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The greatest common divisor (GCD) of integers a and b, at least one of which is nonzero, is the greatest positive integer d such that d is a divisor of both a and b; that is, there are integers e and f such that a = de and b = df, and d is the largest such integer.
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Visualisation of using the binary GCD algorithm to find the greatest common divisor (GCD) of 36 and 24. Thus, the GCD is 2 2 × 3 = 12.. The binary GCD algorithm, also known as Stein's algorithm or the binary Euclidean algorithm, [1] [2] is an algorithm that computes the greatest common divisor (GCD) of two nonnegative integers.
If gcd(a, b) = 1, then a and b are said to be coprime (or relatively prime). [4] This property does not imply that a or b are themselves prime numbers. [5] For example, 6 and 35 factor as 6 = 2 × 3 and 35 = 5 × 7, so they are not prime, but their prime factors are different, so 6 and 35 are coprime, with no common factors other than 1.
For example, the integers 4, 5, 6 are (setwise) coprime (because the only positive integer dividing all of them is 1), but they are not pairwise coprime (because gcd(4, 6) = 2). The concept of pairwise coprimality is important as a hypothesis in many results in number theory, such as the Chinese remainder theorem .
The following table shows how the extended Euclidean algorithm proceeds with input 240 and 46. The greatest common divisor is the last non zero entry, 2 in the column "remainder". The computation stops at row 6, because the remainder in it is 0. Bézout coefficients appear in the last two columns of the second-to-last row.
The first in decimal: 4, 6, 8, 9, 12, 18, 20, 22, 24, 26, 28, 30 (sequence A046760 in the OEIS). An economical number has been defined as a frugal number, but also as a number that is either frugal or equidigital.
For example, 1 / 4 , 5 / 6 , and −101 / 100 are all irreducible fractions. On the other hand, 2 / 4 is reducible since it is equal in value to 1 / 2 , and the numerator of 1 / 2 is less than the numerator of 2 / 4 . A fraction that is reducible can be reduced by dividing both the numerator ...