Bertie

As yet I don't know how to prove this, but a formula that seems to work is

$$\displaystyle p=2\left(n-2^{\bmod\left(\frac{(\log(n)-0.001)}{\log(2)}\right)\right)-1.$$

The logs are to base 10 and mod (I hope, otherwise I need to find another operator,) delivers the integer part of the expression within the brackets.

p is measured from the first person to be eliminated, being the number of positions further on around the circle.

For example n = 6 will return a value of p = 3. So, from the first person eliminated count to the third position further round the circle.

Start by sending a photographer and a cannibal across with the photographer returning.

That produces

CCPPP.............................C     (where the dots indicate the river)

Now two cannibals across (PPP............CCC) with one returning,

CPPP...............................CC

Next, two photographers across (CP....................PPCC) and a photographer and a cannibal returning

CCPP...............................CP

Next, two photographers across (CC...................CPPP) with a cannibal returning

CCC...............................PPP

Now two cannibals over (C.............CCPPP)

with one returning to collect the last cannibal.

The same sort of routine can be achieved if two cannibals go over to begin with.

The graphical method is fine if you have access to a suitable plotter, if you haven't, having only a pocket calculator, your stuck.

Here's the mathematicians method of solution.

Begin with a substitution using the trig identity

$$\cos2A=2\cos^{2}A-1$$ ,   from which    $$2\cos^{2}A=\cos2A+1.$$

(in order to get the angles on both sides the same).

That gets you

$$3\sin2x= \cos2x+2,$$

which is rewritten as

$$3\sin2x-\cos2x = 2.$$

Now consider the trig identity

$$R\sin(2x-\alpha)=R\sin2x\cos\alpha-R\cos2x\sin\alpha.$$

Comparing with the LHS of the equation,

$$R\cos\alpha=3, \text{ and }R\sin\alpha=1,$$

so, squaring and adding, $$R^{2}=3^{2}+1^{2}=10, \text{ so }R=\sqrt{10},$$

and dividing, $$\tan\alpha=1/3.$$

That gets us

$$\sqrt{10}\sin(2x-\alpha)=2,$$

$$\sin(2x-\alpha)=2/\sqrt{10},\quad 2x-\alpha=\sin^{-1}(2/\sqrt{10}),$$

$$2x-\alpha = 39.23,\quad 140.77,\quad 399.23,\quad 500.77, \dots \text{ deg},$$

and with alpha = 18.43 deg, that leads to (0 - 360 deg, 2dp),

$$x = 28.83, \quad 79.60, \quad 208.83, \quad 259.60, \dots \text{ deg.}$$

The sum of the individual digits of PACIFIC is 45.

Alan is much too polite.

Consider the number

$$\displaystyle N=30.29.28.(36.27!+25)=36.30!+30.29.28.25$$

Working mod 31,

$$\displaystyle 36\equiv5,\quad \text{and}\quad 30!\equiv -1,$$  

( the second of those being Wilson's theorem with p = 31, )

so, multiplying,

$$\displaystyle 36.30! \equiv 5.(-1)=-5.\quad ...........(1)$$

Also,

$$30 \equiv (-1), \: 29 \equiv (-2), \: 28\equiv (-3)\: \text{and}\: 25\equiv (-6),$$

so, again multiplying,

$$\displaystyle 30.29.28.25\equiv (-1).(-2).(-3).(-6)=36\equiv 5.\quad ...............(2)$$

Adding (1) and (2),

$$\displaystyle N\equiv -5+5\equiv 0,$$

which says that N is divisible by 31.

Looking back at the definition of N, since 30.29.28 is not divisible by 31 (which is prime), it follows that the expression within the brackets must be divisible by 31.

Unless we are told something to the contrary, unknowns in equations such as this are always assumed to be integers.

If  d is the greatest common divisor (gcd) of a and n, the linear congruence  

$$\displaystyle ax\equiv b \bmod n$$  

has a solution only if b is also divisible by d and in that case there will be an infinite number of solutions given by

$$\displaystyle x=x_{0}+\frac{nt}{d},$$

where   $$x_{0}$$   is any solution whatever, and t is an integer, (positive or negative), or zero.

