**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 prime, we
need not 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.