This puzzle replaces all numbers with other symbols.

Your job, as the title suggests, is to find what number fits in the place of $\bigstar$.

All symbols abide to the following rules:

  1. Each 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 $\beta$ which are in the same puzzle, $\alpha\neq\beta$.
  3. The following equations are satisfied (this is the heart of the puzzle): $$ \text{I. }\alpha\times\alpha=\alpha \\ \space \\ \text{II. }\alpha+\alpha=\beta \\ \space \\ \text{III. }\beta+\alpha=\gamma \\ \space \\ \text{IV. }\gamma\times\beta=\delta \\ \space \\ \text{V. }\delta\times\gamma=\varepsilon \\ \space \\ \text{VI. }\varepsilon-\beta=\bigstar $$

What is a Solution?

A solution is an integer 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 this function, satisfies all given equations.

What is a Correct Answer?

An answer is considered correct if you can prove that a certain value for $\bigstar$ is a solution. This can be done easily by getting a function from every symbol in the puzzle to the correct integers (that is, find an example for $f:S_1\to\Bbb Z$).

An answer will be accepted if it is the first correct answer to also prove that the solution is the only solution. In other words, there is no other possible value for $\bigstar$.

Good luck!

Next Puzzle

  • $\begingroup$ NOTE: once an answer is accepted, the next one in the series will be posted shortly after $\endgroup$
    – NODO55
    Commented Jan 17, 2018 at 17:15
  • $\begingroup$ side note: to get $\bigstar$ use \bigstar in MathJax. This could come in handy when writing an answer to this $\endgroup$
    – NODO55
    Commented Jan 17, 2018 at 17:24
  • 1
    $\begingroup$ × is just multiplication? $\endgroup$
    – phflack
    Commented Jan 17, 2018 at 17:26
  • $\begingroup$ yes. I do not know the MathJax symbol for dot multiplication, so that was what I used. $\endgroup$
    – NODO55
    Commented Jan 17, 2018 at 17:27
  • 2
    $\begingroup$ The LaTeX (MathJax) symbol for dot multiplication is \cdot, by the way. $\endgroup$ Commented Jan 17, 2018 at 17:44

6 Answers 6


I. α×α=α

The only values that satisfy this are 0 and 1

II. α+α=β

β=2α, but since 2.Each symbol represents a unique number, α and β both cannot be 0, so α=1 and β=2

III. β+α=γ

Plugging in our known values, γ=2+1=3

IV. γ×β=δ

Plugging in more values, δ=3*2=6

V. δ×γ=ε

Plugging in more values, ε=6*3=18

VI. ε−β=★

Plugging in more values, ★=18-2=16


This all relies on α=1
$$α*α=α$$ $$α*α-α=0$$ $$α^2-α=0$$ Using the quadratic formula... $$aα^2+bα+c=0$$ $$a=1, b=-1, c=0$$ $$α=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ The two values for α... $$α=\frac{-(-1)+\sqrt{(-1)^2-4*1*0}}{2*1}=\frac{1+1}{2}=1$$ $$α=\frac{-(-1)-\sqrt{(-1)^2-4*1*0}}{2*1}=\frac{1-1}{2}=0$$ But since $β=2α$, and symbols are unique, α and β cannot be 0

  • 1
    $\begingroup$ Typing up the proof, copying/pasting in symbols, and then fixing the formatting takes a lot longer than expected. Fun puzzle though $\endgroup$
    – phflack
    Commented Jan 17, 2018 at 17:46
  • $\begingroup$ Perhaps consider reformatting your answer so people will see it a bit more clearly? $\endgroup$
    – NODO55
    Commented Jan 17, 2018 at 17:59
  • $\begingroup$ Trying to figure out the formatting, I have a hard enough time with bullet points on StackOverflow $\endgroup$
    – phflack
    Commented Jan 17, 2018 at 18:01
  • 8
    $\begingroup$ The quadratic forumal feels like overkill for this. You could have just factorised into $(α - 1 ) α = 0$ and thus concluded the possible values for alpha. $\endgroup$
    – Chris
    Commented Feb 25, 2018 at 15:30

My answer is that:

$\bigstar = 16$


Equation I can be rewritten as a quadratic equation: $$\alpha^2 - \alpha = 0$$ This quadratic factors to $$\alpha (\alpha - 1) = 0 $$ With roots at $\alpha = 0$ and $\alpha = 1$. This means 0 and 1 are the only two values for $\alpha$ that satisfy the first equation. If $\alpha = 0$, then from equation II, we can see that $\beta = \alpha + \alpha$ must also equal $0.$ But we know that each symbol has a unique value! So, $\alpha = 1$.

From there, each subsequent symbol only has one possible value, determined through simple arithmetic operations:
$$\beta = \alpha + \alpha = 1 + 1 = 2$$ $$\gamma = \beta + \alpha = 2 + 1 = 3 $$ $$ \delta = \gamma \times \beta = 3 \times 2 = 6 $$ $$\epsilon = \delta \times \gamma = 6 \times 3 = 18$$ $$ \bigstar = \epsilon - \beta = 18 - 2 = 16$$
Because each equation only has one possible solution after equation I, this must also be the only possible value for $\bigstar$.

