5
$\begingroup$

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.

All symbols follow these 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/inequalities are satisfied (this is the heart of the puzzle): $$ \begin{array}{lc} \text{I. }&\qquad a\ @\ a<a\ \#\ a\ \#\ a \\\space\\ \text{II. }&\qquad b\ $\ c=c\ $\ b\ @\ b \\\space\\ \text{III. }&\qquad d\ @\ d\ \#\ a<c\ $\ b \\\space\\ \text{IV. }&\qquad e\ \%\ a\ \%\ e=e\ \%e\ \%\ c\ @\ a\ @\ b \\\space\\ \text{V. }&\qquad c\ \%\ f=a\ \%\ f\ \#\ b\ \%\ f \\\space\\ \text{VI. }&\qquad g\ $\ g\ @\ d\ \#\ e<f\ \%\ f \\\space\\ \text{VII. }&\qquad c\ $\ c\ $\ c\ $\ c\ $\ c\ \%\ d\ $\ d\ $\ d<h\ \#\ b\ \#\ h \\\space\\ \text{VIII. }&\qquad a\ \#\ i=a\ $\ i\ $\ i\ \#\ a \\\space\\ \text{IX. }&\qquad e\ \#\ h\ $\ e\ \#\ h<\bigstar<e\ $\ c\ $\ c\ @\ a \end{array} $$

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

$\endgroup$
8
  • $\begingroup$ 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? $\endgroup$
    – NODO55
    Aug 4, 2023 at 22:03
  • $\begingroup$ 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 *.) $\endgroup$
    – Gareth McCaughan
    Aug 5, 2023 at 0:04
  • $\begingroup$ @GarethMcCaughan Yes, the usual rules apply $\endgroup$
    – NODO55
    Aug 5, 2023 at 6:39
  • 1
    $\begingroup$ Should we assume < and = have their usual meaning? $\endgroup$
    – msh210
    Aug 5, 2023 at 18:02
  • 1
    $\begingroup$ @hexomino since the usual order of operations applies, $a\text^b\text^c=a\text^(b\text^c)$, $a-b+c=(a-b)+c$, $a*b\text^c = a*(b\text^c)$ $\endgroup$
    – NODO55
    Aug 8, 2023 at 18:33

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ You missed something that restricts exactly what you need. It is a bit sneaky, but consider consulting the original equations $\endgroup$
    – NODO55
    Aug 9, 2023 at 19:11
  • $\begingroup$ Also, in case 1.i.a, have you considered $f\in \mathbb{Z}_{even}$? $\endgroup$
    – NODO55
    Aug 9, 2023 at 19:15
  • $\begingroup$ 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 $\endgroup$
    – NODO55
    Aug 9, 2023 at 19:50
  • $\begingroup$ @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. $\endgroup$
    – hexomino
    Aug 9, 2023 at 20:28
  • 1
    $\begingroup$ @NODO55 What about $n=14$? That's a whole integer, even better. Gives $-1$ as one of the roots. $\endgroup$
    – hexomino
    Aug 9, 2023 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.