mathematics - Find the value of $bigstar$: Puzzle 5 - Every little Symbol

This puzzle replaces all numbers (and operations) with other symbols.

Your job, as the title suggests, is to find what value fits in the place of $\bigstar$.

All symbols abide to the following rules:

1. 
   
   
   
   • Each numerical symbol represents integers and only integers. This means fractions and irrational numbers like $\sqrt2$ are not allowed. However, negative numbers and zero are allowed.
   
   
   • Any symbol that is NOT numerical must be one of the following operations: $\{+,-,\times,\text{^}\}$. Notice how all operation are binary operations. This means that all operation symbols must have a number on their left and on their right. Use that fact to your advantage!
   
   
   
   
   


2. Each symbol represents a unique number/operation. This means that for any two symbols $\alpha$ and $\beta$ which are in the same puzzle, $\alpha\neq\beta$.


3. The following equations are satisfied (this is the heart of the puzzle): $$ \text{I. }a\,@\,b=a\,@\,c \\ \space \\ \text{II. }a\,\#\,a=d \\ \space \\ \text{III. }d\,\#\,d=c \\ \space \\ \text{IV. }b=c\,\$\,d \\ \space \\ \text{V. }e\,\$\,e\,\%\,e\,\$\,b=c\,\%\,d\,@\,c \\ \space \\ \text{VI. }\bigstar=e\,\$\,d\,\%\,a $$

What is a Solution?

A solution is a value for $\bigstar$, such that, for the group of numerical symbols in the puzzle $S_1$ and for the operational symbols in the puzzle $S_2$ there exists a one-to-one function $f:S_1\to\Bbb Z$ and another one-to-one function $g:S_2\to\{+,-,\times,\text{^}\}$ which, after replacing all provided symbols using these functions, satisfies all given equations.

What is a Correct Answer?

An answer is considered correct if you can prove that a certain value for $\bigstar$ is a solution. This can be done easily by getting a function from every symbol in the puzzle to the correct values (that is, find an example for $f:S_1\to\Bbb Z$ and $g:S_2\to\{+,-,\times,\text{^}\}$).

An answer will be accepted if it is the first correct answer to also prove that the solution is the only solution. In other words, there is no other possible value for $\bigstar$.

Good luck!

Previous puzzles in the series:

#1 #2 #3 #4

Next Puzzle

Answer

$\bigstar$ is:

$11$

With

$@ = \text^$
$\# = +$
$\$ = \times$
$\% = -$
$a = 1$
$b = 8$
$c = 4$
$d = 2$
$e = 6$

---

Explanation:

$\text{I. }a\,@\,b=a\,@\,c$

There are 2 ways this could be true with all distinct symbols:
Possibility 1:
$@$ is $\times$ operator, and $a$ is 0.
$@ = \times$
$a = 0$

Possibility 2:

$@$ is $\text^$ operator, and $a$ can be either $0$ or $\pm1$.
$@ = \text^$
$a = 1\space or\space 0\space or -1$

$\text{II. }a\,\#\,a=d$
$\text{III. }d\,\#\,d=c$

From this operation, we can see that $a$ cannot be 0 since no matter the operation, there will be a duplicate between $a$, $c$, and $d$. As such, only possibility 2 is valid:
$@ = \text^$
$a = 1\space or\space -1$

Building from that, $\#$ is not multiplication since $d$ and $c$ would be $\pm1$ too, and there would be at least a duplicate between $a$, $b$ and $c$.

$\#$ is not subtraction either, cause then $d$ would be 0, and $c$ would be equal to $d$.
Therefore:
$\# = +$
$a = 1\space or -1$
$d = 2\space or -2$
$c = 4\space or -4$

$\text{IV. }b=c\,\$\,d$

if $\$$ is $-$, $b$ would be the same as $d$.

Therefore:
$\$ = \times$
$b = 8$

By elimination:
$\% = -$

$\text{V. }e\,\$\,e\,\%\,e\,\$\,b=c\,\%\,d\,@\,c$

$e * e - e * b = c - d ^ c$
$e^{2} - 8e = (\pm4) - 2^{(\pm4)}$

If $a$, $b$ and $c$ are negative:
$e^{2} - 8e = -4 - (-2^{-4})$

$e^2 - 8e = -4 + 1/16$
No integer solution for e, so this is invalid.

Thus $a$, $b$ and $c$ are positive:
$e^{2} - 8e = 4 - 2^4$
$e(e-8) = -12$
$e = \{2, 6\}$

Since $d$ is already 2, $e$ cannot be 2 too.
Therefore:
$e = 6$

$\text{VI. }\bigstar=e\,\$\,d\,\%\,a$

$6 * 2 - 1 = 11$
This should leave no other possibilities, since we've exhausted all possibilities for a-e and the symbols.