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): $$ \text{I. }a\,@\,a=b\,@\,b \\ \space \\ \text{II. }a\,\#\,a=a\,\$\,a \\ \space \\ \text{III. }b\,\%\,b=a \\ \space \\ \text{IV. }c=a\,\#\,a\,\#\,a\,@\,a\,\#\,a\,@\,b \\ \space \\ \text{V. }d\,\$\,(e\,@\,b)=(c\,@\,e)\,\$\,(e\,\#\,a) \\ \space \\ \text{VI. }d=e\,\#\,a\,\%\,f\,\#\,a \\ \space \\ \text{VII. }d=g\,\$\,a \\ \space \\ \text{IIX. }e\,\$\,(e\,\%\,f)\,@\,a\,\%\,c=\bigstar $$

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 #5

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
    Commented Jan 18, 2018 at 9:26
  • 1
    $\begingroup$ is a^a^a = a^(a^a) or (a^a)^a ? $\endgroup$ Commented Jan 18, 2018 at 10:19
  • 1
    $\begingroup$ @NODO55 (a^a)^a = a^(a^2). But you are right that a^a^a = a^(a^a) $\endgroup$
    – Kruga
    Commented Jan 18, 2018 at 12:09
  • 1
    $\begingroup$ You labbeled the now 8th equation with the roman 9 $\endgroup$ Commented Jan 19, 2018 at 8:27
  • 2
    $\begingroup$ Roman 8 is not written this way. $\endgroup$ Commented Jan 19, 2018 at 11:21

3 Answers 3


The solution is (were ?)

$a=2, b=1, c=11, d=36, e=6, f=0, g=18, \bigstar=45$.
$@$ is $-$, $\%$ is $+$, $\#$ is $\text^$, and $\$$ is $\times$.

and (before equation $\text{VII. }$ is inserted)

$a=2, b=1, c=3, d=32, e=-5, f=21, \bigstar=152587890626$.
$@$ is $-$, $\%$ is $+$, $\#$ is $\times$, and $\$$ is $\text^$.

The newly inserted $\text{VII. }$ invalidates the latter solution because

$32=d=g^2$, and $g$ is not an integer.

My reasoning is that

From $\text{I. }$, $@$ cannot be $+$ since $a\neq b$, $@$ cannot be $\text^$ because of the following,

First it should be noted that $0^0$ is undefined, if we consider $f(n)=\left|n^n\right| (n\in \mathbb Z)$ , then for $n<-1$, $f(n)$ is strictly increasing with $n$ but strictly smaller than $1$, and for $n>1$ ,$f(n)$ is strictly increasing with $n$ and strictly larger than $1$, and of course, $1^1\neq (-1)^{-1}$ and $f(1)=f(-1)=1$.

So $@$ can only be $-$ or $\times$. For the latter case, $a=-b$.

For $\text{II. }$, the possible values that the LHS and RHS can take form the set $\{2a, 0, a^2, a^a\}$, and two of them are equal, enumerating all possibilities result in $a=0$ ,$a=1$, or $a=2$.
From $\text{III. }$, $b\%b=a$, if $a=2$ and $b=-a=-2$, there is no choice for $\%$ to make this equation true.
If $a=1$ and $b=-a=-1$, then $\%$ must be $\times$ and $@$ cannot be $\times$.
If $a=0$ and $b=-a=0$ then $a=b$ and it is not allowed.
One way or another, $@$ cannot be $\times$ so $@$ is $-$.

Now if $a=1$, then $\{\#,\$\}=\{\times,\text^\}$, so $\%$ is $+$, but $b\%b=a$ means $2b=1$ and $b$ is not an integer,
if $a=0$, then $b\%b=a$ either means $\%$ is $-$(which is already taken by $@$), or $b=0=a$, neither of them are allowed.
Hence $a=2$.

Now $b\%b=2$, $\%$ cannot be $\text^$ or $\times$, the only choice is $+$, and $b=1$.

$\text{IV. } $ now becomes $c=2\#2\#2-2\#2-1$. If $\#$ is $\times$, then $c=3$, if $\#$ is $\text^$, then $c=11$.

Now we have 2 choices, each leading to a valid solution to the original puzzle.