  • $\begingroup$ why are there only two numbers that satisfy Eq.1 ? $\endgroup$
    – NODO55
    Commented Jan 17, 2018 at 17:44
  • $\begingroup$ @NODO55 Edited the first part of my proof to show the algebra behind the two possible values for \alpha $\endgroup$
    – DqwertyC
    Commented Jan 17, 2018 at 17:50
  • $\begingroup$ Yup. Just that phflack beat you to it. $\endgroup$
    – NODO55
    Commented Jan 17, 2018 at 17:51
  • $\begingroup$ Definitely way better formatting, mine just turned into a bit of a mess $\endgroup$
    – phflack
    Commented Jan 17, 2018 at 17:53
  • $\begingroup$ @phflack You can make yours a lot cleaner using MathJax. Just add a dollar sign on either end of your equation, then use \alpha to make the α character, or \beta for β and so on. Or, add two dollar signs to either end to center the equation on its own line. After that, MathJax will take care of the real formatting for you. $\endgroup$
    – DqwertyC
    Commented Jan 17, 2018 at 18:00

Equation I implies that

$\alpha = 1$ or $\alpha = 0$, since these are the only two numbers (integers or otherwise) for which $\alpha^2 = \alpha$. To see this, note that the equation is equivalent to $\alpha^2 - \alpha = 0$, or $\alpha (\alpha - 1) = 0$, which implies that either $\alpha = 0$ or $\alpha -1 = 0$.

Combining this with Equation II, we conclude that

$\alpha = 1$, since if $\alpha$ were 0 we would have $\alpha + \alpha = \alpha$, and we know that $\alpha \neq \beta$. This means that $\beta = 2$.

With these assignments in mind, it is easy to calculate the rest of the symbols:

$\gamma = 2 + 1 = 3$; $\delta = 3 \times 2 = 6$; $\epsilon = 6 \times 3 = 18$; and $\bigstar = 18 - 2 = 16$.


The numbers for the greek letters, alphabetically, are 1, 2, 3, 6, 18, respectively. $\bigstar$ stands for 16.

The reasoning is that,

From I. $\alpha^2=\alpha$, so $\alpha=0$ or $\alpha=1$ From II. $\beta=0$ if $\alpha=0$, and $\beta=2$ if $\alpha=1$. The former case is against the rule that $\alpha\neq\beta$, so we have a unique soluttion. Substitute the values of $\alpha$ and $\beta$ to the remaining equations to get the complete solution.

  • $\begingroup$ awesome, you proved you found the solution! Now, can you prove it is the only solution? $\endgroup$
    – NODO55
    Commented Jan 17, 2018 at 17:41
  • $\begingroup$ @NODO55 Proven. See edited answer. $\endgroup$ Commented Jan 17, 2018 at 17:55
  • $\begingroup$ Kind of on the edge on this answer. I say its fine though. However, some peeps beat ya to it. Maybe next puzzle? $\endgroup$
    – NODO55
    Commented Jan 17, 2018 at 17:58

Because it's easier, I'm just using the English A->F:

The first equation is A x A = A. There are only two integers that satisfy that criteria, and they are 0 and 1. If A was 0, then the second equation would be 0 + 0 = B, and B would be 0 as well. Since A cannot equal B, A cannot be 0, and A must be 1.



A x A = A (1 x 1 = 1)


A + A = B (1 + 1 = 2; B=2)


B + A = C (1+2=3; c=3)


C x B = D (2 x 3 = 6; D=6)


D x C = E (6 x 3 = 18; E=18)


E - B = ★ (18-2 = 16; ★ = 16)

The final answer is:




$\bigstar = 16$


the possible values of $\alpha$:

given the equation $\alpha\times\alpha = \alpha$ we can solve it by transforming it like so: $\alpha\times\alpha-\alpha=0$ and then $\alpha(\alpha-1)=0$, therefore either $\alpha=0$ or $\alpha-1=0$ - which yields $\alpha=1$ - must be true

The correct value of $\alpha$:

moving forward with $\alpha=0$ gives $\beta=\alpha+\alpha=0+0=0$, making, two variables have the same value, namely, $\alpha=\beta$; this is forbbiden by the problem so does not lead to a correct answer

The final stretch:

now let us go ahead with $\alpha=1$, if we plug that into the equation for $beta$, and solve all the subsequent equations accordigly, we get: $$\beta=1+1=2$$ $$\gamma=\beta+\alpha=2+1=3$$ $$\delta=\gamma\times\beta=3\times 2=6$$ $$\epsilon = \delta\times\gamma = 6\times 3 = 18$$ and finally $$\bigstar=\epsilon-\beta = 18-2 = 16$$

NOTE: $\alpha$ can be found using the general solution of a quadratic equation but the method used here is simpler and more straightforward


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