1
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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 down, I recommend you solve Puzzle 1 first.

All symbols abide to the following 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 $\beta$ which are in the same puzzle, $\alpha\neq\beta$.
  3. The following equations are satisfied (this is the heart of the puzzle): $$ \text{I. }a^a=a \\ \space \\ \text{II. }a+a+a=b \\ \space \\ \text{III. }c<d<c^b\times(b-c) \\ \space \\ \text{IV. }a\times(c+d)=e\times e \\ \space \\ \text{V. }f^g=g^f \\ \space \\ \text{VI. }c\times c+d\times d<(f+g)\times e\times e \\ \space \\ \text{VII. }c-b<h<(e+a)^f \\ \space \\ \text{IIX. }h<\bigstar <e-h $$

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.

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 values (that is, find an example for $f:S_1\to\Bbb Z$).

A complete answer is a correct answer which also proves that the solution is the only solution. In other words, there is no other possible value for $\bigstar$.

How is an Answer Accepted?

After the puzzle is asked, a one day grace period will be given, in which no answer will be accepted. After that day passes, the complete answer which makes the least assumptions will be accepted. If no complete answer will appear within the grace period, the first complete answer that appears after the grace period will be accepted.

Good luck!

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


Previous puzzles in the series:

Initial pack: #1 #2 #3 #4 #5 #6 #7

Inequality pack: #8

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We must have

$\bigstar=1$.

Proof:

V is true when f=g, of course, but that won't do; also when {f,g}={2,4} or when {f,g} = {-2,-4}; and those are all the integer solutions to V. So f,g are 2,4 in some order or -2,-4 in some order. The latter is impossible because VI says positive < (f+g) . positive, which is impossible if f+g is negative.

Now

I is true only when a is -1 or +1. (You might argue for 0, but II forbids that since then we'd have b=a.) By II, b is then -3 or +3.

Let's look first at the case where

a=1, b=3. Then III says c < d < c^3(3-c). Can we have c>0? No! The LHS is positive so the RHS had better be too, so c<3. But 1,2,3 are already taken, contradiction. Can we have c=0? No! We get 0 < d < 0. What if c<0? Then the bounds on d are the wrong way around, because c^3(3-c) is always more negative than c. (Because it's c times c^2(3-c); the latter is the product of two positive factors, at least one of which must be >1.)

So

a=-1, b=-3. Now III says c < d < c^-3.(-3-c). If c>0 then that RHS is negative, so we have positive < something < negative, contradiction; if c=0 then it says 0 < d < 0, contradiction, so c<0. Then either c=-2 or c<-3. In the former case we have -2 < d < -1/8.(-3 - -2) = 1/8 which (since -1 is already taken) requires d=0. Then IV says -1 . (-2+0) = e^2 which is impossible. So c<-3. Then c < d < 1/c^3 . (-3-c) and the RHS always lies between -1 and 0. So d is negative (and must in fact be <= -2, since -1 is already taken) but less negative than c. IV then says that -(c+d) = e^2.

Note now that

f+g=6 so VI says c^2+d^2 < 6e^2 = -6(c+d). That's quite constraining. In particular it implies -(c+d) < 12. (Because c^2+d^2 >= (c+d)^2/2, so (c+d)^2/2 < -6(c+d); dividing by the positive quantity -(c+d)/2 we get -(c+d) < 12.) This thing has to be a square. Which square? Well, c <= -4 and d <= -2, so 6 = 2+4 <= -(c+d) <= 12. The only square in this range is 9.

But

the only ways to make 9 with our given constraints so far are (-7,-2) and (-5,-4). We can readily check that the former violates VI. So c=-5 and d=-4. And we have e^2=9 so e must be 3. (-3 is already taken.)

Our remaining conditions now look like

-2 < h < 2^f and h < $\bigstar$ < 3-h. The latter implies 3-h >= h+2 or h <= 1/2, so in fact we must have h=0 and then $\bigstar$ is either 1 or 2 -- but 2 is already taken, so $\bigstar$ is 1.

We have nailed down

specific values for all variables except f,g, which could be either 2,4 or 4,2.

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  • $\begingroup$ You got the right value, which is nice, but some things are missing: 1. "those are all the solutions to Eq.V". Are you sure about that? 2. on your first case with Eq.III why must LHS be positive? 3. the second row of the "Note now that" block is confusing me. How did you get those conclusions? $\endgroup$ – NODO55 Feb 25 '18 at 15:48
  • $\begingroup$ 1 and 2 are genuine oversights; I'm in the process of fixing them. 3 could use another few words of explanation, which I'm adding at the same time. [EDITED to add:] Now done. $\endgroup$ – Gareth McCaughan Feb 25 '18 at 18:21
  • $\begingroup$ this is a great improvement, however I still don't get how you came to the conclusion of the now third row of the "Note now that" block (it was on the second row but oh well). Good job though! $\endgroup$ – NODO55 Feb 25 '18 at 18:39
  • $\begingroup$ I've made it more explicit; how do you find it now? (Perhaps I've misidentified the bit that wasn't clear, in which case please accept my apologies.) $\endgroup$ – Gareth McCaughan Feb 25 '18 at 19:36
  • $\begingroup$ The bit you edited was very clear. The bit I don't understand is how you concluded that -(c+d)>=-(-4-2) $\endgroup$ – NODO55 Feb 25 '18 at 19:41

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