Find the value of $\bigstar$: Puzzle 11 - No such thing as equality

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}-ad^{e} \\ \space \\ \text{VI. }e\times c

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

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

2 Answers

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 $$(c-d)^2. If $$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 ]]

• I've beaten you to it. Oct 13, 2022 at 4:30
• 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
– Prem
Oct 13, 2022 at 5:17
• 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 $a$? Oct 13, 2022 at 6:51
• 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 .....
– Prem
Oct 13, 2022 at 8:01
• .... & good effort in making this Puzzle , @NODO55 .... [[ +1 ]] to both of you !
– Prem
Oct 13, 2022 at 8:05