**Prime and Composite numbers**

Factors of every number can be written as explained below-

In product form 1 can be written as 1×1, thus has only one factor(1).

2 can be written as 2×1, and it has two factors(1,2)

3 can be written as 3×1, and it has two factors(1,3)

4 can be written as 1×4,2×2,4×1 and has three factors(1,2,4)

5 can be written as 5×1, and it has two factors(1,5)

6 can be written as 1×6,2×3,3×2,6×1, and it has four factors(1,2,3,6)

7 can be written as 1×7, and it has two factors(1,7)

8 can be written as 1×8,2×4,4×2,8×1, and it has four factors(1,2,4,8)

From above it is clear that number 1 has only one factor and that is 1 itself, whereas all other numbers either have two or more factors. In case of two factors, the factors are 1 and number itself. In case of more

than two factors, the factors are 1, number itself and few more number in between 1 and number itself.

*All the numbers other than 1, whose only factors are 1 and the number itself, are called **Prime numbers**.*

*Numbers having more than two factors are called **Composite numbers**.*

Consider number 1 to 100.

1 is not a prime number as it has only one factor. It is also known as Unique number.

2 is even number and it has 2 factor( 1 and number itself). It is called even Prime Number.

3 is odd number and has two factors( 1 and number itself). It is prime number.

4 is even number and has three factors( 1, 2 and number itself). It has more than two factors therefore, it is Composite number. Here it is important to note that all even numbers have a distinct factor 2, other than the normal factors( 1 and number itself), therefore all even numbers other than 2 are composite numbers.

By examining all other odd numbers, it is concluded that numbers 5,7,11,13,17,19,23,29,31,37,41,47,53,59,61,67,71,73,79,83,89 and 97 are prime numbers less than 100. Prime numbers of > 100 can also be listed.

**Divisibility test**

By examining a given number, it can be worked out, if it is divisible by some numbers e.g. 2,3,4,5,6,8,9,10,11.

All even numbers are divisible by 2.

All numbers ending with a 0 are divisible by 10 as any multiple of 10 has 0 at unit place. Such numbers are also divisible by 2 and 5 as these numbers are even numbers and also multiple of 5. It is clear from here that if a number is divisible by any particular number then it is also divisible by all factors of this number.

All numbers ending with 0 or 5 are divisible by 5, as any multiple of 5 ends with 5 or 0 at unit place.

All numbers having sum of their digits divisible by 3, are divisible by 3.

All even numbers divisible by 3, are also divisible by 6, because being even number, these are divisible by 2. It is clear from here that if a number is divisible by two numbers, it is also divisible by their product.

If last two digits of any number, with their places unchanged, is divisible by 4, all such numbers are divisible by 4. For example, in number 824 last two digits form a number 24; and 24 is divisible by 4; so 824 is divisible by 4.

If last three digits of any number, with their places unchanged, is divisible by 8, all such numbers are divisible by 8. For example, in number 1128 last three digits form a number 128; and 128 is divisible by 8; so 1128 is divisible by 8.

If sum of all digits of a number is divisible by 9, the number is also divisible by 9.

If the difference of, sum of all alternate digits of a number from highest place value side and sum of all other remaining digits, is 0 or multiple of 11, then such numbers are divisible by 11. For example in number 12012, the sum of alternate digits from highest place value side is 1+0+2=3 and sum of remaining digits is 2+1=3. The difference of these sums of digits is 3-3=0, hence the number 12012 is divisible by 11.

**Common factors**

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

Factors of 24 are 1, 2, 3, 4, 6, 8,12, 24

Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Factors 1, 2, 3, 6 are common for 18, 24 and 48. Similarly common factors for any set or group of numbers can be listed.

**Multiples**

Multiples of 3 are 3, 6, 9, 12, 15, 18 21, 24, 27, 30, 33, 36……

Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40….

12, 24,36,… are common multiples of 3 and 4 together.

**Prime Factors**

When a number is written as a product of its factors in form of prime numbers, all the factors are called prime factors of that number e.g.

72=2×36 or 2x2x18 or 2x2x2x9 or 2x2x2x3x3

So 2x2x2x3x3 is factorization of 72 in form of prime factors.

**Highest common factor(HCF) **

For a given set of numbers, the highest value factor which is common for all given numbers, is known as HCF e.g. for numbers 24,36,48 factors are, 24=2x2x2x3; 36=2x2x3x3; 48=2x2x2x2x3

Here 2x2x3 are the common factors of 24, 36, 48. So, HCF of these numbers is 2x2x3=12

Important thing to note that lowest HCF is 1 and highest value of HCF cannot be more than the lowest number of the given set of numbers.

To calculate HCF, workout all the prime factors of each given number, and then pick only those factors that appear in the list of factors of all the numbers. Now multiply all the picked factors, this is the required HCF.

**Lowest common multiple**

For a set of numbers 12, 18, 72; their multiples are

12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144……

18, 36, 54, 72, 90, 108, 126, 144, 162, 180…..

72, 144, 216…

From above it is clear that 72, 144… are common multiples of 12, 18, 72 and 72 is lowest multiple. So 72 is called LCM.

**The Lowest Common Multiple (LCM) of two or more given numbers is the lowest (or smallest or least) of their common multiples. LCM is always either equal to highest number or any multiple of it.**

To find out LCM of a given set of numbers, first work out their prime factors. Now select all the prime factors with maximum number of times they appear in any of the numbers. Product of all these selected prime numbers is LCM of given set of numbers. Example

Find LCM of 20, 25, 30, 35

Factors of 20 are 2x2x5

Factors of 25 are 5×5

Factors of 30 are 2x3x5

Factors of 35 are 5×7

In the above, the prime factors appearing are 2, 3, 5, and 7. Here it is seen that 2 is appearing maximum 2 times in 20, 5 is appearing maximum 2 times in 25, and 3 and 7 are appearing only once. So

LCM of given set of numbers is ( 2×2)x3x(5×5)x7=2100

While calculating HCF sometimes the given numbers(3,4 or 5 digit numbers) are big enough to work out their prime factors. In that case apply division method. Divide biggest number by any smaller number and find out remainder; if it is 0, then divide the third number with same divisor and find out remainder; if it is 0, then the divisor is HCF of three given numbers. Same division method can be continued for more given numbers. If remainder is not 0, then divide the divisor with this remainder and find the next remainder; if remainder is not 0, divide the last divisor with this remainder and continue this process till remainder is 0. Now apply division test on other numbers with last divisor. The smallest divisor so obtained is HCF of all numbers.

Example: Find HCF of 3104,3316 and 6644

Divide 6644 by 3104; remainder is 436; divide 3104 by 436; remainder is 52; divide 436 by 52; remainder is 20;divide 52 by 20; remainder is 12; divide 20 by 12; remainder is 8; divide 12 by 8; remainder is 4; divide 8 by 4; remainder is 0. Last divisor is 4. Now divide next number 3316 by last divisor i.e. 4; remainder is 0. HCF of given numbers is 4.

**Exercise:**

1.What is maximum weight that can measure 69 and 75 kg in minimum number of weighings.

2.What is maximum size of a measure to fill tanks of 403, 434 and 465 liter capacity tanks with minimum no. of pourings.

3.What is smallest three digit number which is exactly divisible by 6, 8, 12

and what is largest three digit number which is exactly divisible by these three nos.

4.what is smallest number which when divided by 8,12 and 15 leaves a remainder of 1 in each case.