Question: Pick seven numbers so that every operation (addition, subtraction, multiplication, but not division) gives the same result.

* For subtraction, you may take the result's absolute value.
* Numbers cannot be $0$.

> 
> Example: 
> 
> 
> * $a, b, c, d, e, f, g$ are the numbers.
> * $a + b + c + d + e + f + g = x$
> * $| a - b - c - d - e - f - g | = x$ or $ a - b - c - d - e - f - g = x$
> * $a \times b \times c \times d \times e \times f \times g = x$
> 
> 
> 

How should I choose the numbers?

Answer: 
> 
>  If we choose the values so that the absolute value is necessary for the subtraction, then:
>  $$a+b+c+d+e+f+g = -(a-b-c-d-e-f-g)\\2a=0$$
>  Which makes $a$ zero, which is not allowed.  
> 
>  So we have
>  $$a+b+c+d+e+f+g = a-b-c-d-e-f-g\\b+c+d+e+f+g=0$$
>  and
>  $$a+b+c+d+e+f+g = a\cdot b \cdot c \cdot d \cdot e \cdot f \cdot g\\ a = a\cdot b \cdot c \cdot d \cdot e \cdot f \cdot g\\ b \cdot c \cdot d \cdot e \cdot f \cdot g=1$$
>  There are many solutions to these two equations for $b,c,d,e,f,g$. We can choose $b=c$ and $d=e=f=g$ to reduce it to two equations with two variables.  
> 
>  (Side remark: I tried putting $c=d=e=f=g$ first, but that leads to complex numbers, as did trying $b=c=d, e=f=g$.)  
> 
>  So we now get:
>  $$2b+4d=0$$
>  and 
>  $$b^2\cdot d^4=1$$
>  Solving this gives
>  $$ 4d^2 \cdot d^4 = 1\\ d=2^{-\frac{1}{3}} \text{ and } b=-2\cdot 2^{-\frac{1}{3}}$$
>  So one solution is:  
> 
>  $a$ has any value  
> 
>  $b=c= -2\cdot 2^{-\frac{1}{3}}$  
> 
>  $d=e=f=g=2^{-\frac{1}{3}}$ 
> 
> 
>
From the first two equations we get

> 
>  $$\begin{align}b+c+d+e+f+g &= -(b+c+d+e+f+g) \\ \Rightarrow \, b+c+d+e+f+g &= 0\end{align}$$
>  [As Athin already posted in his answer](https://puzzling.stackexchange.com/a/61201/40914 "deleted") the second equation written with the absolute values would imply $a = 0$ and is therefore not allowed.
> 
> 
> 

In order to also fulfill the third equation we need

> 
>  $b \cdot c \cdot d \cdot e \cdot f \cdot g = 1$, which is solved for example by setting $b = c = d = e = \frac{1}{A}$ and $f = g = A^2$.
> 
> 
> 

Together with the condition for the first two equations we then have

> 
>  $$\begin{align}4 \cdot \frac{1}{A} + 2 \cdot A^2 &= 0 \\ \Rightarrow \, A &= -\sqrt[3]{2}\end{align}$$
> 
> 
> 

Altogether, one of the many solutions is

> 
>  $a = 42 \\ b = c = d = e = -2^{-\frac{1}{3}} \\ f = g = 2^{\frac{2}{3}}$   
>  so $x = 42$.
> 
> 
>