How to find the number of divisors of a positive integer

In this post, we're going to be talking about the positive integers{Prime factors} which divide any composite number. Integer division has a rule that any positive integer greater than or equal to 2, has Prime divisors. We have been learning how to divide integers, since our early years, but in this article, you shall be introduced to a whole new way of finding the number of distinct dividers of any composite number, using the methods learned in school about how to divide positive and negative integers.

To begin this post first of all you would be introduced to the notion of prime power factorization. Now what does the term 'prime power factorization' mean; it basically means to express any positive integer or natural number as a product of powers of prime numbers. That is breaking down, that natural number into principal components which are further irreducible. First of all, consider the prime power factorization of the first 10 natural numbers:-

• 1 = 1X1 .
• 2 = 2X1 .
• 3 = 3X1 .
• 4 = 2X2 .
• 5 = 5X1 .
• 6 = 2X3 .
• 7 = 7X1
• 8 = 2X2X2 .
• 9 = 3X3 .
• 10 = 2X5 .

Now, let's try to break down 75, using this prime factorization method:- The key to finding prime factors of any positive integer is that you figure out the smallest number which divides it completely and leaves out the quotient; As the value for the next step. Then you keep on doing so until the number is further irreducible( that is it consists of prime integers only ).

75 = 25X3 ......(I).

Since 25 = 5X5, let's plug it in eqn(I)=>75 = 5X5X3; now writing it down in the form of a product of powers notation:- 75=5^2X3.

Now, the distinct factors of 75 are:-

• 1 = 3^0X5^0 .
• 3 = 3^1X5^1 .
• 5 = 3^0X5^1 .
• 15 = 3X5 .
• 25 = 5^2X3^0 .
• 75 = 5^2X3 .

Subscribe to our channel:-

https://www.youtube.com/channel/UClGrlBH3WAqhtixdFRtTG8Q?sub_confirmation=1

That is these are six distinct integer factors of 75; and the exponents of the prime factors; in the prime factorization are 2 and 1, hence:- the relationship between the number of prime factors of 75 and the powers of prime numbers found, by means of its prime power factorization method is:-

6 = (2+1)*(1+1) .

Similarly, let's consider the case of 500.

The prime power factorization of 500 is given as:-

500 = 2*250 = 2*2*125 = 2*2*5*25 = 2*2*5*5*5 = 2^2*5^3 .

Now, the distinct integer factors of 500 are:-

• 500 = 2^2*5^3 .
• 250 = 5^3*2 .
• 125 = 5^3 .
• 100 = 2^2*5^2 .
• 50 = 5^2*2 .
• 25 = 5^2 .
• 20 = 2^2*5 .
• 10 = 2^1*5^1 .
• 5 = 5^1*2^0 .
• 4 = 2^2 .
• 2 = 2^1*5^0 .
• 1 = 2^0*5^0 .

Now, there are 12 distinct factors of 500, and '12' could be expressed in terms of the powers of prime factors, found in the factorization of 500{that is - 2 and 3}; are:-

12 = (2+1)*(3+1) ;

Just Consider the image below to get a better idea via visualization:-

Here is actually the formula for finding the number of divisors, via knowing the prime power factorization:-

N = a1^n1*a2^b2*a3^b3.......ak^bk implies that the number of distinct positive integer divisors of N = (n1+1)*(n2+1)*(n3+1)*........*(nk+1) .