Rom

That's some keen insight there Melody! 

\(f'(x) = -3 - 20x^4 - 15x^2\\ \text{this should pretty clearly be less than zero for all }x\)

The vertices sum to 84 for every possible ordering of the face numbers.

\(\text{there are only so many pairs of integers that satisfy the second equation}\\ \text{They are }(\pm 1, \pm 7),(\pm 7,\pm 1),(\pm 5,\pm 5)\\ \text{each of these produces 4 ordered pairs }(a,b) \text{ thus there are 12 total} \)

\(\text{consider the polynomial } \\ q(x) = p(x) - (d-1-x)\\ q(x) = 0,~\forall x = k=0,1,\dots ,(d-1)\\ q(x) = \prod \limits_{k=0}^{d-1}~(x-(d-1-k))\\ \text{now let }d=2008 \\ q(x) = \prod \limits_{k=0}^{2007}~(x-2007+k)\\ p(2008) = q(2008) +(2007-2008) = q(2008)-1\\ q(2008) = \prod \limits_{k=0}^{2007}~k+1 = 2008!\\ p(2008)=2008!-1\)

Many thanks to Walagaster over at MHF for the idea of q(x)

The answer is 2008!-1

I don't know why yet.

I'll assume the sine wave is 0 at t=0, i.e. zero phase shift.

\(A \sin(2\pi (700)(25\times 10^{-6}))=53.3V\\ A \sin\left(\dfrac{7\pi}{200}\right) = 53.3V\\ A = \dfrac{53.3}{\sin\left(\dfrac{7\pi}{200}\right)} =485.719V\)

\(A \sin(2\pi (700\times (50\times 10^{-6})) = 105.956V\)

\(p(x)=(x-1)(x-2)(x-3)\\ r(x) = f(x) - p(x)q(x),~\text{ where }q(x) \text{ is the quotient of }f(x) \text{ and }p(x)\\ \text{note that }p(1)=p(2)=p(3) = 0\\ r(1) = f(1) - p(1)q(1) = f(1) = 2\\ r(2) = f(2) = 3\\ r(3) = f(3) = 5\)

\(\text{we have 3 equations so we are looking for a polynomial with 3 degrees of freedom}\\ \text{i.e. a 2nd degree polynomial }a x^2 + b x+ c\\ \text{plugging things in we get 3 equations}\\ a+b+c = 2\\ 4a+2b+c=3\\ 9a+3b+c=5\\ \text{This can be solved to obtain }\\ a=\frac{1}{2},b= -\frac{1}{2},c=2, \text{ and thus the remainder polynomial is}\\ r(x) = \dfrac 1 2(x^2 - x + 4)\)

\(450 \cdot (450)^{-1} = 1 \pmod{3599}\\ \text{something to immediately check is does 450 divide 3600?}\\ 3600=450\cdot 8 \\ 8\cdot 450 = 1 \pmod{3599}\\ 8 = (450)^{-1} \pmod{3599}\)

not sure where the triangle fits into things

xxxxxxxx