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To test for divisibility by D, where D ends in 1, 3, 7, or 9, the following method can be used. [12] Find any multiple of D ending in 9. (If D ends respectively in 1, 3, 7, or 9, then multiply by 9, 3, 7, or 1.) Then add 1 and divide by 10, denoting the result as m. Then a number N = 10t + q is divisible by D if and only if mq + t is divisible ...
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 ...
Proof of Bertrand's postulate; Proof that the sum of the reciprocals of the primes diverges; Cramér's conjecture; Riemann hypothesis. Critical line theorem; Hilbert–Pólya conjecture; Generalized Riemann hypothesis; Mertens function, Mertens conjecture, Meissel–Mertens constant; De Bruijn–Newman constant; Dirichlet character; Dirichlet L ...
These tests typically require factorization of n + 1, n − 1, or a similar quantity, which means that they are not useful for general-purpose primality testing, but they are often quite powerful when the tested number n is known to have a special form. The Lucas test relies on the fact that the multiplicative order of a number a modulo n is n ...
6: an even number that passes the divisibility test for 3. 7: sum of all the digits is a multiple of 7. 5: successive subtraction of final two digits from all the other digits yields a multiple of 5. 12: an even number that passes the divisibility test for 5. Base 11 (a prime base, for comparison): 2: sum of all the digits is a multiple of 2.
If p is an odd prime and p − 1 = 2 s d with s > 0 and d odd > 0, then for every a coprime to p, either a d ≡ 1 (mod p) or there exists r such that 0 ≤ r < s and a 2 r d ≡ −1 (mod p). This result may be deduced from Fermat's little theorem by the fact that, if p is an odd prime, then the integers modulo p form a finite field , in which ...
There are infinitely many Fermat pseudoprimes to any given basis a > 1. [1]: Theorem 1 Even worse, there are infinitely many Carmichael numbers. [2] These are numbers for which all values of with (,) = are Fermat liars. For these numbers, repeated application of the Fermat primality test performs the same as a simple random search for factors.
Digit sums and digital roots can be used for quick divisibility tests: a natural number is divisible by 3 or 9 if and only if its digit sum (or digital root) is divisible by 3 or 9, respectively. For divisibility by 9, this test is called the rule of nines and is the basis of the casting out nines technique for checking calculations.