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It is divisible by 2 and by 11. [6] 352: it is divisible by 2 and by 11. 23: Add 7 times the last digit to the rest. (Works because 69 is divisible by 23.) 3,128: 312 + 8 × 7 = 368: 36 + 8 × 7 = 92. Add 3 times the last two digits to the rest. (Works because 299 is divisible by 23.) 1,725: 17 + 25 × 3 = 92. Subtract 16 times the last digit ...
Furthermore, if b 1, b 2 are both coprime with a, then so is their product b 1 b 2 (i.e., modulo a it is a product of invertible elements, and therefore invertible); [6] this also follows from the first point by Euclid's lemma, which states that if a prime number p divides a product bc, then p divides at least one of the factors b, c.
Two properties of 1001 are the basis of a divisibility test for 7, 11 and 13. The method is along the same lines as the divisibility rule for 11 using the property 10 ≡ -1 (mod 11). The two properties of 1001 are 1001 = 7 × 11 × 13 in prime factors 10 3 ≡ -1 (mod 1001) The method simultaneously tests for divisibility by any of the factors ...
The first condition is the Fermat primality test using base 2. In general, if p ≡ a (mod x 2 +4), where a is a quadratic non-residue (mod x 2 +4) then p should be prime if the following conditions hold: 2 p−1 ≡ 1 (mod p), f(1) p+1 ≡ 0 (mod p), f(x) k is the k-th Fibonacci polynomial at x.
A strong divisibility sequence is an integer sequence () such that for all positive integers m, n, gcd ( a m , a n ) = a gcd ( m , n ) . {\displaystyle \gcd(a_{m},a_{n})=a_{\gcd(m,n)}.} Every strong divisibility sequence is a divisibility sequence: gcd ( m , n ) = m {\displaystyle \gcd(m,n)=m} if and only if m ∣ n {\displaystyle m\mid n} .
Using fast algorithms for modular exponentiation and multiprecision multiplication, the running time of this algorithm is O(k log 2 n log log n) = Õ(k log 2 n), where k is the number of times we test a random a, and n is the value we want to test for primality; see Miller–Rabin primality test for details.
In any case, proving that Pomerance's counterexample does not exist is far from proving Carmichael's conjecture. However if it exists then infinitely many counterexamples exist as asserted by Ford. Another way of stating Carmichael's conjecture is that, if A ( f ) denotes the number of positive integers n for which φ ( n ) = f , then A ( f ...
This completes the proof that = is prime. The certificate of primality for N = 27457 {\displaystyle N=27457} would consist of the two ( p , a p ) {\displaystyle (p,a_{p})} pairs (2, 5) and (3, 2). We have chosen small numbers for this example, but in practice when we start factoring A we may get factors that are themselves so large their ...