· A
Prime Number has exactly two factor.

[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]

**-----Total 25 between 1 to 100**
·
A
Composite Number has more than two factors.

[4,6,8,9,10………………………….] ------Total 74 between
1 to 100

Let
us take a number, 15.

Factors
of 15 are 1,3,5,15

·
Total
no of factors = 4

·
Total
no of Prime Factors = 2

·
Total
no of different Prime factors = 2

Now,
we will learn how to find out no of factors of a no.

**N = A**

^{p}× B^{q}………………..(1)**No of Factors = (p+1)(q+1)**

Any
real number can be written in the form of Equation (1)

Example:

15 = 5

^{1}×3^{1}
Hence
of factors of number 15 are (1+1)×(1+1)= 4

No
of prime factors = 2 (5 & 3)

60 = 2×2×3×5

= 2

^{2}×3^{1}×5^{1}
Hence
of factors of 60 are (2+1)(1+1)(1+1) =12

No
of prime factors = 4 (2,2,3,5)

No
of different prime factors = 3(2,2,3,5)

**Points to Remember:**

·
To
find out number of Odd factors, consider only Odd prime no

Example:

60 = 2×2×3×5

= 2

^{2}×3^{1}×5^{1}
No
of Odd factors = (1+1)(1+1) = 4

Here
we have considered only odd prime numbers i.e. 3 & 5.

No
of Even Factors = Total factors – Odd Factors

**Power of Prime no.**

·
Power
of P in n! = n/p

^{1}+n/p^{2}+n/p^{3}+ ……..
·
Only
integer part of each term is considered.

·
Terms
are considered up to the place results are greater than 1 .

Example:

Power
of 2 in 4!

= 4/2 + 4/2

^{2}+4/2^{3}
= 2 + 1 + 0 (since the value of
third term is less than 1)

= 3

Example:

Power
of 3 in 80!

= 80/3

^{1}+80/3^{2}+80/3^{3}+ 80/3^{4}
= 26.667+8.889+2.9629+0.9876

= 26+8+2+0 (Only integer parts are
considered)

=36

**Problem:**

How
many numbers between 300 and 700 , which are exactly divisible by 2,3 & 7
together.

Solution:

Between
300 to 700 i.e from 301 to 699

Step
1:

Take
LCM of 2,3 & 7 that comes out to be 42

Step
2:

Divide
First and second number by 42

Step
3:

Ignore
the remainder and subtract the result of both division. This will give you the
no of numbers.

Hence
the no of such numbers between 300 & 700 are 9