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matefkr
who gives a fuck? and why for? no need to resolve unresolved problems for free, epsecially not for some bastard sickrat police.


08 Jan 2014, 23:59 

tthsqe
the answer is "clearly" 49413. Seriously, though, obviously restraints need be places on the function in order that this puzzle has a unique solution. As it currently is stated it is meaningless.


09 Jan 2014, 03:43 

nop
the correct answer is 1698  8691 = 6993 which would be dificult for a procesor without neurons


09 Jan 2014, 23:20 

MHajduk
Yeah, this brainteaser is slightly misleading since we have here rather a play with graphical forms of the numbers that have to be transformed by rotation of 180 degrees (or composition of horizontal and vertical reflection) and then we have to subtract the transformed number from the original one. Such a play with mirrors. Note that not every number on the left side of the equations can be an object of the aforementioned transformation. We assume that 1 is represented by I character and the number has to be composed of 0, 1, 6, 8 and 9 digits only (otherwise it would be hard to keep the proper notation of those numbers).


09 Jan 2014, 23:26 

typedef
nop wrote: the correct answer is 1698  8691 = 6993 which would be dificult for a procesor without neurons Since a computer can solve palindromes, yes it's possible. The last step is just vertical palindrome or whichever orientation comes last. 

10 Jan 2014, 04:39 

pelaillo
The post asked for a mathematical formula and then tthsqe had given a right answer.
The given numbers form the points in a parabola in the form of y=ax2+bx+c So, the post should have asken for an ingenious solution, not a mathematical function.


10 Jan 2014, 12:59 

MHajduk
Accordingly to your formula in the attached image
0.0012*(1698^2)19.798*1698+79474 = 49316.8408 http://www.wolframalpha.com/input/?i=0.0012%2A%281698%5E2%2919.798%2A1698%2B79474 The result is not an integer and differs from the tthsqe solution (49413). 

10 Jan 2014, 13:12 

pelaillo
Quote:
You are right. There is no integer solution for this point. 

10 Jan 2014, 18:00 

tthsqe
Your coefficients are not accurate enough  if you calculate the coefficients more accurately and apply round at the end, you have the function:
Code: f(x) = round( 3*(48693*x^2792741017*x+3182103935224)/119995010 ) http://www.wolframalpha.com/input/?i=Round%5BInterpolatingPolynomial%5B%7B%7B6061%2C4152%7D%2C%7B8016%2C1092%7D%2C%7B9608%2C1512%7D%7D%2C1698%5D%5D 

10 Jan 2014, 20:19 

MHajduk
Well, I wouldn't call it a "simple" function. It's not a continuous also since you use the ceiling function to round up its value to the integer.
What I expected at the first moment when I started to think about this puzzle, was a strict Diophantine equation, defined by a continuous polynomial function that for those integer arguments will give us exactly integer values but unfortunately it seems that it wasn't the main idea of the authors. 

10 Jan 2014, 20:58 

tthsqe
OK  well lets try to find a polynomial of minimal degree that interpolates the given three points and takes only integer values on the integers, This is a perfectly reasonable problem. It might not have a solution, though.
EDIT: of course there is such a polynomial, but the degree bound I got is fairly high. Maybe bitrake can crack this one with his project Euler training. 

10 Jan 2014, 21:03 

MHajduk
I have a following idea but be aware that it isn't fully tested yet.
Let be given a function f(x) = k1*(xa)(xb)(xc) + k2*(xa)(xb)(xd) + k3*(xa)(xc)(xd) + k4*(xb)(xc)(xd) where a, b, c, d are our integer arguments: a = 6061 b = 8016 c = 9608 d = 1698 For any of the arguments listed above three of the four addends will be equal to zero and only one will be non zero (we assume that coefficients k1, k2, k3 and k4 are non zero). We want to know what values should have those coefficients to make this whole thing working. Let's see f(a) = k4*(ab)(ac)(ad) so k4 = f(a) /[(ab)(ac)(ad)] Analogously we obtain that k1 = f(d)/[(da)(db)(dc)] k2 = f(c)/[(ca)(cb)(cd)] and k3 = f(b)/[(ba)(bc)(bd)] All differences between numbers a, b, c, d are non zero, so coefficients should be non zero rational numbers. 

10 Jan 2014, 21:38 

tthsqe
yesyes, but there is no guarantee that the resulting polynomial function is integervalued on the integers.


10 Jan 2014, 21:43 

MHajduk
tthsqe wrote: yesyes, but there is no guarantee that the resulting polynomial function is integervalued on the integers. 

10 Jan 2014, 21:46 

alessandro95
Since you can costruct a lagrange polynomial (http://en.wikipedia.org/wiki/Lagrange_polynomial) passing throught any set of given points you can just pick whatever number you like as solution and then build an adeguate polynomial, so there need to be constraints on what "a function" is for this problem because is meaningless stated like that
Edit: this is the least degree polynomial passing throught all of the 4 points (8057606818 x^3)/31147645143227535(101950689058369 x^2)/20765096762151690+(345583408377147071 x)/12459058057291014427782522336379659964/10382548381075845 Last edited by alessandro95 on 10 Jan 2014, 22:14; edited 2 times in total 

10 Jan 2014, 21:54 

tthsqe
MHajduk, But that is too easy, and f(1698) can be whatever you want. I tried to make it interesting by saying that f(x) should be an integer for any integer x, and the degree should be as small as possible. This should eliminate several possibilities for f(1698).
There are lots of integervalued polynomials that satisfy f(6061) = 4152 f(8016) = 1092 f(9608) = 1512 but I have no idea what the minimum degree such polynomials is. Last edited by tthsqe on 10 Jan 2014, 22:00; edited 1 time in total 

10 Jan 2014, 21:56 

MHajduk
alessandro95 wrote: Since you can costruct a lagrange polynomial (http://en.wikipedia.org/wiki/Lagrange_polynomial) passing throught any set of given points you can just pick whatever number you like as solution and then build an adeguate polynomial, so there need to be constraints on what "a function" is for this problem because is meaningless stated like that 

10 Jan 2014, 21:57 

MHajduk
tthsqe wrote: MHajduk, But that is too easy, and f(1698) can be whatever you want. I tried to make it interesting by saying that f(x) should be an integer for any integer x, and the degree should be as small as possible. This should eliminate several possibilities for f(1698). 

10 Jan 2014, 21:58 

alessandro95
Edit: isn't there a way to delete a post?
Last edited by alessandro95 on 10 Jan 2014, 22:16; edited 1 time in total 

10 Jan 2014, 22:01 

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