# Find the value of $\bigstar$: Puzzle 12 - Not enough variables

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 $$\bigstar$$. To get the basic idea, I recommend you solve Puzzle 1 and Puzzle 5 first.

• 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/inequalities are satisfied (this is the heart of the puzzle): $$\begin{array}{lc} \text{I. }&\qquad a\ @\ a

## What is a Solution?

A solution is a value for $$\bigstar$$, such that, for the set of numerical symbols in the puzzle $$S_1$$ and the set of operator symbols in the puzzle $$S_2$$ there is a subtitution $$f:S_1\to\Bbb Z$$ and $$g:S_2\to\{+,-,\times,\text{^}\}$$ that satisfies all given equations.

Can you prove that there is only one possible value for $$\bigstar$$, and find that value?

Some nice symbols for your solution:

• $$\bigstar:\;$$ $\bigstar$
• $$\text^:\;$$ $\text^$
• $$\#:\;$$ $\#$
• $$\%:\;$$ $\%$
• $$\mapsto:\;$$ $\mapsto$

# Good luck!

Previous puzzles:

Introduction: #1 #2 #3 #4 #5 #6 #7

Inequalities: #8 #9 #10 #11

• In the making of this puzzle I also created a function (in Mathematica) that converts the symbols into all the possible configurations. Should I share it? Aug 4, 2023 at 22:03
• Should we assume that the usual rules of operator precedence apply? (So that, e.g.,, a @ b & c means (a @ b) & c if @ is * and & is +, but it means a @ (b & c) if @ is + and & is *.) Aug 5, 2023 at 0:04
• @GarethMcCaughan Yes, the usual rules apply Aug 5, 2023 at 6:39
• Should we assume < and = have their usual meaning? Aug 5, 2023 at 18:02
• @hexomino since the usual order of operations applies, $a\text^b\text^c=a\text^(b\text^c)$, $a-b+c=(a-b)+c$, $a*b\text^c = a*(b\text^c)$ Aug 8, 2023 at 18:33

I think the solution is as follows

$$\# \mapsto \text^$$
$$@ \mapsto +$$
$$\\\ \mapsto \times$$
$$\% \mapsto -$$
$$a = -1$$
$$b = 0$$
$$c = -2$$
$$d = -3$$
$$e=3$$.
$$f$$ can be anything.
$$|g| < 6$$
$$h=1$$
$$i$$ is an odd integer.
$$\bigstar =10$$

What follows is the explanation of how I found this. Apologies for the length, it is essentially a lot of case-bashing. The solution is found at the very end. Apologies also for any mistakes, there are probably a few in here.

Case 1

$$\# \mapsto -$$

This would mean that the right hand side of the first inequality is $$-a$$.

If $$@ \mapsto \times$$ then the left hand side is non-negative (a square) and so $$a$$ must be strictly negative. But since $$a^2 < -a$$, it can only be that $$-1 < a< 0$$ and this is not allowed.

If $$@ \mapsto +$$ then the left hand side is $$2a$$ and it must be that $$a$$ is strictly negative, we call this case 1.(i).

If $$@ \mapsto \text^$$ then the left hand side is $$a^a$$ and again $$a$$ must be negative, we call this case 1.(ii).

Case 1.(i)

If $$\\\ \mapsto \times$$ then, by equation II, $$b \times c = c \times b + b$$ and $$b=0$$.
This would mean $$\% \mapsto \text^$$ and equation V gives $$c^f = a^f - b^f$$ but since $$b=0$$ and $$f \neq 0$$ it forces $$c=-a$$ and $$f$$ must be even (thanks OP).
Equation VIII becomes $$a-i = ai^2 - a$$ or $$a = \frac{i}{2-i^2}$$ The right hand side here is an integer only when $$i=-2,-1,0,1$$ and $$2$$ and is different to $$a$$ only when $$i=-2$$ (where $$a=1$$) or $$i=2$$ (where $$a=-1$$) and since we've established that $$a$$ is negative, it must be that $$a=-1$$ and $$i=2$$ and thus $$c=1$$.
Looking at equation IV, the right hand side is $$e^e - 1$$ and the left hand side is either $$e$$, if $$e$$ is even or $$1/e$$, if $$e$$ is odd. The only integer solution here is $$e=0$$ but we already have $$b=0$$ so there is no overall solution in this case.

Alternatively, if $$\\\ \mapsto \text^$$ then equation II gives $$b^c = c^b + b$$.
A little bit of work can show us that this equation admits just two integer solutions, namely $$(b,c) = (0,0)$$ (disallowed for being equal) or $$(b,c) = (1,0)$$. This also means $$\% \mapsto \times$$ and equation V gives $$0 = af - f$$ and since $$a<0$$, it must be that $$f=0=c$$ which is not allowed so there are no possible solutions in case 1.(i).

