**Numbers - Prime numbers and Applications**

**Prime Factor:**

A prime
number which exactly divides a number is the prime factor of that number.

Factors of
42 : 1,2,3,6,7,14,21,42

Prime
factor of 42 : 2,3,7

**Prime Factorization:**

**Fundamental Theorem of Arithmetic :**

**It**states that every natural number greater than 1 can be written as a unique product of prime numbers. This theorem is also called the unique factorization

**theorem**or the unique-prime-factorization

**theorem**.

How to write a number as product of prime factors :

If the given number is a prime, we
need not write the number as product of prime factors. Because it has no any other
prime number as its factor.

If the number is not a prime number,

**A**. Find a smallest prime number that exactly divides the given number.

**B.**Divide the given number by the prime to obtain a quotient.

**C.**Divide the quotient with the prime number and if the quotient is not divisible by the prime number taken, take next prime number which exactly divides it

**D**

**.**Repeat the process with the quotient until the quotient is itself a prime number

**E.**The prime factorization is given by the product of the primes used in the division process and the final prime quotient.

Example : 180 = 2 x 2 x 3 x 3 x 5

__Number of factors of a given number:__

**How to find the number of factors of a number**

Write the given
number as product of prime factors.

Consider the number
N=a

^{p }× b^{q }× c^{r }×… where a, b, c are prime numbers and p, q, r are positive integers.
Then number of
factors of ‘N’ is

N= (p+1)(q+1)(r+1)…..

Factors of 18 :
1,2,3,6,9,18

**Example :Find the number of factors of 18?**

Solution : 18=2

^{1}×3^{2}
Number of factors=
(1+1) (2+1)

= (2×3) =6

__Sum of the factors of a given number____:__

**Finding the sum of the factors of a number:**

Write the given
number as product of prime factors

Consider the number
of N= a

^{p }× b^{q }× c^{r }×… where a, b, c are prime numbers and p, q, r are positive integers.
Then sum of the
factors= $ (\frac{a^{p+1}\text{–}1}{a-1})\times
)(\frac{b^{q+1}\text{–}1}{b-1})\times(\frac{c^{r+1}\text{–}1}{c-1})
)$

**Example : Find the sum of the factors of 120**

Write 120 as
product of prime factors => 120=2

^{3 }× 3^{1 }x 5^{1}
Sum of factors = $ (\frac{2^{3+1}\text{–}1}{2-1})\times
)(\frac{3^{1+1}\text{–}1}{3-1})\times(\frac{5^{1+1}\text{–}1}{5-1}
)$

= $ (\frac{2^{4}\text{–}1}{1})\times
)(\frac{3^{2}\text{–}1}{2})\times(\frac{5^{2}\text{–}1}{4})
)$

= ($\frac{15}{1}$
) ×($ \frac{8}{2} $
) x ($ \frac{24}{4}$
) =15 x 4 x 6 = 360

__Number of ways to express a number as product of 2 factors____:__

**To find the number of ways a number can be expressed as product of 2 factors:**

Express given
number into product of prime factors.

Consider N= a

^{p }× b^{q }× c^{r }×… where a, b, c are primes and p, q, r are positive integers.
Number of factors =
(p+1) (q+1) (r+1)….

Product of two
factors = $\frac{Number\:of\:factors}{2}$

**In how many ways you can express 24 as a product of two of its factors?**

24 = 2

^{3 }x 3^{1}
Number of factors
of 24 = (3 + 1) (1 + 1)= 8

So, number of ways
to express 24 as product of two factors = $\frac{8}{2}$
=4

__Number of ways to express a number as a product of two co-primes :__
First write down
the give number as product prime factors.

Say N= a

^{p }× b^{q }× c^{r }×… where a, b, c are primes and p, q, r are positive integers.
Then number of ways
N can be written as product of two relatively primes = 2

^{k-1}where k is the number of prime factors.**Example: In how many ways 180 can be written as product of two of its co-prime factors ?**

Solution : Write
180 as product of prime factors

180 = 2

^{2}x 3^{2}x 5^{1}.
The total number of
unique prime factors are 3.

Hence required
number of ways = 2

^{3-1 }= 4.**Number of co-primes that are less than given number:**

To find the number
of relative prime numbers that are less than the given number, express the
given number as product of prime factors .

N= a

^{p }× b^{q }× c^{r }×… where a, b, c are primes and p, q, r are positive integers
Then the number of
co-primes less than N

= n (1- $\frac{1}{a}$ ) (1- $\frac{1}{b}$)(1-$\frac{1}{c})$

= n (1- $\frac{1}{a}$ ) (1- $\frac{1}{b}$)(1-$\frac{1}{c})$

**Example : Find the number of co-primes to 180 that are less than 180?**

Solution : Expressing
180 as product of prime factors => 180 = 2

^{2}x 3^{2}x 5^{1}
So number of coprimes
that are less than 180 and co prime to 180

= 180 x (1 – $\frac{1}{2}$) (1- $\frac{1}{3}$) (1 -$\frac{1}{5}$) =48.

**Product of Factors of a number :**

**How to find the product of all the factors of a number :**

To find the product
of all divisors of the number N, write N as product of prime factors.

N= a

^{p }× b^{q }× c^{r }×… where a, b, c are primes and p, q, r are positive integers
Find the number of
factors of N n(f) = (p+1)(q+1)(r+1)

Then, product of
all the factors = N

^{$\frac{n(f)}{2}$}**Example: Find the product of all the factors of 120?**

Solution : Write
the 120 into product of prime factors

120 = 2

^{3}x 3 x 5
Number of factors
of 120 n(f)= (3+1)(1+1)(1+1) = 4 x 2 x 2 =16

Product of factors
of 120 = 120

^{$\frac{16}{2}$}= 120^{8}**Finding the Unit digit :**

**How to find the unit digit in a product or an expression**

__Unit digit rules:__

__In this there are four rules to find out the unit digit of a given number.__

__Rule (1):__**The digits which are equal to the unit digit of power anything.**

The digits in this rule are 0, 1, 5, 6

0

^{1}=0 1^{1}=1 5^{1}=5 6^{1}=6
0

^{2}=0 1^{2}=1 5^{2}=20 6^{2}=36
0

^{3}=0 1^{3}=1 5^{3}=125 6^{3}=216
------ ------ ------- --------

------ ------ ------- --------

__Rule (2):__**Odd power unit digits and even power unit digits.**

The digits in this rule are 4, 9.

Ã¨ Odd
powers of 4, the unit digit is 4

4

^{1}=4
4

^{3}=64
……….

……….

Ã¨ Even
powers of 4. The unit digit is 6

4

^{2}=16
44=256

………..

Ã¨ Odd
powers of 9. The unit digit is 9.

9

^{1}=9
9

^{3}=729
…………

Ã¨ Even
powers of 9. The unit digit i.e 1

9

^{2}=81
9

^{4}=6561
………….

__Rule (3):__**This rule is in cyclic form of length ‘4’. The digits in this rule are 2, 3, 7, 8….**

2

^{1}=2 3^{1}=3 7^{1}=7
2

^{2}=4 3^{2}=9 7^{2}=49
2

^{3}=8 3^{3}=27 7^{3}=343
2

^{4}=16 3^{4}=81 7^{4}=2401
2

^{5}=32 3^{5}=343 7^{5}=16807
8

^{1}=8
8

^{2}=64
8

^{3}=512
8

^{4}=4096
8

^{5}=32768
Find the last digit
of 2

^{343}
Solution: Here
power is 343.

So divide 343 by 4
we will get remainder 3 so unit digit is same as that of 2

^{3}= 8.