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$\newcommand{\d}{\mathbin{\triangle}}$How do you solve this question; $$1 \mathrel\triangle 0 = 3$$ $$1 \d 2 = 12$$ $$0 \d 2 = 5$$ $$1 \d 5 = 2 \d 3 = 96$$ Find $$2 \d 4 = \text{?}$$ Here it is the answer to the equation(please keep in mind that i need the method);

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Kitiara
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Questions like this are really exercises in guessing what the asker was thinking. It is not possible to determine the answer mathematically. Instead, you have to come up with some definition of $a \mathbin\triangle b$ that agrees with the information given. There are many such definitions. Here is one:

Let us represent the pair $(a,b)$ by $2^a 3^b$. Using the points $(2^1 3^0,3)$, $(2^1 3^2,12)$, $(2^0 3^2,5)$, $(2^1 3^5,96)$ and $(2^2 3^3, 96)$, we find the unique interpolating polynomial of minimal degree: $$ f(x) = \frac{150671}{272276484480}x^4 - \frac{19120391}{54455296896}x^3 + \frac{15349819}{378161784}x^2 - \frac{7010527}{56023968}x + \frac{72147351}{23343320}.$$ Now define $a \mathbin\triangle b = f(2^a3^b)$. We obtain $2 \mathbin\triangle 4 = -540445677/333476$.

Note that the polynomial $f$, called the Lagrange polynomial, can be found algorithmically, and so we have a general technique to automatically produce answers to questions like these.

Eric M. Schmidt
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Let us write $f(x,y)=x \mathbin\triangle y$. Then by interpolation there is a polynomial $f(x,y)$ satisfying the conditions $$f(1,0)=3$$ $$f(1,2)=12$$ $$f(0,2)=5$$ $$f(1,5)=96$$ $$f(2,3)=96$$ for example

$$ f(x,y)=\frac{1}{110}(708x^3 + 621x^2y - 295xy^2 + 116y^3 - 378). $$ Then we have $$f(2,4)=\frac{6603}{55}.$$ This is not the only possibility, of course.

Dietrich Burde
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  • Nice try, the closest answer so far – Kitiara Aug 18 '17 at 19:21
  • @Kitiara Are you implying that you created this problem and thus had an answer in mind? If so, this is a question for the puzzling page. I say that because mathematically there are an infinite number of valid solutions. – gen-ℤ ready to perish Aug 18 '17 at 19:28
  • No, i know the answer but not the way you solve. – Kitiara Aug 18 '17 at 19:30
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    @Kitiara This question is thus not suited for the mathematics site. You say you know the answer, but as these mathematicians have stated and rigorously demonstrated, there are many many more than one answer. – gen-ℤ ready to perish Aug 18 '17 at 19:35
  • No, this one has a unique answer. – Kitiara Aug 18 '17 at 19:38
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    @Kitiara you may as well post the answer. As has been shown there are multiple answers consistent with the conditions of problem. Posting the answer gives an additional data point which would likely eliminate a number of proposed answers. With the answer posted, there's a somewhat better chance someone can answer your question "How?". (Note that there will still be an infinite number of polynomials, as well as other formulations, even taking the answer into consideration). – Χpẘ Aug 18 '17 at 20:18
  • Alright, but i need a solution with the current information. (instead of reverse engineering) – Kitiara Aug 18 '17 at 20:28
  • @Χpẘ I modified my first post and added the answer there. – Kitiara Aug 18 '17 at 20:39
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    @kitiara there is no other choice than "reverse engineering". When you start with the results you can only go backwards to find "implementation". And there are infinitely many "implementations" that give same result. The art in this is to find an implementation that is somehow elegant. – Χpẘ Aug 18 '17 at 23:26
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Essentially, the task is to fill in the question mark in the following $\triangle$ operation table.

\begin{align} \triangle \quad&\quad 0&\quad 1&\quad\quad\quad2&\quad3&\quad\quad\quad4&\quad5\\ \hline\\ 0\quad&\quad*&*&\quad\quad\quad5&\quad*&\quad\quad\quad*&\quad*\\ 1\quad&\quad3&*&\quad\quad\quad12&\quad*&\quad\quad\quad*&\quad96\\ 2\quad&\quad*&*&\quad\quad\quad*&\quad96&\quad\quad\quad?&\quad*\\ \end{align}

Without further information about the $\triangle$ operation, one may pretty much fill in the missing spaces with random entries.

For example, one might add information such as:

"The operation $m\triangle n$ acts on non-negative integers $m$ and $n$ using a combination of the four operations of arithmetic plus exponentiation."

But even with that addendum, there might not be a unique answer.

John Wayland Bales
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  • There is a unique answer, we just don't know the way to solve. – Kitiara Aug 18 '17 at 19:39
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    @Kitiara As several people have stated, there are multiple ways to formulate $\triangle$ to match the stated conditions. So no, there's not a unique answer. However, there is probably an intended answer which would likely involve an "elegant", somewhat simple solution. – Χpẘ Aug 18 '17 at 19:59
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A possible, somewhat simple, partial formulation is $a\triangle b = 2(a\triangle (b-1))$. This formulation is consistent with the three cases of $1\triangle b$. Using this formulation would give $2\triangle 4=2(2\triangle 3)=192$

Χpẘ
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