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:

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

  • $\begingroup$ NOTE: once an answer is accepted, the next one in the series will be posted shortly after. Side Note: to get $\bigstar$ use \bigstar in MathJax, and to get $\text^$ use \text^. This could come in handy when writing an answer to this. $\endgroup$ – NODO55 Jan 18 '18 at 7:29
  • $\begingroup$ Would it be possible for you give this series more descriptive titles? Perhaps use something that's different about this puzzle e.g. the missing operators. $\endgroup$ – boboquack Jan 18 '18 at 8:49
  • $\begingroup$ Then what do you do when there isn't anything too special about the puzzle, or if the special part of the puzzle is practically a giveaway? $\endgroup$ – NODO55 Jan 18 '18 at 9:11
  • $\begingroup$ Side note to solution: apparently I completely missed a possible value for a... but surprisingly it worked out in the end, so good job solvers! $\endgroup$ – NODO55 Jan 18 '18 at 9:37

$\bigstar$ is:



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


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

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 $d$.
$\# = +$
$a = 1\space or -1$
$d = 2\space or -2$
$c = 4\space or -4$

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

if $\$$ is $-$, $b$ would be the same as $d$.
$\$ = \times$
$b = 8$

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.
$e = 6$

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

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

  • $\begingroup$ with Eq.1, why cant $@$ be $*$ ? $\endgroup$ – NODO55 Jan 18 '18 at 8:06
  • $\begingroup$ Ah, my bad. It is possible for @ to be * if a = 0, but eq.2 quickly shows that if a = 0, d and c will be, too. Thus 0 is out of the question. I'll rework my explanation accordingly. $\endgroup$ – votbear Jan 18 '18 at 8:09
  • $\begingroup$ Another note: why cant $\#$ be $*$? a=-1 seems to work with Eq.2 $\endgroup$ – NODO55 Jan 18 '18 at 8:12
  • $\begingroup$ Yep, just mentioned that case too in my edit. I think that made it easier to read/follow too, thank you for the tips! $\endgroup$ – votbear Jan 18 '18 at 8:16
  • $\begingroup$ You've got a slight problem left, if a=0 and # = ^, then 0^0=1, so you need to go one step further and say that in that case, you would have d = 1 and c = 1^1 = 1 which is a contradiction. $\endgroup$ – Florian Bourse Jan 18 '18 at 9:53

$a=1, b=8, c=4, d=2, e=6, \bigstar=11$.


From $\text{I. }$, $@$ cannot be $+$ or $-$. If $@$ is $\times$, $a$ is $0$ and from $\text{II. }$, $d$ will also be $0$. so $@$ is $\text{^}$, so $a\text{^}(b-c)=1$ (we already know that $a$ is not $0$), so $a=\pm1$.

$\text{#}$ cannot be $-$, otherwise from $\text{III. }$, $c=0$, and from $\text{IV. }$, if $\$ $ is $\times$, $b=0$, if $\$$ is not $\times$, $b=d$, neither of these cases are acceptable.

if $\text{#}$ is $\times$, then from $\text{II. }$, $d=1$ and $a=-1$, and from $\text{I. }$, $b$ and $c$ are of same parity. So from $\text{IV. }$,$\$$ can neither be $+$ or $-$. there are nothing left to use for $\$$.

Hence $\text{#}$ must be $+$, and from $\text{II. }$, $d=2a$, from $\text{III. }$, $c=2d=4a$

From $\text{IV. }$, $\$$ cannot be $-$ otherwise $c=d$. So $\$$ is $\times$ and from $\text{IV. }$, $b=cd=8a^2$. Now $\%$ is minus. Plug everything into $\text{V. }$

$e\times e-e\times b=c-d\text{^}c$

We know from the very beginning that $a=\pm 1$, So LHS is clearly an integer, and if $a=-1$ the RHS will not be an integer.

Hence $a=1, b=8, c=4, d=2$, plug them to the RHS of $\text{V. }$ to get $-12$. Solve a quadratic equation the get $e=2$ or $e=6$. $d=2$, so $e=6$.

Now plug everything to the final formula to get $\bigstar=6\times2-1=11$

  • $\begingroup$ Still struggling with type setting. Not sure what I have done wrong but can you take a look? @NODO55 $\endgroup$ – Weijun Zhou Jan 18 '18 at 8:44
  • $\begingroup$ You need to add 2 spaces at the end of the sentence for it to register as a newline. $\endgroup$ – votbear Jan 18 '18 at 8:50
  • $\begingroup$ Thank you for helping me out! You saved my afternoon. @Votbear $\endgroup$ – Weijun Zhou Jan 18 '18 at 8:52
  • $\begingroup$ Seems fine to me. $\endgroup$ – NODO55 Jan 18 '18 at 9:35

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