John Von Neumann Puzzle?
452296
The fly has only 1 minute to fly at 60 mph regardless of the back and forth before the trains collide, therefore 1 mile is the distance it will cover.
1172101
Doesn't sound like you know what ID means let alone irreducible complexity or paley's watchmaker argument to back it up at all and what the hell is your question?  Seems to just be a ignorant claim without any good supporting evidence or logic to back it up.  Keep reading and maybe you'll learn something.
755203
Dear friend,\n\nYou're proposing us the famous Basel Problem, solved by Leonhard Euler in 1735. You should try Euler's approach to the problem, you can see it in the following link:\n\nhttp://en.wikipedia.org/wiki/Basel_problem#Euler_attacks_the_problem\n\nHope I've helped on something, and wish you a nice day!
73554
ZERO (O)  NO RECTANGLES..\n\n\nCHESS BOARD HAVE ONLY SQUARES..
36132
http://www.websudoku.com/\n\nThere is a print option below the puzzle.
599042
I'll solve it for you. Send it over.\n\nstill\nshell\ngrill\ncliff\n give us a clue
7918
That would be a "random variable". So I guess other words of the puzzle might be wrong.
752655
There is so many number theory and algebra behind it but the main concept is to multiply the message matrix with a key matrix and decrypt with the inverse of the key matrix.\n\nYou can also look at Hill Cipher
315067
You seem to have pasted some sentence fragments together.  Is it a puzzle?
239035
Independence....\n\nAnd what =0\n\nBig I
997490
many, many discrete logic states...not just 0 or 1
373971
No it's not... but sometimes with computers even the impossible is possible with a certain amount of dedication!\n\nThere is a group who are interested in arcade games that have been wrestling with this very problem, and have started to come up with some solutions it would appear.\n\nLinks to their pages are below - maybe amongst their codes and notes there is something that will help you out.
689151
New puzzle: Three words.\n\nY_u ~  nee_ ~  Je_e_!
964234
Take a closer look at the ASCII set in Hexadecimal.  A is 41H and a is 61H.   It makes perfect sense to a GEEK.\n\nGood luck and Happy Computing!
973961
a) i +(1/3)(j-i)\n\n  b)  \n1-12 =>1-4 & 5-12\n\n... 1-4=>1 & 3-4\n.......3-4=>3 & 4\n\n...5-12=> 5-6 & 7-12\n......5-6=> 5 & 6\n......7-12=>7-8 & 9-12\n..........7-8=>7 & 8\n..........9-12=> 9 & 10-12\n...............10-12=>10 & 11-12\n....................11-12=>11 & 12\n\n c) 1/3 of the computations lead to a subset of the previous set\nwhich is 1/2 the size of the other subset which is lead to twice\nas often. So it averages out to the same number of\ncomputations as a conventional binary search.
978758
There's a mathematical riddle that claims to "prove" 1 = 2, but if you examine it closely enough, it turns out to be dividing by 0.  Sorry I csn't remember the riddle.
934917
C is at the end, so its just the other 8 positions to play with.\n\nCount how many permutations of those 8 there are.\n\nThen count how many there are where A and B stand side by side.\n\nThen subtract those out.\n\nEdit:  Poster below me is way off.  without any restrictions, 9 men could be arranged in over 360000 ways.   Not 81.\n\nBasically, you can take 'C' out of it.  There are 8 spots left, and 8 men.  Spot 1 can hold any of the 8.  spot 2 any of the remaining 7 and so on, giving 8X7X6X5X4X3X2 which is 8! possibilities.\n\nThat's 40320.\n\nOf those 40320, some involve A and B being side by side.  How many?  Well, they could be in slot 1&2, 2& 3 .... up to 7&8.  And they could be A-B or B-A.  That's 14 possibles.  For each of those 14, the other 6 spots could be filled 6! ways.\n\nSo there are 14 X 6! = 14 X 720 = 10800 combos with C in the 9 spot and A & B side by side.\n\nTherefor, of the 40320 with C in the 9 spot, 40320-10800 = 30240 possible arrangements have C at the end, but A and B NOT side by side.