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  2. Division algorithm - Wikipedia

    en.wikipedia.org/wiki/Division_algorithm

    Long division is the standard algorithm used for pen-and-paper division of multi-digit numbers expressed in decimal notation. It shifts gradually from the left to the right end of the dividend, subtracting the largest possible multiple of the divisor (at the digit level) at each stage; the multiples then become the digits of the quotient, and the final difference is then the remainder.

  3. Square number - Wikipedia

    en.wikipedia.org/wiki/Square_number

    Squares of odd numbers are odd, and are congruent to 1 modulo 8, since (2n + 1) 2 = 4n(n + 1) + 1, and n(n + 1) is always even. In other words, all odd square numbers have a remainder of 1 when divided by 8. Every odd perfect square is a centered octagonal number. The difference between any two odd perfect squares is a multiple of 8.

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  5. Division (mathematics) - Wikipedia

    en.wikipedia.org/wiki/Division_(mathematics)

    For example, 20 apples divide into five groups of four apples, meaning that "twenty divided by five is equal to four". This is denoted as 20 / 5 = 4, or ⁠ 20 / 5 ⁠ = 4. [2] In the example, 20 is the dividend, 5 is the divisor, and 4 is the quotient.

  6. Long division - Wikipedia

    en.wikipedia.org/wiki/Long_division

    Next, the 1 is multiplied by the divisor 4, to obtain the largest whole number that is a multiple of the divisor 4 without exceeding the 5 (4 in this case). This 4 is then placed under and subtracted from the 5 to get the remainder, 1, which is placed under the 4 under the 5.

  7. Mental calculation - Wikipedia

    en.wikipedia.org/wiki/Mental_calculation

    Note that if n 2 is the closest perfect square to the desired square x and d = x - n 2 is their difference, it is more convenient to express this approximation in the form of mixed fraction as . Thus, in the previous example, the square root of 15 is 41 8 . {\displaystyle 4{\tfrac {-1}{8}}.}

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  9. Polynomial long division - Wikipedia

    en.wikipedia.org/wiki/Polynomial_long_division

    x 3 has been divided leaving no remainder, and can therefore be marked as used by crossing it out. The result x 2 is then multiplied by the second term in the divisor −3 = −3x 2. Determine the partial remainder by subtracting −2x 2 − (−3x 2) = x 2. Mark −2x 2 as used and place the new remainder x 2 above it.