Apart from the regular numbers, now we will reform the numbers and call them peculiar numbers. We will consider the numbers in range of [1-9].

This is how the peculiar numbers are different from regular numbers.

  • Every peculiar number is processed in this way;

  • x is any number and it's peculiar form is ( x + x - x * x )^x (You take care of operation priority)

Now the problem is how to reshape the operations so that results become same as in regular numbers.


  • 3 + 2 = 5
  • Peculiar(3) + Peculiar(2) must be 5
  • Here, we have to reform the + so that the peculiar operation gives 5 again.

But here we have to think about all numbers between [1-9] and these operations +, -, x, /. An example solution will be like;

The new form of addition sign is like

  • a + b = a^b + b*a - b

closed as unclear what you're asking by JMP, Rand al'Thor, Beastly Gerbil, user58, athin Mar 3 '18 at 18:40

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  • 2
    $\begingroup$ not Peculiar(5)? $\endgroup$ – JMP Mar 3 '18 at 8:56
  • $\begingroup$ did you prove by putting values? I do not know the answer $\endgroup$ – InaccurateWeatherReport Mar 3 '18 at 9:01
  • $\begingroup$ are we going for Peculiar(2)+Peculiar(3)=Peculiar(5) or Peculiar(2)+Peculiar(3)=5? the first looks more interesting... $\endgroup$ – JMP Mar 3 '18 at 9:10
  • $\begingroup$ it will result same as in regular numbers so answer will be 5 not peculiar(5) $\endgroup$ – InaccurateWeatherReport Mar 3 '18 at 9:13
  • 5
    $\begingroup$ Do you actually have an answer for this? I ask because I see no reason for any explicit expression for these operators to exist (except for stupid polynomial constructions that are essentially the same as a table look-up). $\endgroup$ – Jaap Scherphuis Mar 3 '18 at 10:32

I'll use $p_1$ to $p_9$ to mean the values Peculiar(1) to Peculiar(9).

Consider the following function:

$$D_k(x) = \prod_{i\neq k} \frac{x-p_i}{i-p_i}$$

So for example

$$D_2(x) = \frac{(x-p_1)(x-p_3)(x-p_4)(x-p_5)(x-p_6)(x-p_7)(x-p_8)(x-p_9)}{(2-p_1)(2-p_3)(2-p_4)(2-p_5)(2-p_6)(2-p_7)(2-p_8)(2-p_9)}$$

This function is constructed such that

$$D_k(p_i) = \begin{cases} 1, & i = k \\ 0, & i \ne k \end{cases}$$

You can now construct any peculiar operator as follows.

$$a+_{peculiar}b = \sum_{i=1}^9 \sum_{j=1}^9 D_i(a)\cdot D_j(b)\cdot(i+j)\\ a\times_{peculiar}b = \sum_{i=1}^9 \sum_{j=1}^9 D_i(a)\cdot D_j(b)\cdot(i\cdot j)$$


For any peculiar numbers $a$ and $b$, only one of the terms in the double sum is non-zero, and it contributes exactly the value that you want the operator to have for those $a$ and $b$. It is basically a glorified look-up table encoded into a complicated expression. This does not use any particular properties of the peculiar numbers except for the fact that they are all distinct.


Let's denote the set of peculiar numbers $P = \{ (2n-n^2)^n : n \in \{ 1,2, \dots ,9 \} \} $
Our goal is to define functions:
$$ +_P:P^2 \to \mathbb{N} $$ $$ -_P:P^2 \to \mathbb{N} $$ $$ \cdot_P:P^2 \to \mathbb{N} $$ $$ /_P:P^2 \to \mathbb{N} $$

We're working in finite space, so we can define a function by values
$dp: P \to \mathbb{N} $
$$ dp(a) = \begin{cases} 1, & a = 1 \\ 2, & a = 0 \\ 3, & a = -27 \\ \vdots \\ 9, & a = 28179280429056 \end{cases} $$

Then it's very easy to define required functions (note that operators on the left hand side are from $P$, while the ones on the right are standard operators in $\mathbb{N}$
$$ p +_P q = dp(p) + dp(q) $$
$$ p -_P q = dp(p) - dp(q) $$
$$ p \cdot_P q = dp(p) \cdot dp(q) $$
$$ p /_P q = dp(p) / dp(q) $$


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