Math Is Fun Forum

P(n+1) is (n^3 + 3n^2 + 3n + 1)%6 = (n+1)%6 
and assume we start with n^3%6 = n%6, where % means mod (like in C language).
(n^3 + 3n^2 + 2n)%6 + (n+1)%6 = (n+1)%6
(n^3 + 3n^2 + 2n)%6 = 0
Substitute n%6 in place of n^3%6 and get:
(3n(n+1))%6=0
For even numbers, 3n is a multiple of 6, so that works.
For odd numbers, the (n+1) part is even, so that works.

Another example is 9.5 in base-2, should it go to 10 or 9?
1001.1
1010 or
1001
Note that now you have only three values in a row:    #.0 (whole number)       #.1 (half)       #.0 (whole number)

I think that .50 should not be rounded in either direction because 1/2 is exactly 1/2 away from the adjacent whole numbers.
0       1234      5     6789        0

I've decided to give one more hint... 
If the measurement from a pointy corner to the center is one meter,
then how tall is the tetrahedron?

Google "monty hall problem".  It might give you some insight into this.  At least that's my guess.
Also, if I was to hasard a guess, I would guess .6 / (.55 + .6) which is 0.5217 which is not your answer, oh well.

Those are two very nice calculators!  I'm impressed.  I started writing my large digit calculation program yesterday, so I'll continue with that.  Plus, I'm not very good at examining programs written by others, and I like writing things from scratch.

The following web page says you can use any value of x in the cosine summation and the sine too.
Your recommendation is a good one to get the angle within the four quadrants.  You might even
be more accurate if you stay with the first quadrant and alter the signs as needed.

http://euclideanspace.com/maths/algebra … /index.htm

Compare 3p and 4p.   Okay.  Then compare 2q and -6q.  Oh, 2 times 3 is 6, so that will be the easy way to solve.
We will get rid of the q and find p.   Multiply 3p +2q = 9 by 3 and add it to the other equation.
So 9p + 6q = 27.   Add it to second equation.  13p + zero q = 52
p = 4.  Put 4 back into any equation and get q.  q = -3/2 

Another way is to graph the two lines on graph paper and see where they intersect.

I wrote this in Just BASIC v1.0.
It computes the cosine of angles 0, 5, 10, 15, 20, 25, 30, ... to 90 degrees.
The computation of cosine of 90 degrees comes out about 10^-15, not exactly zero though.

'Note: dangle is angle in degrees and rangle is angle in radians
for dangle = 0 to 90 step 5
pi = 3.141592653589793238462643383279502884197169399375105820974944
rangle=(pi*dangle)/180
'Note:  cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...
rangleSqrd=rangle * rangle
diver = 0
nextTerm = 1
cosineNow = 1

for iter=1 to 10
for d1 = 1 to 2
  diver = diver + 1
  nextTerm = nextTerm / diver
next d1
nextTerm = 0 - nextTerm
nextTerm = nextTerm * rangleSqrd
cosineNow = cosineNow + nextTerm
next iter
print "Cosine( ";dangle;" ) = ";cosineNow
next dangle