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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 ...
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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 ...
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
The non-negative integers partially ordered by divisibility. The division lattice is an infinite complete bounded distributive lattice whose elements are the natural numbers ordered by divisibility. Its least element is 1, which divides all natural numbers, while its greatest element is 0, which is divisible by all natural numbers.
The set {,,,,,} partially ordered by divisibility is not a lattice. Every pair of elements has an upper bound and a lower bound, but the pair 2, 3 has three upper bounds, namely 12, 18, and 36, none of which is the least of those three under divisibility (12 and 18 do not divide each other).
In fact, there is a unique largest divisible subgroup of any group, and this divisible subgroup is a direct summand. [8] This is a special feature of hereditary rings like the integers Z : the direct sum of injective modules is injective because the ring is Noetherian , and the quotients of injectives are injective because the ring is ...
If we order the integers in the interval [1, 2n] by divisibility, the subinterval [n + 1, 2n] forms an antichain with cardinality n. A partition of this partial order into n chains is easy to achieve: for each odd integer m in [1,2 n ], form a chain of the numbers of the form m 2 i .