--%>

Explain Factorisation by trial division

Factorisation by trial division: The essential idea of factorisation by trial division is straightforward. Let n be a positive integer. We know that n is either prime or has a prime divisor less than or equal to √n. Therefore, if we divide n in turn by the primes 2, 3, 5,..., going possibly as far as [√n], we will either encounter a prime factor of n or otherwise be able to infer that n is prime. Repeating this process as often as necessary we will be able to nd all the prime
factors of n.

We can re fine this idea a little. If we fi nd on division by the prime p that p is a factor of n, then we can recommence trial divisions, but now dividing into the integer n/p rather than n. Also, the divisions can start with the prime p rather than restarting with 2, since we know that n, and hence n/p, has no prime factors smaller than p.

Further, we now need only carry out trial divisions by primes up to [√n/p]. Similarly for later steps.

An obvious difficulty with trial division is that we need either to store or to generate the primes up to [√n], and it may be better simply to divide by all the integers from 2 up to [√n], or, for example, by 2 and then all the odd numbers up to [√n].

Other improvements are possible too, but even with a few improvements the trial division algorithm is inefficient , and the algorithm is unsuitable for all but fairly small initial values of n.

Despite this, the trial division algorithm is in practical use. It is often used as a preliminary phase in a factorisation algorithm to nd the `small' prime factors of a number, and the remaining unfactorised part, containing all the `large' prime factors, is left to later phases.

Most numbers have some small prime factors. For example, it is not hard to show that about 88% of positive integers have a prime factor less than 100 and that about 91% have a prime factor less than 1000, and trial division will be good at finding these factors.

On the other hand, most numbers also have large prime factors. It can be shown (though not so easily) that a random positive integer n has a prime factor greater than √n with probability ln 2, or about 69% of the time, and of course if n is large, then trial division will not be of any help in nding such a factor.

   Related Questions in Mathematics

  • Q : Problem on Nash equilibrium In a

    In a project, employee and boss are working altogether. The employee can be sincere or insincere, and the Boss can either reward or penalize. The employee gets no benefit for being sincere but gets utility for being insincere (30), for getting rewarded (10) and for be

  • Q : Use MS Excel to do the computations

    Select a dataset of your interest (preferably related to your company/job), containing one variable and atleast 100 data points. [Example: Annual profit figures of 100 companies for the last financial year]. Once you select the data, you should compute 4-5 summary sta

  • Q : Probability assignments 1. Smith keeps

    1. Smith keeps track of poor work. Often on afternoon it is 5%. If he checks 300 of 7500 instruments what is probability he will find less than 20substandard? 2. Realtors estimate that 23% of homes purchased in 2004 were considered investment properties. If a sample of 800 homes sold in 2

  • Q : Research Areas in Medical Mathematical

    Some Research Areas in Medical Mathematical Modelling:1. Modeling and numerical simulations of the nanometric aerosols in the lower portion of the bronchial tree. 2. Multiscale mathematical modeling of

  • Q : Probability and Stochastic assignment

    Introduction to Probability and Stochastic Assignment 1: 1. Consider an experiment in which one of three boxes containing microchips is chosen at random and a microchip is randomly selected from the box.

  • Q : Formal logic2 It's a problem set, they

    It's a problem set, they are attached. it's related to Sider's book which is "Logic to philosophy" I attached the book too. I need it on feb22 but feb23 still work

  • Q : First-order formulas over the

    Consider the unary relational symbols P and L, and the binary relational symbol On, where P(a) and I(a) encode that a is apoint and a (sraight) line in the 2-dimensional space, respectively, while On(a,b) encodes  that a is a point, b is a line, and o lies on b.

  • Q : Competitive equilibrium 8. Halloween is

    8. Halloween is an old American tradition. Kids go out dressed in costume and neighbors give them candy when they come to the door. Spike and Cinderella are brother and sister. After a long night collecting candy, they sit down as examine what they have. Spike fi

  • Q : Problem on budgeted cash collections

    XYZ Company collects 20% of a month's sales in the month of sale, 70% in the month following sale, and 5% in the second month following sale. The remainder is not collectible. Budgeted sales for the subsequent four months are:     

  • Q : Area Functions & Theorem Area Functions

    Area Functions 1. (a) Draw the line y = 2t + 1 and use geometry to find the area under this line, above the t - axis, and between the vertical lines t = 1 and t = 3. (b) If x > 1, let A(x) be the area of the region that lies under the line y = 2t + 1 between t