This puzzle replaces all numbers 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 first.

All symbols follow these 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.
  2. Each symbol represents a unique number. This means that for any two symbols $\alpha$ and $\beta$ in the puzzle, $\alpha\neq\beta$.
  3. The following equations inequalities are satisfied (this is the heart of the puzzle): $$ \text{I. }a^{a}-a<a\times a \\ \space \\ \text{II. }b\times b-a^{a}\times b<a \\ \space \\ \text{III. }b^{c}<a\times a\times c \\ \space \\ \text{IV. }d\times d-c\times d+c\times c<c\times d+a \\ \space \\ \text{V. }a\times e^a\times c>d^{e} \\ \space \\ \text{VI. }e\times c<a\times b-a\times d \\ \space \\ \text{VII. }e^{a}<\bigstar<a\times b^{a} $$

What is a Solution?

A solution is a value for $\bigstar$, such that, for the set of symbols in the puzzle $S_1$ there is a subtitution $f:S_1\to\Bbb Z$ that satisfies all given equations.

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

Good luck!

Side Note: to get $\bigstar$ use $\bigstar$, and to get $\text^$ use $\text^$

Previous puzzles:

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

Inequalities: #8 #9 #10

Next puzzle

  • 1
    $\begingroup$ Should we assume that only letters and ★ are integers, and all other symbols have their usual meanings? $\endgroup$
    – msh210
    Oct 12, 2022 at 22:38
  • $\begingroup$ in this one, yes. I usually say that the symbols are unknown in the puzzle description, and I use different symbols to avoid confusion $\endgroup$
    – NODO55
    Oct 13, 2022 at 5:03

2 Answers 2


Solving (1) shows that $a$ is either a negative integer (if $a\le-2$, $a^a$ is some number with magnitude less than $1$ and $a^2+a$ is positive; the inequality also holds when $a=-1$) or $1$ or $2$. $0$ is excluded because $0^0=1$.

If $a$ is a negative integer, the magnitude argument on $a^a$ shows that the left-hand side of (2), a quadratic in $b$, is always nonnegative when $b$ is an integer, which leads to a contradiction. Simple casework then shows that $(a,b)$ can only take one of five possibilities: $$\{(1,0),(2,0),(2,1),(2,3),(2,4)\}$$

We jump to (4), which can be rearranged to $(c-d)^2<a$. If $a=1$, since everything is an integer $(c-d)^2$ must be $0$, so $c=d$ which violates the "unequal values" constraint. Thus $a$ is forced to be $2$ and $|c-d|=1$ (which way we don't know yet) and $b\in\{0,1,3,4\}$.

Now (3), which is $b^c<4c$, solves as follows: $b=4$ leads to no solution, $b=3$ forces $c=1$, $b=0$ or $b=1$ merely constrains $c\ge1$.

(5) and (6) are now $2ce^2>d^e$ and $ec<2(b-d)$ respectively. If $c\ge3$ then $d\ge3$ as well (it can't assume the value of $a$), but then $b$ is $0$ or $1$ so the right-hand side of (6) is at most $2(1-3)=-4$. Then $e$ must be negative. We now look at (7), which is $e^2<\bigstar<0$ if $b=0$ – impossible – so $b=1$ and $e^2<\bigstar<2$, forcing $e=0$ and $\bigstar=1$ which violates the unequal values constraint.

Hence $c\ge3$ leads to a contradiction in any case, so $c=1$ and $d=0$, forcing $b=3$. A $0^e$ term now appears in (5), so $e$ is nonnegative; (6) becomes $e<6$, so $e=4$ or $e=5$. The latter choice gives $25<\bigstar<18$ which is impossible; the former gives $16<\bigstar<18$ and we finally have $\bigstar=17$.


EQ (1) gives $a=0,1,2$ but

EQ (4) has $(c-d)^2$ which eliminates $a=0,1$ & hence $a=2$

EQ (2) has $b^2-4b+4$ which makes $b=2,1,3,0,4$

EQ (7) has $2b^2$ which is $0,2,8,18,32$
This must be a little larger than a Square, which give $2,8,18$.
We get $e=0,1,2,3,4$

Looks like $e=4$ & $e^2=16$
which makes $b=3$ & $2b^2=18$
leaving $\bigstar=17$

[[ I will complete the analysis , figuring out the other letters , if my thinking is correct ]]

  • $\begingroup$ I've beaten you to it. $\endgroup$ Oct 13, 2022 at 4:30
  • $\begingroup$ Oh Ok ! No Problem ! Observation : Assuming that both our Answers are correct , I was earlier by 7 minutes , in giving the final answer & then pausing while waiting for feedback from OP , @ParclyTaxel $\endgroup$
    – Prem
    Oct 13, 2022 at 5:17
  • $\begingroup$ Hi, the answer is correct, now all that's left is to show it is the only possible answer. In particular, in eq.1 did you find all the possible values for $a$? $\endgroup$
    – NODO55
    Oct 13, 2022 at 6:51
  • $\begingroup$ I had Heuristically eliminated Negative values for $a$ because , in that Case , EQ (7) will have Positive less than Negative (IMPOSSIBLE) & I had also used EQ (4) to conclude that $C$ & $D$ Differ by 2 , but that was not changing the outcome that $\bigstar = 17$ , hence I did not list it in the Answer. Overall : good effort in solving this Puzzle , @ParclyTaxel ..... $\endgroup$
    – Prem
    Oct 13, 2022 at 8:01
  • $\begingroup$ .... & good effort in making this Puzzle , @NODO55 .... [[ +1 ]] to both of you ! $\endgroup$
    – Prem
    Oct 13, 2022 at 8:05

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