Case 1.(ii)

If $$\\\ \mapsto \times$$ then equation II tells us that $$bc = cb^b$$ which means that either $$c=0$$ or $$b=-1$$ or $$1$$.
This would also mean that $$\% \mapsto +$$ and equation IV gives $$e+a+e = e+e+c^{a^b}$$ or $$a = c^{a^b}$$. This rules out $$c=0$$ as $$a$$ would be $$0$$. If $$b=1$$ or $$-1$$ then $$c = a^{1/a}$$ or $$a^a$$ but $$a=-1$$ gives $$c=-1$$ and other negative values of $$a$$ make $$c$$ non integral.

Alternatively, if $$\\\ \mapsto +$$ the equation II becomes $$b+c = c+b^b$$ which means $$b=1$$ or $$-1$$. In this case, $$\% \mapsto \times$$ and equation IV tells us that $$c = a^{1/a}$$ or $$a^a$$ (which runs into the same problems as previous) or that $$e=0$$.
If $$e=0$$, it means that $$f \neq 0$$ and we can factor it out of equation V to get $$c = a-b$$. Plugging this into equation III, gives us $$d^d = 2a$$ but this isn't possible since $$a$$ is a negative integer and the minimum possible for $$d^d$$ is $$-1$$. Hence, there are no possible solutions in case 1.(ii)

Case 2

$$\# \mapsto +$$

This puts the right hand side of equation I as $$3a$$.
If $$@ \mapsto -$$ then the left hand side of I is 0 so we have $$a>0$$. We call this case 2.(i).

If $$@ \mapsto \times$$ then we have $$a^2 < 3a$$ so that $$a = 1$$ or $$2$$. We call this case 2.(ii).

If $$@ \mapsto \text^$$ then we have $$a^a < 3a$$ so that $$a=1$$ or $$2$$. We call this case 2.(iii)

Case 2.(i)

If $$\\\ \mapsto \times$$ then equation II gives $$b=0$$. But then equation III gives $$d^d + a < 0$$ which is not possible with $$a > 0$$.

Alternatively, if $$\\\ \mapsto \text^$$, equation II becomes $$b^c = c^b - b$$ which has integer solutions $$(b,c) = (0,0)$$ (forbidden) and $$(b,c) = (1,2)$$. Looking at equation III, we have $$a < c^b$$ which is $$2$$ which puts $$a=1=b$$. Hence there are no solutions in Case 2.(i).

Case 2.(ii)

If $$\\\ \mapsto -$$ then equation VIII gives $$3i = a$$ which makes $$i$$ non-integral as $$a=1$$ or $$2$$.

Alternatively, if $$\\\ \mapsto \text^$$ then equation VIII gives $$i = a^{i^i}$$ which has no real solutions for $$i$$ when $$a=2$$ and gives $$i=1$$ when $$a=1$$. Hence, there are no solutions in case 2.(ii)

Case 2.(iii)

Again $$\\\ \mapsto -$$ gives us the same issue with $$a$$ as in 2.(ii).

Alternatively, if $$\\\ \mapsto \times$$, equation VIII gives $$i=ai^2$$ which only has integer solutions $$i=1=a$$ (disallowed) or $$i=0$$. However, equation II gives $$b=0$$ so this does not work. Hence, there are no solutions in case 2.(iii)

Case 3

$$\# \mapsto \times$$

If $$@ \mapsto +$$ then it must be that $$a>1$$, we call this case 3.(i)

If $$@ \mapsto -$$ then we must have $$a>0$$, we call this case 3.(ii)

If $$@ \mapsto \text^$$ then the only possibility is $$a=2$$, we call this case 3.(iii)

Case 3.(i)

If $$\\\ \mapsto -$$ then equation VIII becomes $$2ai = a-i$$ or $$i = \frac{a}{2a+1}$$ which is non-integral for $$a>1$$.

Alternatively, if $$\\\ \mapsto \text^$$ then equation VIII becomes $$i = a^{i^i}$$ which has no solutions for $$a>1$$. Hence there are no solutions in case 3.(i).

Case 3.(ii)

If $$\\\ \mapsto +$$ then equation II gives $$b=0$$. Equation V gives $$a^f = c^f$$ and since $$f \neq 0$$ and $$a \neq c$$ it forces $$f$$ to be even and $$a=-c$$. However, equation VIII gives directly $$a=-i$$, forcing $$c=i$$ which is not allowed.

Alternatively, if $$\\\ \mapsto \text^$$ then VIII gives $$i = a^{i^i}$$ which, as before, does not lead to a valid solution. Hence there are no solutions in case 3.(ii)

Case 3.(iii)

If $$\\\ \mapsto +$$ then equation II gives $$b = b^b$$ meaning that $$b=1$$ or $$-1$$.
Equation IV then gives us $$a = c^{a^b}$$ which makes $$c$$ irrational if $$b=1$$, but if $$b=-1$$ then $$c=4$$.
However, equation V then tells us that $$4-f = (2-f)(-1-f)$$ or $$f^2 = 6$$ which makes $$f$$ irrational and therefore, no solution in this case.