For the equation in question,

$$\displaystyle 7x = 1\bmod 26,\qquad d = 1,$$

so the general solution will be

$$\displaystyle x = x_{0}+\frac{26t}{1} = x_{0}+26t.$$

Finding a suitable   $$x_{0}$$   is a trial and error thing, run through the multiples of 7 and 26 looking for the first pair for which the multiple of 7 is 1 greater than the multiple of 26 That happens to be 15 and 4, so the general solution is

$$\displaystyle x = 15 + 26t, \quad t\:\in \:Z.$$  

The result given earlier comes from the theory of linear Diophantine equations.

If

$$\displaystyle d = \gcd(a,b),$$

the equation

$$\displaystyle ax+by = c$$

has integer solutions only if c is also divisible by d, and in that case there will be an infinite number of solutions given by

$$\displaystyle x = x_{0}+\frac{bt}{d},\quad y=y_{0}-\frac{at}{d} \quad (t\: \in Z),$$

and where  $$\displaystyle \; x_{0},y_{0}$$   is any particular solution.

(For large values of a and b, $$\displaystyle \; x_{0},y_{0}$$   and d are usually found using Euclid's algorithm.)

( The equation

$$\displaystyle ax\equiv b \bmod n$$

is equivalent to

$$\displaystyle ax-b=kn \Rightarrow ax-kn=b,$$

which, with a change of letters and a change of sign, is the Diophantine equation above.)

The best method for dealing with powers and roots of complex numbers, is to switch them to polar form and then to use De Moivre's theorem.

So here,

$$\displaystyle \frac{-7+21\sqrt{3}\imath}{2}=-\frac{7}{2}(1-3\sqrt{3}\imath)=-\frac{7}{2}\sqrt{28}\angle\tan^{-1}(-3\sqrt{3})$$

(ask for details if you need them)

$$\displaystyle = -7\sqrt{7}\angle 280.89339\deg.$$

Raise that to the power one third and you have,

$$(-7\sqrt{7}\angle(-280.89339+360k))^{1/3}$$

$$\displaystyle = -\sqrt{7}\angle(93.63113+120k), \quad k=0,1,2.$$

That gets you

$$\displaystyle k=0:\quad -\sqrt{7}\angle 93.63113 = -\sqrt{7}(\cos93.63113+\imath\sin93.63113)$$

$$\displaystyle =0.16752-2.64044\imath.$$

and the other two values of k,

$$\displaystyle 2.20291+1.46533\imath \;\text{ and } -2.37047+1.17511\imath \quad\text{respectively}.$$

Thanks for the cake Melody, it looks delicious.

I'm not going to ask you how many candles were on it.

I feel a bit of a fraud coming in on this one after the introductory work done by the three of you, providing the ideas and so on, however ,

We need

$$\displaystyle \frac{x}{x^{2}+x+1}\quad \text{to equal} \quad 2.$$

Turn that upside down, as suggested by Chris, and we have

$$\displaystyle x+1+\frac{1}{x}=\frac{1}{2}\quad \dots(1).$$

Now look at the rhs, suppose that's equal to $$k$$ .

Turn that upside down and we have

$$\displaystyle x^{2}+1+\frac{1}{x^{2}}=\frac{1}{k}.$$

Squaring (1),

$$\displaystyle x^{2}+1+\frac{1}{x^{2}}+2x+2+\frac{2}{x}=\frac{1}{4},$$

so

$$\displaystyle x^{2}+1+\frac{1}{x^{2}}= \frac{1}{4}-2\left(x+1+\frac{1}{x}\right)=\frac{1}{4}-\left(2\times\frac{1}{2}\right)=-\frac{3}{4},$$

in which case

$$\displaystyle \frac{1}{k}=-\frac{3}{4},\quad \text{so} \qquad k=-\frac{4}{3}.$$

A grape en route to a better existence ?