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Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if ...
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
For example, for division by 3, the factors 1/3, 2/6, 3/9, or 194/582 could be used. Consequently, if Y were a power of two the division step would reduce to a fast right bit shift. The effect of calculating N/D as (N·X)/Y replaces a division with a multiply and a shift. Note that the parentheses are important, as N·(X/Y) will evaluate to zero.
For example, there are six divisors of 4; they are 1, 2, 4, −1, −2, and −4, but only the positive ones (1, 2, and 4) would usually be mentioned. 1 and −1 divide (are divisors of) every integer. Every integer (and its negation) is a divisor of itself. Integers divisible by 2 are called even, and integers not divisible by 2 are called odd.
As an illustration of this, the parity cycle (1 1 0 0 1 1 0 0) and its sub-cycle (1 1 0 0) are associated to the same fraction 5 / 7 when reduced to lowest terms. In this context, assuming the validity of the Collatz conjecture implies that (1 0) and (0 1) are the only parity cycles generated by positive whole numbers (1 and 2 ...
The simplest way of viewing division is in terms of quotition and partition: from the quotition perspective, 20 / 5 means the number of 5s that must be added to get 20. In terms of partition, 20 / 5 means the size of each of 5 parts into which a set of size 20 is divided.
For instance, 6 has proper divisors 1, 2 and 3, and 1 + 2 + 3 = 6, so 6 is a perfect number. The next perfect number is 28, since 1 + 2 + 4 + 7 + 14 = 28. The first four perfect numbers are 6 , 28 , 496 and 8128 .
The number of points (n), chords (c) and regions (r G) for first 6 terms of Moser's circle problem. In geometry, the problem of dividing a circle into areas by means of an inscribed polygon with n sides in such a way as to maximise the number of areas created by the edges and diagonals, sometimes called Moser's circle problem (named after Leo Moser), has a solution by an inductive method.