phil1882

if n is composite, break it into its prime factors, if n is prime, place < > around n and replace n with the nth prime its is. the first 20 numbers then are... 1 <> 2 <<>> 3 <><> 4 <<<>>> 5 <><<>> 6 <<><>> 7 <><><> 8 <<>><<>> 9 <><<<>>> 10 <<<<>>>> 11 <><><<>> 12 <<><<>>> 13 <><<><>>14 <<>><<<>>> 15 <><><><> 16 <<<><>>> 17 <><<>><<>> 18 <<><><>> 19 <><><<<>>> 20 hypothesis: 1) there is no general method for adding recursive numbers. 2) numbers that differ by 1 wont differ in recursive representation by more than 2 brackets 3) symmetric recursive numbers, recursive numbers that can be represented in such a way that they are a mirror image at the middle, grow at a logarithmic rate somewhat similar to the primes. can you confirm or disprove any of these?

it seems to me what you really need is an encryption method that can be reversed if necessary but would take longer to reverse than to solve the sudoku yourself and get the same encryption. what do you think of this idea? it may not even require a computer, depending on how difficult the encryption method is.

doorenownly is the best i could com up with.

i'd say its

assuming you also had scissors...

no clue on this one, you can divide a single line segment to any measurement, but showing any point m to construct a triangle of any parameter? much much harder i think.

alright sorry, i tried to make it such that there was only 1 solution but looks like i failed. thanks for finding all 3.

yeap, or more specifically a power of 3.

i'm not going to list how to make each weight, however

this solution doesn't work, as tim and fred both end in 2. the numbers 1-5 should appear once in each digit place between men, as hint #2 tries to explain.

Five men, one of which is Greg each selected 5 unique 3 digit numbers from a hat. given the clues, can you tell what the numbers are and the men who selected them? 1. Each digit for the number chosen by each man is unique. (that is, if the number begins with 1, 1 wont appear anywhere else in the number.) 2. Each digit for each place between men is unique. (that is, if a number begins with 1, no other number begins with 1.) 3. Only the digits 1-5 are used. 4. Andy has the smallest number, and it's evenly divisible by 5. 5. Both Bob's and Fred's numbers are evenly divisible by 3. 6. Bob's number is smaller than Fred's. 7. Tim's number is evenly divisible by 2, and a prime. 8. Tim's number is greater than Bob's. 9. Fred's middle most number is prime. 10. Bob's middle and last number are odd. 11. the middle digit of Tim's number is twice as big as Andy's. good luck and enjoy!

i was thinking it would be false if both were born on he same day, but even here it would be true, because they can't be born at exactly the same time.

1. if all borks are grogs, and all grogs are snorts, are all snorts borks? 2. prove that GCD(A,B) = GCD(A+B,A) = GCD(A+B,B) 3. how much water, approximately, would you need to cover all land on earth in 3 feet of snow? 4. whats the shortest distance between any two points on a globe? A) straight line B) curved line 5. and just to be silly: what's the speed of dark?

if you really wanna see something bizarre, the Riemann zeta function tells us that the sequence 1 +2 +3 +4 +5 .... = -1/12.

the equation 1/1^2 +1/2^2 +1/3^2 ... is well known to be pi^2/6, and is equivalent to 1/(1-1/2^2) *1/(1-1/3^2) *1/(1-1/5^2) ...

i'd take the bet.

i had an idea of looking at the last three digits of powers of 5 and 3, but for 3 its not very good, its quite a long list. i'd say this problem is close to intractable. or at least i dont see any good non mass calcuation solution to it for large powers.

i have no idea how you could solve this without computing at least a few decimal points of sqrt(5). for example one possiblility might be multipying by 3 -sqrt(5), but then i would have to divide by this number at the end of the problem. the fastest way i know of, would be to compute about 1000 decimal points of 3+sqrt(5), and then use the fact that a^n * a^n = a^(2*n). ie something like v = 3 +sqrt(5) while n != 0: if n&1 == 0: n -= 1 v *= (3+sqrt(5)) else: n >>= 1 v *= v print v

it was a goood run y-san hosted some fun ones. i'm not particuarly skilled at hosting forum games but if we had more poeple i probably could.

9*x*(sin x) +4/(x*sin x) if x = pi/2: 16.68 if x = pi/4: 12.2 if x = pi/8: 27.97 if x = 3*pi/8: 13.47 if x = 3*pi/16: 15.17 so it looks like its pi/4.

in a 20x25 rectangle, 500 unit squares can be fitted. spacing the 120 squares out such that slightly less than a unit square is between them, gives roughly 240 area that the circle cannot be. rotating the squares in an alternating pattern of diamond and square gives roughly 324 area that the circle cannot enter. moving the squares out of sync such that they are in a somewhat hexagonal pattern maximizes the area that the circle cannot be, giving roughy 411 area the circle connot enter. this still leaves plenty of space for the circle, even in the worst case.