The first one, which is invalidated by the newly added $\text{VII. }$, is

$\textbf{A.}$ $c=3$, $\#$ is $\times$, $\$$ is $\text^$ $\text{V. }$ now becomes,
$$d\text^(e-1)=(3-e)\text^(e\times2)$$ $$d^{e-1}=(3-e)^{2e}$$
$\text{VI. }$ becomes, $$d=e\times2+f\times2$$ $$d=2e+2f$$ Combine them to make, $$(2e+2f)^{e-1}=(3-e)^{2e}$$
$$(2e+2f)^{e-1}=(e^2-6e+9)\times(e^2-6e+9)^{e-1}$$ $(e^2-6e+9)$ is clearly an integer, and it is the $e-1$ power of a rational number(call it $k$), which must also be an integer or its reciprocal. Let $l=k$ if $k$ is an integer, and $l=\frac1k$ if $k$ is the reciprocal of an integer.
$$\left|e-3\right|^2=(e-3)^2=l^{\left|e-1\right|}\ge l^{\left|e-3\right|-2}$$
Consider $g(n)=\frac{\left|l\right|^{n-2}}{n^2} (n\in\mathbb Z^+)$. For $\left|l\right|>1$, $g(n)$ increases monotonically and is strictly larger than 1 when $n>8$, which contradicts the inequality above. So we are left with only a few cases to consider.

if $\left|l\right|=0$, then $e=3=c$ and is ruled out.
If $\left|l\right|=1$, then $e=4$, $l=1$, but $l=1$ means $e^2-6e+9=2e+2f$, and this can be ruled out by parity.

Now the only thing left unchecked is $\left|e-3\right|<9$, now come back to
$$(e-3)^2=l^{\left|e-1\right|}$$ Enumerating $e$ to find out that $e\in\{-5, -1, 0, 2, 3, 4\}$, otherwise $l$ is not an integer.
In the set above, $2$ and $3$ are already used.
if $e=0$, then $l=9$, $9(2e+2f)=e^2-6e+9$. This can be ruled out by parity.
if $e=-1$, then $l=\pm4$, $\pm4(2e+2f)=e^2-6e+9$, $f=3=c$ or $f=-1=e$, neither is acceptable.
if $e=4$, then $l=1$, $2e+2f=e^2-6e+9$. This can be ruled out by parity.
if $e=-5$, then $l=\pm2$, $\pm2(2e+2f)=e^2-6e+9$, $f=21$ or $f=-11$, and $d=32$ or $d=-32$, respectively.

Now plug in everything to $\text{VIII. }$, if $f=-11$, $\bigstar$ is not an integer. If $f=21$, we have a solution, which was valid before the equation $\text{VII. }$ was inserted (there was no variable $g$ in the original puzzle). $$(-5)\text^(-5+21)-2+3=152587890626$$.

The other choice is

$\mathbf{B. } $ $c=11$ and $\#$ is $\text^$ and $\$$ is $\times$,
$\text{V. }$ now becomes,
$$d\times(e-1)=(11-e)\times(e\text^2)$$ $$d(e-1)=(11-e)e^2$$
$\text{VI. }$ becomes, $$d=e\text^2+f\text^2$$ $$d=e^2+f^2$$ Combine them to make $$(e^2+f^2)(e-1)=(11-e)e^2$$ $$f^2(e-1)=(12-2e)e^2$$
Now $f^2$ is nonnegative and $e^2$ is also nonnegative ,so $e<7$ and $e>0$. For every $e$ calculate the corresponding $f$. $1,2,3$ are already used, if $e=4$ or $e=5$, $(12-2e)e^2$ is not a multiple of $e-1$. For $e=6$ we have the solution $f=0$.
Finally, plug in everything to $\text{VIII. }$ to get $$6\times(6+0)-2+11=45$$ From the newly added $\text{VII. }$, the value of the newly added variable $g$ is,
$$d=g\times2, g=18.$$

  • $\begingroup$ well done. A proof much more complex than my version however. You are the first person to outright prove the singularity of the solution other than myself. $\endgroup$
    – NODO55
    Commented Jan 18, 2018 at 18:23
  • $\begingroup$ Thank you. Finally a case that I am not beaten by fellow answerers by speed... I am happy, no, eager, to learn the much simpler version. $\endgroup$ Commented Jan 19, 2018 at 1:16

my suggestion is


based on



@ -> -
# -> ^
$ -> x
% -> +

as we then have

I.2-2 = 1-1
II. 2^2 = 2x2
III. 1+1 = 2
IV. 11 = 2^2^2 - 2^2 - 1
V. 36 x (6-1) = (11-6)x(6^2)
VI. 36 = 6^2 + 0^2
VII. 6x(6+0) - 2 + 11 = 45

