Find the value of $\bigstar$: Puzzle 1 - Evaluation

Question

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

Answer 1

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

Proof

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

Answer 2

My answer is that:

$\bigstar = 16$

Proof:

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 $\bigstar$.

Answer 3

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 $\bigstar = 18 - 2 = 16$.

Answer 4

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.

Answer 5

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.

then

1

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

2

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

3

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

4

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

5

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

6

E - B = ★ (18-2 = 16; ★ = 16)

The final answer is:

16

Answer 6

Solution

$\bigstar = 16$

proof:

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