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The GCD is a multiplicative function in the following sense: if a 1 and a 2 are relatively prime, then gcd(a 1 ⋅a 2, b) = gcd(a 1, b)⋅gcd(a 2, b). gcd( a , b ) is closely related to the least common multiple lcm( a , b ) : we have
Since b ≥ φ N−1, then N − 1 ≤ log φ b. Since log 10 φ > 1/5, (N − 1)/5 < log 10 φ log φ b = log 10 b. Thus, N ≤ 5 log 10 b. Thus, the Euclidean algorithm always needs less than O divisions, where h is the number of digits in the smaller number b.
For example, if you had two types of coins valued at 6 cents and 14 cents, the GCD would equal 2, and there would be no way to combine any number of such coins to produce a sum which was an odd number; additionally, even numbers 2, 4, 8, 10, 16 and 22 (less than m=24) could not be formed, either.
The assumption is that the loop must be normalized – written so that the loop index/variable starts at 1 and gets incremented by 1 in every iteration. For example, in the following loop, a=2, b=3, c=2, d=0 and GCD(a,c)=2 and (d-b) is -3. Since 2 does not divide -3, no dependence is possible.
In other words, it is the number of integers k in the range 1 ≤ k ≤ n for which the greatest common divisor gcd(n, k) is equal to 1. [2] [3] The integers k of this form are sometimes referred to as totatives of n. For example, the totatives of n = 9 are the six numbers 1, 2, 4, 5, 7 and 8.
For every pair of elements x, y of a GCD domain R, a GCD d of x and y and an LCM m of x and y can be chosen such that dm = xy, or stated differently, if x and y are nonzero elements and d is any GCD d of x and y, then xy/d is an LCM of x and y, and vice versa.
Therefore, equalities like d = gcd(p, q) or gcd(p, q) = gcd(r, s) are common abuses of notation which should be read "d is a GCD of p and q" and "p and q have the same set of GCDs as r and s". In particular, gcd( p , q ) = 1 means that the invertible constants are the only common divisors.
The set {3,19} generates the group, which means that every element of (/) is of the form 3 a × 19 b (where a is 0, 1, 2, or 3, because the element 3 has order 4, and similarly b is 0 or 1, because the element 19 has order 2).
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