Prime factorization

Prime factorization is a method of representing a number in terms of multiples of prime numbers which is more powerful in identifying co-primes, LCM, HCF, number of its factors.

prime factors


What are prime numbers?
These are the numbers which have no factors other than one and itself.

Divide and rule was
one of the policies approached by many rulers in ruling  a kingdom thus they are divided into small parts as they can and finally they have power on every single part, prime factorization is also a similar method if you are unable to find solution for a big number, divide it into small pieces as multiples of prime numbers and play with it.
prime factors

How to find the number of factors of a given number?
If the number given is a small one, we can write all factors of it and answer but if the number is too big we should be smart enough to find the solution using prime factorization thus we can save more time in the competitive exam.

Factors:
If the prime factorization of a number N is 2^b 3^c..............
then number of factors of N = (b+1)(c+1)(d+1).............
Number of odd factors of N = (c+1)(d+1)....................   [remove all powers of 2
Number of even factors of N = (b)(c+1)(d+1)..............   [put a 2 aside so that when multiplied, it makes all the other factors even]

The numbers which have exactly 3 factors are the squares of prime numbers.
e.g. 3^2 = 9 has 1, 3, 9 as its factors.
*********************************************************************************
Example1: Prime factorize 4950 and find the total number of its factors.
solution:
                       2|  4950
                       -------------------
                         5|   2475
                       -------------------
                            5 | 495
                      --------------------
                              3 | 99
                       -------------------
                                3 | 33
                       -------------------
                                   | 11

So 4950=(2^1)*(3^2)*(5^2)*(11^1)   (all 2,3,5,11 are prime factors of 4950)
No.of factors of 4950=(power of 2+1)(power of 3+1)(power of 5 +1)(power of 11 +1)
                                  =(1+1)(2+1)(2+1)(1+1)
                                  =2*3*3*2
                                  =36
all prime factors seperated

Example 2: Prime factorize 24 and find no.of its factors.
solution:
              2| 24
             -------------
               2| 12
            --------------
                 2 |6
            --------------
                     3

24=(2^3)*(3^1)
No.of factors=(power of 2 +1)(power of 3 +1)
                     =(3+1)(1+1)
                     =4*2
                     =8
Justification: 1*24,2*12,3*8,4*6

To speak in relation to selection/combinations: we can select 2 in 4 ways(no two, one two, two twos, three twos) and we can select three in 2 ways(no three, one three)

Hope you have liked the trick, leave a comment if you have any doubts.

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