  • $\begingroup$ Please wrap your answers / partial answers in >! Spoilers $\endgroup$
    – QBrute
    Commented Jan 18, 2018 at 11:18
  • $\begingroup$ That looks right. Now try to prove that value for $\bigstar$ is the only possible value. $\endgroup$
    – NODO55
    Commented Jan 18, 2018 at 11:55

Partial answer:

From $\text{I. }a\,@\,a=b\,@\,b$ we can see that

$@$ is neither $+$ or $*$, because that would imply $a=b$, so $@\in\{-, \text^\}$

If $@=-$, then $a\in\Bbb Z$ and $b\in\Bbb Z$ with $a\neq b$
If $@=\text^$, then $a\in\{0,1\}$ and $b\in\{0,1\}$ with $a\neq b$ (here I assume $0^0=1=1^1$, see here)

From $\text{II. }a\,\#\,a=a\,\$\,a$ we can either deduce

$a=2$, if $\#\in\{+,*,\text^\}$ and $\$\in\{+,*,\text^\}$, because $2+2=2*2=2^2$ or
$a=0$, if either
$\#=-$ and $\$\in\{+,*\}$ or
$\$=-$ and $\#\in\{+,*\}$ because
$a^a=0^0=1\neq 0 = a$


$a=0\lor a=2$

In $\text{III. }b\,\%\,b=a$ we can see that

$\%\notin \{*,\text^\}$ because neither $b*b$ nor $b^b$ have solutions in $\Bbb Z$ so that $b\,\%\, b = a$ with $a\in\{0,2\}$ and $a\neq b$
$\Rightarrow\,\%\in \{+,-\}$
If $\%=\,-$, then $b\,-\,b\,=\,a\,=\,0$, but $a=0$ only if either $\#=-$ or $\$=-$

Combining $\text{I. }$ and $\text{III. }$

$b\,+\,b\,=\,2b=\,a$ with $a\neq b \neq 0$
$\Rightarrow 2a\,@\,2a\,=\,a\,@\,a$
we get $@\neq \text^$, because the equation will only hold if $a=0$
$\Rightarrow @\,=\,-$
$\Rightarrow\#\in\{*,\text^\}$ and $\$\in\{*,\text^\}$

Now looking at $\text{IV. }c=a\,\#\,a\,\#\,a\,@\,a\,\#\,a\,@\,b$

With $\#\,=\,\text^$
$\Rightarrow \$\,=\,*$

Plugging into $\text{V. }d\,\$\,(e\,@\,b)=(c\,@\,e)\,\$\,(e\,\#\,a)$ gives

Which gives integer solutions $(d\,,\,e)$:
- $(-162, -9)$
- $(-48, -4)$
- $(-6, -1)$
- $(0, 0)$ This we can ignore because $d\neq e$
- $(0, 11)$

Plug into $\text{VI. }d=e\,\#\,a\,\%\,f\,\#\,a$

$d\,=e^2+f^2$ We can see that $d\ge 0$, so the only pair $(d\,,\,e)$ that would fit is $(0, 11)$
In our case this can't be used because
$0\,=\,11^2\,+\,f^2\rightarrow f\,=\pm\sqrt{-121}\notin\Bbb Z$
All solution pairs from equation $\text{V. }d\,\$\,(e\,@\,b)=(c\,@\,e)\,\$\,(e\,\#\,a)$ has been ruled out
$\Rightarrow \#\neq \text^$
$\Rightarrow \#\,=\,*\land\$\,=\text^$
$\Rightarrow c=2\,*\,2\,*\,2\,-\,2\,*\,2\,-\,1\,=8-4-1=3$

Updating the remaining equations

$\text{V. }d\,^{e\,-\,1}=(3\,-\,e)\,^{2e}$
$\text{VI. }d=e\,*\,2\,+\,f\,*\,2=2\,(\,e\,+\,f\,)$
$\text{VII. }e\,^{e\,+\,f}\,-\,2\,+\,3=e\,^{e\,+\,f}\,+1=\bigstar$

To be continued...

  • $\begingroup$ for your conclusions on statement II, why only 2 or 0 work, and why with that sort of operation assignment? For example, for a=0, $\#=+$, $\$=*$ might work... Providing explanation to your conclusions is important since you might miss more things. $\endgroup$
    – NODO55
    Commented Jan 18, 2018 at 12:56

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