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Prime Factorization of Natural Numbers – Lucid Explanation of the Method of Finding Prime Factors

Prime Factors (PF):

The factors of a natural number which are prime numbers are called the PFs of that natural number.

Examples:

The factors of 8 are 1, 2, 4, 8.

Of these, only 2 is the PF.

Also 8 = 2 x 2 x 2;

The factors of 12 are 1, 2, 3, 4, 6, 12.

Of these, only 2, 3 are PFs

Also 12 = 2 x 2 x 3;

The factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30.

Of these, only 2, 3, 5 are PFs

Also 30 = 2 x 3 x 5;

The factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42.

Of these, only 2, 3, 7 are PFs

Also 42 = 2 x 3 x 7;

In all of these examples here, each number is expressed as a product of PF

In fact, we can do this for any natural number ( ≠ 1).

Multiplicity of SPs:

For a PF ‘p’ of a natural number ‘n’, the multiplicity of ‘p’ is the greatest exponent ‘a’ for which ‘p^a’ divides ‘n’ exactly.

Examples:

We have 8 = 2 x 2 x 2 = 2^3.

2 is the PF of 8.

The multiplicity of 2 is 3.

Also, 12 = 2 x 2 x 3 = 2^2 x 3

2 and 3 are the SP of 12.

The multiplicity of 2 is 2 and the multiplicity of 3 is 1.

Prime factorization:

Expressing a given natural number as the product of PF is called prime factorization.

or Prime factorization is the process of finding all PFs, as well as their multiplicity for a given natural number.

The prime factorization of a natural number is unique, except for order.

This statement is called the fundamental theorem of arithmetic.

Method of prime factorization of a given natural number:

STEP 1 :

Divide the given natural number by its smallest PF

2ND STEP :

Divide the quotient obtained in step 1, by its smallest FP.

Continue dividing each of the following quotients by their smallest PF, until the last quotient is 1.

STEP 3:

Express the given natural number as the product of all these factors.

This becomes the prime factorization of the natural number.

The steps and method of presentation will be made clear by the following examples.

Solved example 1:

Find the prime factorization of 144.

Solution:

2 | 144

———-

2 | 72

———-

2 | 36

———-

2 | 18

———-

3 | 9

———-

3 | 3

———-

end | 1

See the presentation method given above.

144 is divided by 2 to get the quotient of 72 which is again

divided by 2 to get the quotient of 36 which is again

divided by 2 to get the quotient of 18 which is again

divided by 2 to get the quotient of 9 which is again

divided by 3 to get the quotient of 3 which is again

divided by 3 to get the quotient of 1.

See how PFs are presented to the left of the vertical line

and the quotients to the right, below the horizontal line.

Now 144 must be expressed as the product of all PFs

which are 2, 2, 2, 2, 3, 3.

So, prime factorization of 144

= 2 x 2 x 2 x 2 x 3 x 3. = 2^4 x 3^2 Rep.

Solved example 2:

Find the prime factorization of 420.

Solution:

2 | 420

———-

2 | 210

———-

3 | 105

———-

5 | 35

———-

7 | 7

———-

end | 1

 

See the presentation method given above.

420 is divided by 2 to get the quotient of 210 which is again

divided by 2 to get the quotient of 105 which is again

divided by 3 to get the quotient of 35 which is again

divided by 5 to get the quotient of 7 which is again

divided by 7 to get the quotient of 1.

See how PFs are presented to the left of the vertical line

and the quotients to the right, below the horizontal line.

Now 420 must be expressed as the product of all PFs

which are 2, 2, 3, 5, 7.

So, prime factorization of 420

= 2 x 2 x 3 x 5 x 7 = 2^2 x 3 x 5 x 7. Rep.

Sometimes you may need to apply the divisibility rules to know the least PF with which we have to perform the division.

Let’s see an example.

Solved example 3:

Find the prime factorization of 17017.

Solution :

The number given = 17017.

Obviously, it’s not divisible by 2. (the last digit is not even.)

Sum of digits = 1 + 7 + 0 + 1 + 7 = 16 is not divisible by 3

and therefore the given number is not divisible by 3.

Since the last digit is neither 0 nor 5, it is not divisible by 5.

Let’s apply the divisibility rule of 7.

Twice the last digit = 2 x 7 = 14; remaining number = 1701;

difference = 1701 – 14 = 1687.

Twice the last digit of 1687 = 2 x 7 = 14; number remaining = 168;

difference = 168 – 14 = 154.

Twice the last digit of 154 = 2 x 4 = 8; remaining number = 15;

difference = 15 – 8 = 7 is divisible by 7.

Thus, the given number is divisible by 7.

Divide by 7.

17017 ÷ 7 = 2431.

Divisibility by 2, 3, 5 being excluded,

divisibility by 4, 6, 8, 9, 10 is also excluded.

Let’s apply the rule of divisibility by 11.

Sum of alternate digits of 2431 = 2 + 3 = 5.

Sum of the remaining digits of 2431 = 4 + 1 = 5.

Difference = 5 – 5 = 0.

Thus, 2431 is divisible by 11.

2431 ÷ 11 = 221.

As divisibility by 2 is excluded, divisibility by 12 is also excluded.

Let’s apply the divisibility rule of 13.

Four times the last digit of 221 = 4 x 1 = 4; remaining number = 22;

sum = 22 + 4 = 26 is divisible by 13.

So 221 is divisible by 13.

221 ÷ 13 = 17.

Let’s introduce all these divisions below.

7 | 17017

———-

11 | 2431

———-

13 | 221

———-

17 | 17

———-

end | 1

 

Thus, the prime factorization of 17017

= 7 x 11 x 13 x 17. Rep.

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