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.
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:-
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) .
Now, there are certain rules; Which are used to check whether a positive integer, no matter how large it may be is divisible by 7 and 11. If you want to know more in this regard, then please continue the read below:-
To check the divisibility by 7:-
We use -2 as the osculator, that is:- we do repeated subtractions of the product of the units
digit, with -2; from the truncated portion of the number, until finally a positive integer is obtained which could be explicitly determined that it is a multiple of seven or not.
Example_1:-let's
Let's try 52,871 as our test case:-
Step-1:-let's separate out the units digit =1. Then multiply this 1 with minus two and add it to the truncated part of the integer that is 5287, that is:- units digit = 1 and truncated portion = 5287.
Step-2:-Now let's do the conventional method to get the new number, that is the new number = 5287 - 2X1 =5285.
Now let's repeat the same step to get 528-2X5 = 518 = new number.
Now repeat again to get 51-2X8 = 35.
And that's it we all know 35 is a multiple of seven, that is:- 35/7 = 5. And although you can repeat the previous steps in this case as well to further verify that:- 3-2X5 = -7; which is the same as 0 when speaking of remainders with division.
Example 2:-let's consider a much easier case, since the entire process has already been
explained through a more complicated example. That is:- let's consider
119. The following are the steps:-
New number = 11-2x9 =11-18 = -7.
-7+7 = 0 ; (you can add any integral multiple of the divisor, to come down to a number which shall be the same as that of the original number in terms of the remainder with the divisor ).
Result:- since the remainder, in this case, = 0, hence it could be concluded that 119 is an integral multiple of 7 and that is verified using the fact that:- 7*17 = 119.
Tell that is:- 37,800
divisible by 7 or not, by commenting in the comments section.
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To check the divisibility by 11:- in this case, what you got to do is that:- starting from the extreme right, you take the sum of the digits which occur at the odd positions and subtract from it the sum of the digits which occur at the even positions. Now, if this difference of sums comes out to be zero or an integral multiple of 11, then the given integer is divisible by 11 otherwise indivisible.
Example_1:- let's verify whether 14,641 is an integral multiple of 11 or not.
The following are the steps:-
Step-1:-Take the sum of the alternating digits at the odd indices =1+6+1 = 8.
Step-2:-Take the sum of the alternating digits at the even indices =4+4 = 8 and the sum of alternating digits at the odd indices =1+6+1 = 8.
Step-3:-Compute the difference of sums = 8-8 = 0
Result:- since the difference is 0 hence according to the rule14,641 is an integral multiple of 11 and that could be verified by means of the fact that 14641/11 = 1331.
Tell whether is 16105100 divisible by 11 or not, by commenting in the comments section.
