Search results
Results from the WOW.Com Content Network
For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1 ≡ 4 (mod 6). Equivalently, n − 1 / 3 ≡ 1 (mod 2) if and only if n ≡ 4 (mod 6). Conjecturally, this inverse relation forms a tree except for the 1–2–4 loop (the inverse of the 4–2–1 loop of the unaltered function f defined in the Statement of the problem section of ...
For odd square, since there are (n - 1)/2 same sided rows or columns, there are (n - 1)(n - 3)/8 pairs of such rows or columns that can be interchanged. Thus, there are 2 (n - 1)(n - 3)/8 × 2 (n - 1)(n - 3)/8 = 2 (n - 1)(n - 3)/4 equivalent magic squares obtained by combining such interchanges. Interchanging all the same sided rows flips each ...
The Basel problem is a problem in mathematical analysis with relevance to number theory, concerning an infinite sum of inverse squares. It was first posed by Pietro Mengoli in 1650 and solved by Leonhard Euler in 1734, [1] and read on 5 December 1735 in The Saint Petersburg Academy of Sciences. [2] Since the problem had withstood the attacks of ...
15 puzzle. To solve the puzzle, the numbers must be rearranged into numerical order from left to right, top to bottom. The 15 puzzle (also called Gem Puzzle, Boss Puzzle, Game of Fifteen, Mystic Square and more) is a sliding puzzle. It has 15 square tiles numbered 1 to 15 in a frame that is 4 tile positions high and 4 tile positions wide, with ...
Goldbach's conjecture is used when studying computation complexity. [36] The connection is made through the Busy Beaver function, where BB (n) is the maximum number of steps taken by any n state Turing machine that halts. There is a 27 state Turing machine that halts if and only if Goldbach's conjecture is false.
The continued fraction for 1024 ⁄ 15625 (0.065536 exactly) is [;15,3,1,6,2,1, 3]; [18] its convergent terminated after the repetend is 313 ⁄ 4776, giving us x 0 =–4776 and y 0 =313. The least value of t for which both N and F are non-negative is 2569, so
Quadratic formula. The roots of the quadratic function y = 1 2 x2 − 3x + 5 2 are the places where the graph intersects the x -axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
Solving an equation symbolically means that expressions can be used for representing the solutions. For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2 (y + 1) – 1, a true statement. It is also possible to take the ...