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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 ...
where f (2k−1) is the (2k − 1)th derivative of f and B 2k is the (2k)th Bernoulli number: B 2 = 1 / 6 , B 4 = − + 1 / 30 , and so on. Setting f ( x ) = x , the first derivative of f is 1, and every other term vanishes, so [ 15 ]
1.4 m – length of a Peel P50, the world's smallest car; 1.435 m – standard gauge of railway track used by about 60% of railways in the world = 4 ft 8 1 ⁄ 2 in; 2.5 m – distance from the floor to the ceiling in an average residential house [118] 2.7 m – length of the Starr Bumble Bee II, the smallest plane
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. Unlike the other basic operations, when dividing natural numbers there is sometimes a remainder that will not go evenly into the dividend; for example, 10 / 3 leaves a remainder of 1, as 10 is not a multiple of 3.
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.
d() is the number of positive divisors of n, including 1 and n itself; σ() is the sum of the positive divisors of n, including 1 and n itselfs() is the sum of the proper divisors of n, including 1 but not n itself; that is, s(n) = σ(n) − n
Division can be performed as in the following example: To divide 6 by 2—that is, to find out how many times 2 goes into 6—note that the length from 0 to 2 lies at the beginning of the length from 0 to 6; pick up the former length and put it down again to the right of its original position, with the end formerly at 0 now placed at 2, and ...
Today, a more standard phrasing of Archimedes' proposition is that the partial sums of the series 1 + 1 / 4 + 1 / 16 + ⋯ are: + + + + = +. This form can be proved by multiplying both sides by 1 − 1 / 4 and observing that all but the first and the last of the terms on the left-hand side of the equation cancel in pairs.