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 are satisfied (this is the heart of the puzzle): $$ \text{I. }a\times a=a \\ \space \\ \text{II. }c<b<a \\ \space \\ \text{III. }b^c<c^b \\ \space \\ \text{IV. }d\times b+b=c \\ \space \\ \text{V. }d\times d+c+c<b\times b<d+d \\ \space \\ \text{VI. }e+b\times b\times e=d \\ \space \\ \text{VII. }a+b\times b+c+d+d+e=\bigstar $$

What is a Solution?

A solution is a value for $\bigstar$, such that, for the group of symbols in the puzzle $S_1$ there exists a one-to-one function $f:S_1\to\Bbb Z$ which, after replacing all provided symbols using these functions, 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

Next Puzzle


3 Answers 3


1. Let's rewrite the equations to the more readable form: $$1.\ a^2=a$$ $$2.\ c<b<a $$ $$3.\ b^c<c^b$$ $$4.\ (d+1)b=c$$ $$5.\ d^2+2c<b^2<2d$$ $$6.\ e(b^2+1)=d$$ $$7.\ a+b^2+c+2d+e=\bigstar$$ 2. Now prove that all numbers (except possibly $a$) are nonzero. Firstly, from (1) we have $a\leqslant 1$ (because either $a=0$ or $a=1$), so $b\leqslant 0$ and $c\leqslant-1$, and $c\ne0$. Next, from (4) it's obvious that $b\ne0$, Now, since $b^2>0$, then $2d$ must be also $>0$ from (5), so $d\ne0$. Finally, from (6) we get $e\ne0$.
3. If we find a solution with $a=1$, we can safely replace it with $a=0$ (and other numbers the same), and it will affect neither the equations (since $a$ is only present in conditions (1) and (2), and it's easy to see that they will still hold when we replace $a=1$ with $a=0$, since we've already proven that $b<0$ (strictly negative)) nor the uniqueness of numbers (since we've proven that $bcde\ne0$), but will change the value of $\bigstar$ (as can be easily seen from (7)). So, to prove the uniqueness of $\bigstar$, we need to prove that $a$ is necessarily $0$ (but this is not sufficient).
4. We already know that $b$ and $c$ are both negative (from (2) and the fact that $b\ne0$), and $d$ and $e$ must be positive from (5) and (6). Now let's introduce $x=-b=|b|$ and $y=-c=|c|$, so all numbers in equations will be positive (of course, except $a$).
5. Now, we get $$d^2+2c=d^2-2y=e^2(x^2+1)^2-2e(x^2+1)x-2x=(e^2x^4-2ex^3+2ex^2-(2e+2)x+e^2) < x^2,$$ or $$e^2x^4-2ex^3+(2e-1)x^2-(2e+2)x+e^2<0$$ ($e$ and $x$ being positive integers). This polynomial (let's designate it $P(x)$) does monotonically increase on $[2;+\infty)$ (here we assume that $x\geqslant 2$, but see below for $x=1$) because its derivative $P'(x)=4e^2x^3-6ex^2+(4e-2)x-(2e+2)$ is positive for $x=2$ and any $e\geqslant1$ (being $P'(2)=32e^2-18e-2$), and $P''(x)=12e^2x^2-12ex+(4e-2)$ is positive for any $x\geqslant2$ and $e\geqslant1$ (so $P'(x)$ keeps its sign, also being positive everywhere for these values of $x$). So, that means that minimum value of $P(x)$ (for $x\geqslant2$) is $P(2)=16e^2-16e+(8e-4)-(4e+4)+e^2=17e^2-12e-8$. It can be less than zero only when $e=1$ (and therefore $a=0$, due to the uniqueness requirement).
6. When $x=1$, we get $e^2-2e+(2e-1)-(2e+2)+e^2<0$, or $2e^2-2e-3<0$, which is true again only for $e=1$.
7. So, $e=1$, $a=0$, $d=x^2+1$, $x\geqslant1$ and $x^4-2x^3+x^2-4x+1<0$ (so either $x=1$ or $x=2$, i.e. $b=-1$ or $b=-2$). If $b=-1$, we get $d=2$ and $c=-3$ (Jens' solution); if $b=-2$, we get $d=5$ and $c=-12$ (Oray's solution). In both cases, $\bigstar=3$.
8. Since we have proven that no other solutions exist, the proof is complete. Q.E.D.



$a=0$, $b=-1$, $c=-3$, $d=2$ and $e=1$ then the equations are fulfilled and $\bigstar = 3$

  • $\begingroup$ is that the only solution though? $\endgroup$
    – NODO55
    Commented Jan 3, 2020 at 18:43
  • $\begingroup$ This answer is exact copy of Oray... can we not just copy others answer and try to solve it ourself $\endgroup$ Commented Jan 4, 2020 at 8:51
  • $\begingroup$ @Ms Designer : Uh...this answer is not the same as Oray's answer and I posted first. $\endgroup$
    – Jens
    Commented Jan 4, 2020 at 10:07

It seems a,b,c,d,e values are not unique; here is another answer;

a=0,b=-2;c=-12,d=5;e=1 but it still makes $\bigstar$=3.

  • $\begingroup$ @nahmid d^2+2c<b2 $\endgroup$
    – Oray
    Commented Jan 3, 2020 at 19:29
  • $\begingroup$ could there be another solution? $\endgroup$
    – NODO55
    Commented Jan 4, 2020 at 0:35

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