Alternatively, if $$\\\ \mapsto -$$, equation VIII gives $$4i = 2-i$$ or $$i=2/5$$ which is not an integer. Hence, there is no solution in case 3.(iii)

Case 4

$$\# \mapsto \text^$$

$$a<-1$$ seems to sometimes create imaginary values on the right hand side of I when $$a$$ is even but there seem to be valid real numbers in play when $$a$$ is odd so we'll restrict to these values.

If $$@ \mapsto -$$ then the left hand side of I is 0 and we just require $$a>0$$. We call this case 4.(i)

If $$@ \mapsto \times$$ then the inequality is satisfied only when $$a>1$$. We call this case 4.(ii)

If $$@ \mapsto +$$ it appears that $$a$$ can take any negative odd value and also any value $$a>1$$. We call this case 4.(iii)

Case 4.(i)

If $$\\\ \mapsto +$$ then equation VIII gives $$a^i = a + i + i^a$$. Any values of $$i < -1$$ will create a non-integer quantity on the left which cannot be rectified by the integer value of the right hand side. After that, we just need to check through a small number of values for $$a$$ and $$i$$ before realising that the largeness of $$a^i - i^a$$ cannot be compensated by $$a+i$$. In this case the solutions are $$(a,i) = (1,0)$$ or $$(2,5)$$.
Equation II implies $$b=0$$ so it must be that $$a=2$$ and $$i=5$$.
Equation V then gives $$cf = 2f$$ which means that $$c=2=a$$ or $$f=0=b$$ so this case doesn't work.

Alternatively, if $$\\\ \mapsto \times$$ the equation II gives $$b=0$$ and equation IV gives $$c=2a$$.
Then equation V gives $$c=a+1$$ so that $$a=1$$ and $$c=2$$.
Inequality III then gives $$0 < 0$$ which is not allowed so there is no solution in case 4.(i)

Case 4.(ii)

If $$\\\ \mapsto +$$ then again equation VIII implies $$(a,i) = (2,5)$$ (as $$(1,0)$$ has already been disallowed).
Equation II implies $$b=b^2$$ so that $$b=0$$ or $$b=1$$.
However, equation IV then gives us $$bc = 1$$ which means $$b=c=1$$.

Alternatively, if $$\\\ \mapsto -$$ then equation VIII gives $$a^i + i^a = a-i$$. If $$a$$ is less than $$-1$$ then the right hand side of this equation is an integer where the left cannot be, similarly when $$i<-1$$. If both numbers are greater than $$1$$, then the left hand side is clearly bigger. The only solution we can get here is thus $$a=1, i=0$$ but this has already been disallowed. Hence there are no solutions in case 4.(ii)

Case 4.(iii)

If $$\\\ \mapsto -$$ then equation VIII gives $$a^i + i^a = a-i$$. As before, the only solution we can get here is $$a=1, i=0$$ but this value for $$a$$ is disallowed by I.

Alternatively, suppose $$\\\ \mapsto \times$$.
Then equation II gives $$b=0$$. Equation V then gives us $$c = a-1$$ and equation IV gives $$c=2a$$. Thus $$a=-1$$ and $$c=-2$$.
Equation VIII then gives $$(-1)^i = -1$$ which tells us $$i$$ is odd.
Inequality III tells us that $$d + d^{-1} < 0$$ and so $$d$$ is negative and necessarily less than $$-2$$ so as not to cross paths with $$a$$ and $$c$$.
In inequality VII, if $$h < -1$$, then the right hand side is $$-\infty$$ (something that I missed before) so $$h > 0$$. Inequality VII then tells us that $$c^5 - d^3 < 1$$ which means that $$d^3 > -33$$ and so $$d=-3$$.
Inequality VI gives $$g^2 + d^e < 0$$ which means that $$e$$ is necessarily odd and $$>1$$ as $$e=1$$ forces $$|g| < 2$$ and all values in this range are already taken.
Finally inequality IX gives $$e^{2h} < \bigstar < ec^2 + a = 4e-1$$.
Since $$h > 0$$ the only value of $$e$$ which can make the left smaller than the right is $$e=3$$ and then we must have $$h=1$$. Then $$e^{2h} = 9$$, $$4e-1 = 11$$ and $$\bigstar=10$$ is the singular value which works.

• You missed something that restricts exactly what you need. It is a bit sneaky, but consider consulting the original equations Aug 9, 2023 at 19:11
• Also, in case 1.i.a, have you considered $f\in \mathbb{Z}_{even}$? Aug 9, 2023 at 19:15
• I also found minor issues in cases 1.i.b, 2.i.b, 4.iii (I don't know any a<-1 such that $a^{a^a}\in \mathbb{R}$) but I think these are trivial enough Aug 9, 2023 at 19:50
• @NODO55 yes, you're right, I have made some mistakes, will fix these up. Regarding $a^{a^a}$, if $a=-3$, for example then this is $(-3)^{-1/27}$ and I think there is a real root here (essentially the negative of the 27th root of 1/3). This works for all negative odd numbers but not evens. Aug 9, 2023 at 20:28
• @NODO55 What about $n=14$? That's a whole integer, even better. Gives $-1$ as one of the roots. Aug 9, 2023 at 21:00