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  2. Basel problem - Wikipedia

    en.wikipedia.org/wiki/Basel_problem

    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]

  3. Josephus problem - Wikipedia

    en.wikipedia.org/wiki/Josephus_problem

    The details of the mechanism used in this feat are rather vague. According to James Dowdy and Michael Mays, [2] in 1612 Claude Gaspard Bachet de Méziriac suggested the specific mechanism of arranging the men in a circle and counting by threes to determine the order of elimination. [3]

  4. Moving sofa problem - Wikipedia

    en.wikipedia.org/wiki/Moving_sofa_problem

    In mathematics, the moving sofa problem or sofa problem is a two-dimensional idealization of real-life furniture-moving problems and asks for the rigid two-dimensional shape of the largest area that can be maneuvered through an L-shaped planar region with legs of unit width. [1] The area thus obtained is referred to as the sofa constant.

  5. Three-body problem - Wikipedia

    en.wikipedia.org/wiki/Three-body_problem

    In the special case of the circular restricted three-body problem, these solutions, viewed in a frame rotating with the primaries, become points called Lagrangian points and labeled L 1, L 2, L 3, L 4, and L 5, with L 4 and L 5 being symmetric instances of Lagrange's solution.

  6. Rhind Mathematical Papyrus - Wikipedia

    en.wikipedia.org/wiki/Rhind_Mathematical_Papyrus

    Problems 1–6 compute divisions of a certain number of loaves of bread by 10 men and record the outcome in unit fractions. Problems 7–20 show how to multiply the expressions 1 + 1/2 + 1/4 = 7/4, and 1 + 2/3 + 1/3 = 2 by different fractions. Problems 21–23 are problems in completion, which in modern notation are simply subtraction problems.

  7. Ancient Egyptian mathematics - Wikipedia

    en.wikipedia.org/wiki/Ancient_Egyptian_mathematics

    In this case 8 times 365 is 2920 and further addition of multiples of 365 would clearly give a value greater than 3200. Next it is noted that ⁠ 2 / 3 ⁠ + ⁠ 1 / 10 ⁠ + ⁠ 1 / 2190 ⁠ times 365 gives us the value of 280 we need. Hence we find that 3200 divided by 365 must equal 8 + ⁠ 2 / 3 ⁠ + ⁠ 1 / 10 ⁠ + ⁠ 1 / 2190 ⁠. [8]

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