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All the numbers below should be assumed to be positive integers. Definition. An abundant number is an integer n whose divisors add up to more than In. Definition. A perfect number is an integer n whose divisors add up to exactly In. Definition. A deficient number is an integer n whose divisors add up to less than In. Example. 12 is an abundant number, because 1 + 2 + 3+ 4 + 6+12 = 28 and 28 > 2x12. However, 14 is a deficient number, because 1 + 2 + 7 + 14 = 24, and 24 < 2 x 14. Your task is to consider the following conjectures and determine, with proofs, whether they are true or false. Conjecture 1. A number is abundant if and only if it is a multiple of 6. Conjecture 2. If n is perfect, then kn is abundant for any k in N. Conjecture 3. If p1 and p2 are primes, then p1/p2 is abundant. Conjecture 4. If n is deficient, then every divisor of n is deficient. Conjecture 5. If n and m are abundant, then n + m is abundant. Conjecture 6. If n and m are abundant, then nm is abundant. Conjecture 7. If n is abundant, then n is not of the form pm for some natural m and prime p.

i belive this site has been posted once before, never the less, i found it to be quite fun. http://tonematrix.audiotool.com/_/10g.840.g40.10g.9.1400.i0.6.o00.i0.0.1409.210.82.84.210 click on the grid and press space to clear all. the above represents a no-three-in-a-line solution to a 16x16 grid. http://tonematrix.audiotool.com/_/10.100.1000.400.20.2.800.1.40.8.g00.g0.800.80.4.g and here's a solution to the n-queens problem.

Hey all, I know I'm new here, but I'd love your opinions! If you don't mind taking three minutes to fill out this survey, I'd really appreciate it. Thanks! (And have your puzzling friends fill it out, too!) http://www.surveymonkey.com/s/MX8KHJN

Can you find Beatles song Titles below ?

The puzzle may have run its course. Probability is ratio of areas, and the triangle area is an hellacious multiple integral. The problem is known as Sylvester's four point problem. I did find the number, as RG gave it, from simulation, not calculation. Uniform point-picking, necessary to form uniformly distributed triangles was an issue. Equally likely radius and angle gives preference to central points. What was interesting to me when I simulated was that I was off by a factor of four that the four points would not be convex. That is, and should be 4 time the probability the 4th dart lands inside the triangle. There are three other areas the 4th dart can land that make a non-convex configuration, and those areas, on average, are equal to the triangle size. Not an intuitive result until you realize any of the darts can be taken to be the 4th dart. So my last question now does not need to be asked, namely, to compare the probabilities of the two cases in the OP. also, that question has been answered in this thread. Also, the mean triangle area was (to me) surprisingly small.

In the all-digital future, X and O are banished from the game of tic-tac-toe. They are replaced by 1 and 0, the the result of such a game might look like this: 1 | 0 | 1 - + - + - 0 | 1 | 1 - + - + - 0 | 1 | 0 Under the usual rules that require getting 3-in-a-row, it would be a draw. But this is the digital age, and there are different rules for winning. If we sum the eight rows of three numbers we get 2, 2, 1 (horizontally) 1, 2, 2 (vertically) and 2, 2 (diagonally). Six of the sums are even, and two are odd. The final parity of the board is thus even, and the game is said to have an even outcome. If there were more odd sums than even, the game would have an odd outcome. If there were four even (and therefore four odd) sums, the game would have a neutral outcome. The game is played as follows: The winner of a fair-coin toss (call him player A) chooses whether to play first or second. The other player (call her player B) decides whether she wants an odd, even, or neutral game outcome. On each turn, a player places his choice of either a 1 or a 0 on any unoccupied place on the grid. As in normal tic-tac-toe, players alternate turns; but here on each turn a player may play either a 0 or a 1. When the places are filled, the board is examined to determine whether it is odd, even or neutral. If the final board parity matches player B's choice, player B wins; otherwise player A wins. The questions to answer are: Is there an advantage to winning the coin toss? Is there a winning strategy for either player?