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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?

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  • $\begingroup$ Should all the numbers be integers? $\endgroup$ – athin Mar 1 '18 at 6:54
  • $\begingroup$ as a=x, there have to be fractions involved, and some of them have to be negative $\endgroup$ – smriti Mar 1 '18 at 6:59
  • $\begingroup$ $a,b,c,d,e,f,g$ should be distinct numbers or not neccessarily? $\endgroup$ – Oray Mar 1 '18 at 7:13
  • $\begingroup$ Not sure if feasible, but if you want to add to the challenge: division, or inverse of division is equal to x. $\endgroup$ – Dennis Jaheruddin Mar 1 '18 at 12:01
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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
$b=c= -2\cdot 2^{-\frac{1}{3}}$
$d=e=f=g=2^{-\frac{1}{3}}$

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  • $\begingroup$ Great solution, the complex numbers arise when the powers (i.e. the number of equal variables) are odd. Other simplifications are possible: eg (b+c+d+e+f)=-1 gives us g. $\endgroup$ – Bee157 Mar 1 '18 at 11:56
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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 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 $x = 42$.

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