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Notice that we can split this up as

3/2  + 8/27 +.......     and

-2/3  - 32/243 +   .......  =  - ( 2/3 + 32/243 + .....)

The geometric ratio of the  first series is  (2/3)^4  =  16/81

So.....the sum of this series  =   (3/2)  / ( 1 - 16/81)  = (3/2) / ( 65/81) =   243/130

The geometric ratio of the terms in the parentheses in the second series is  ( 2/3)^4 = 16/81

So  the sum of the second series is

-  [ (2/3) / ( 1 - 16/81 ) ]  =   -  ( 2/3) / ( 65/81)    =   - (2/3) (81/65) =  - 54 /65 = -108 / 130

So....the sum is   [243 - 108] / 130  =  135 / 130   =  27 /26

editor keeps mangling my answer.

The final answer is

\(\dfrac{t(2t-3)}{4t-2}\leq 0 \Rightarrow t \in (\infty,0] \bigcup \left(\dfrac 1 2, \dfrac 2 3\right]\)

Hmmmmm.....    I still don't get this.....

example      base 4     number      3 3 3 3     is   12 bracelets    ....  there is only 4 total ...plus the possibilty of 0.

    so I think you would have to eliminate any base 4 number that totals more than 4   ....  Yes?    No?

\(\text{well.. let's brute force it and see if we get any insight}\\ z^2+z+1=0\\ z = \dfrac{-1 \pm i \sqrt{3}}{2} = e^{\pm i \frac{2\pi}{3}}\\ \text{so picking either root }\\ z^n + \dfrac{1}{z^n} = e^{i\frac{2n\pi}{3}}+e^{-i\frac{2n\pi}{3}} = 2 \cos\left(\dfrac{2n\pi}{3}\right) \)

\(\text{looking at }a_n = 2\cos\left(\dfrac{2n\pi}{3}\right) \text{ we find that}\\ a_n = \begin{cases} -1 &n \pmod{3} \in \{1,2\}\\ 2 &n \pmod{3} = 0 \end{cases}\)

\(\text{and thus the original sum becomes} \\ \sum \limits_{k=1}^{15}~(1 + 1 + 4) = 15 \cdot 6 = 90\)

Hint: Try to set up variables, solution will come later. 

f(-7)  =   

a(-7)^7 + b(-7)^3 + c(-7) - 5  = 7

a(-1)^7 * (7)^7   +  b (-1)^3 * (7)^3  + c (-1)(7) - 5  = 7

-a(7)^7  - b (7)^3  - c (7) - 5   =  7

- ( a(7)^7 + b (7)^3 + c(7)  )   = 12

a(7)^7 + b(7)^3 + c(7)  =  -12

So

a(7)^7 + b(7)^3 + c(7)  - 5   =  -12 - 5

f (7)   =   -17

\(x = y^4 \text{ and } x = 1-y^2\\ y^4 = 1-y^2\\ y^4 +y^2 -1 = 0\\ y^2 = \dfrac{-1\pm \sqrt{1+4}}{2} = \dfrac{-1\pm \sqrt{5}}{2}\\ y = \pm \sqrt{\dfrac{-1\pm \sqrt{5}}{2}}\)

sorting through these possible solutions we find the only two real ones are

\(y = \pm \sqrt{\dfrac{-1+\sqrt{5}}{2}}\\ \text{both of these values of }y \text{ produce the same value of x}\\ \text{and thus the distance between them is }\\ \Delta y= 2 \sqrt{\dfrac{-1+\sqrt{5}}{2}} = \sqrt{-2 + 2\sqrt{5}}\\ (u,v) = (-2,2)\)

OK....looks like neworleans wins....!!!!

Does this say \(\frac{y+z}{z}\) . If so, try to break it up into \(\frac{y}{z}+\frac{z}{z}\)\(\) , so   \(\frac{y}{z}+1.\)\(\) Therefore, we have    \(4y=3z, y=\frac{3}{4}z\) and \(\frac{3z}{4}*\frac{1}{z}=\frac{3}{4}\)\(\) , so     \(1+\frac{3}{4}=\boxed{1\frac{3}{4}}.\)

This is given by  the sum of a geometric series

sum  =  first term ( 1 - geometric ratio^n) / ( 1 - geometric ratio )

Where   

first term = 1/2

n = 5

geometric ratio = 1/2

So...

