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To test the divisibility of a number by a power of 2 or a power of 5 (2 n or 5 n, in which n is a positive integer), ... 1050 → 0501 (reverse) → 0×1 + 5×3 + 0 ...
Given an integer n (n refers to "the integer to be factored"), the trial division consists of systematically testing whether n is divisible by any smaller number. Clearly, it is only worthwhile to test candidate factors less than n, and in order from two upwards because an arbitrary n is more likely to be divisible by two than by three, and so on.
Since 2 divides , +, and +, and 3 divides and +, the only possible remainders mod 6 for a prime greater than 3 are 1 and 5. So, a more efficient primality test for is to test whether is divisible by 2 or 3, then to check through all numbers of the form + and + which are .
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.
We'll arbitrarily try the value B=7, giving the factor base P = {2,3,5,7}. The first step is to test n for divisibility by each of the members of P; clearly if n is divisible by one of these primes, then we are finished already. However, 187 is not divisible by 2, 3, 5, or 7. Next, we search for suitable values of z; the first few are 2, 5, 9 ...
Dimensional analysis may be used as a sanity check of physical equations: the two sides of any equation must be commensurable or have the same dimensions. A person who has calculated the power output of a car to be 700 kJ may have omitted a factor, since the unit joules is a measure of energy, not power (energy per unit time).
This is the sieve's key distinction from using trial division to sequentially test each candidate number for divisibility by each prime. [2] Once all the multiples of each discovered prime have been marked as composites, the remaining unmarked numbers are primes.
If N = F n > 3, then the above Jacobi symbol is always equal to −1 for a = 3, and this special case of Proth's theorem is known as Pépin's test. Although Pépin's test and Proth's theorem have been implemented on computers to prove the compositeness of some Fermat numbers, neither test gives a specific nontrivial factor.