14) ) + Ck + 2-H +14 ) / (2Ck)
And since Ck, 2-H, and AD all have only a few options, we can see from excel which options give k as an integer. Once we do that all we have to do is calculate G and H and see whether or not GHI = 16. There were only 12 sets of options that gave an integer for k and only one of these gives GHI = 16
>>7865209
Awesome stuff. I would like if somebody posted another puzzle like this to see if I can solve it
>>7865229
Made a slight mistake, the last equation should be
k = ( +/-sqrt( C^2 + (2-H)^2 - 2C(2-H - 2AD + 14) ) + C + 2-H +14 ) / (2C)
>>7864439
You should think about the puzzle a little more. No matter what number you use in the second and third box in the top row, the result of division must be one or greater. When you add that number to the first box in the top row, the result of that addition must be two or greater. Since the last box in the top row is 1, that means that the normal order of operations is not used in this puzzle; rather, the order used in this puzzle is simply first to last.
>>7865309
>You should think about the puzzle a little more. No matter what number you use in the second and third box in the top row, the result of division must be one or greater. When you add that number to the first box in the top row, the result of that addition must be two or greater.
Not true. you are assuming that all of these numbers are positive. There is no solution in the positive integers and no solution if you ignore order of operations. See >>7865123 and >>7865209
Cba