(1/2)  ( 1 - (1/2)^5) / (1 - 1/2)    = 31 / 32  of the pizza

This is just a geometric series, so the answer is:

\(\begin{align*} \frac12+\frac1{2^2}+\frac1{2^3}+\frac1{2^4}+\frac1{2^5} &= \frac{\frac12\left(1-\left(\frac12\right)^5\right)}{1-\frac12}\\ &=1-\left(\frac12\right)^5\\ &=1-\frac1{32}\\ &=\boxed{\frac{31}{32}}. \end{align*}\)

Looks like the correct answer is 105, I'll try to come up with a solution on why.

We'll go with each person gets 0-4 bracelets.

This is just the problem of 4 distinct balls into 4 distinct bins.

Each bracelet can just be tagged with with person it is allotted to i.e.

each arrangement is a 4 digit base 4 number so there are

\(n = 4^4 = 256 \text{ arrangements}\)

Yes, we have to see.

Each bracelet must be given to someone, but....I believe that the person who posted the question probably meant that each person can receive any number of bracelets including no bracelets at all

Everyone has their own take on this.....I'd like to see the "real" solution, myself.....LOL!!!

Does it say that each must be given at least one bracelet?  Not sure....

Here is my 'stab' at a solution....(I almost never get these correct BTW )

For EACH person:

4  bracelets    1 way

3  Bracelets    4 ways

2 bracelets     6 ways

1 bracelet       4 ways

0 bracelet       1 way      Total 16 ways      x    4 people = 64 possible ways    (??)  Some of these may be 'reflections' of the others....so ....    I really do not KNOW !   

Wondering if    f(x) = x          then f(27) = 27

f(f(27) ) = 27

f(f(f(27))) = 27     then k = 27 (which is odd)......    Hmmmmm

Hmm, this looks very tricky.

Since \(k\) is odd, we denote \(f(k)=k+3.\) Any odd number plus \(3\) is even, so \(f(k + 3) = \frac{k + 3}{2}\).

But, if \(\frac{k + 3}{2}\) is odd, therefore \(f \left( \frac{k + 3}{2} \right) = \frac{k + 3}{2} + 3 = 27.\) Then this leads for \(k=45\), and \(f(f(f(45))) = f(f(48)) = f(24) = 12\), so it is even. \(f \left( \frac{k + 3}{2} \right) = \frac{k + 3}{4} = 27\) leads to \(105\), so this is your answer?

Check: \(f(f(f(105))) = f(f(108)) = f(54) = 27.\)

Mmmm....I haven't seen one like this.....but.....here's my best shot....

Let us suppose that the function is

f(k)  = 3k        and let k = 1

So

f(1) =  3 (1)  = 3

f ( f(1) )  = f(3)  =   3(3) =  9

f ( f ( f ( 1 ) )  ) =   f(9)  =  3(9) =   27

So..... k =  1

Anyone else have other thoughts???

\(\dfrac{2z_1+7z_2}{2z_1-7z_2} = \\ \dfrac{2+7\frac{z_2}{z_1}}{2-7\frac{z_2}{z_1}}\\ \text{let }w = 2+ 7\frac{z_2}{z_1}\\ \dfrac{2+7\frac{z_2}{z_1}}{2-7\frac{z_2}{z_1}} = \dfrac{w}{w^*}\)

\(\text{You should know, or you can prove it as an exercise, that }\\ \forall z \in \mathbb{C},~z\neq 0,~\left|\dfrac{z}{z^*}\right| = 1 \text{ so,}\\ \left|\dfrac{2z_1+7z_2}{2z_1-7z_2}\right| = 1\)

Thank you, CPhill!

f(4)  =  -1

f ( f (4) ) = f (-1)  =  1

f ( f ( f (4) ) ) = f (1) =  1

f ( f ( f ( f (4) ) ) )  =  f(1)  =  1

f ( f ( f ( f ( f (4) ) ) ) )  =  f (1)  = 1

Correct, mathtoo...!!!!

xxxx

The number of possible divisions are :

( 6, 2, 1)

(5,  3, 1)

(4, 3,  2)

Each of these sets can be arranged in 3!  =  6 ways

So.....3 possible sets * 6 possible arrangements of each set =   18 ways