Example

1. Factor of 6

1-> Since 1 exactly divides 6

2 -> Since it exactly divides 6

3 -> Since it exactly divides 6

6-> Since it exactly divides 6

2. every number is a factor of itself.

3. every factor of a number is an exact divisor of that number

4. every factor is less than or equal to the given number

5. number of factors of a given number are finite.

Number is 6

Multiple will be

$6 \times 1= 6$

$6 \times 2=12$

$6 \times 3=18$

2. number of multiples of a given number is infinite

3. every number is a multiple of itself

1. 6

The factors of 6 are 1, 2, 3 and 6.

Now, $1+2+3+6 = 12 = 2 \times 6$.

2. 28

All the factors of 28 are 1, 2, 4, 7, 14 and 28.

Now, $1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 \times 28$.

Example:

2, 3, 5, 7, 11 ,13

We can find list of prime numbers till 100 using

All the encircled numbers are prime numbers. All the crossed-out numbers, other than 1 are composite numbers

Example:

4, 6, 8 ,9….

Example

2,4,6,8,10,12,14

Even numbers have 0,2,4,6,8 in it one’s place

Example

1,3, 5,7,9,11......

1. 2 is the smallest prime number which is even.

2. every prime number except 2 is odd.

Number |
Test of divisibility |

2 |
A number is divisible by 2 if it has any of the digits 0, 2, 4, 6 or 8 in its one’s place |

3 |
A number is divisible by 3 if the sum of the digits is a multiple of 3, then the number is divisible by 3. Example 153 - Sum of digit = 1+5+3=9 and 9 /3 =3 So, 153 is divisible by 3 |

4 |
1) For one and two-digit number, just check the divisibility by actual division 2) For number with 3 or more digits is divisible by 4 if the number formed by its last two digits (i.e. ones and tens) is divisible by 4. |

5 |
A number is divisible by 5 if a number which has either 0 or 5 in its one’s place |

6 |
A number is divisible by 6 if a number is divisible by 2 and 3 both |

7 |
It must be checked by actual division |

8 |
1) For one, two-digit number, three-digit and four-digit number, just check the divisibility by actual division 2) For a number with 4 or more digits is divisible by 8, if the number formed by the last three digits is divisible by 8 |

9 |
A number is divisible by 9, if the sum of the digits of a number is divisible by 9 |

10 |
A number is divisible by 10 if a number has 0 in the ones |

11 |
A number is divisible by 11 if the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) of the number is either 0 or divisible by 11, |

1) Common factors of 4,12, and 16

Factors of 4 are 1, 2 and 4.

Factors of 12 are 1, 2, 3, 4, 6 and 12.

Factors of 16 are 1, 2, 4, 8 and 16.

Clearly, 1, 2 and 4 are the common factors of 4, 12, and 16.

Two numbers having only 1 as a common factor are called co-prime number

Example 3 and 4 are co-prime

36, is divisible by 18

Now if we find factors of 18 i.e., 1,2 ,3,4,9,18

So, 36 is also divisible by 1,2 ,3,4,9,18

2) If a number is divisible by two co-prime numbers then it is divisible by their product also

45

It is divisible by 3 (4+5=9)

It is divisible by 5

Since 3,5 are coprime. Now the product is 3 X5=15. Now it is divisible by 15 also

3) If two given numbers are divisible by a number, then their sum is also divisible by that number.

15 is divisible by 3

9 is divisible by 3

Sum = 15+9 =24

We can see that it is also divisible by 3

4) If two given numbers are divisible by a number, then their difference is also divisible by that number

15 is divisible by 3

9 is divisible by 3

Difference = 15 - 9 =6

We can see that it is also divisible by 3

So,

36 = 3×12 = 4 X 9

These form is called Factorisation

Prime Factorisation is expressing the number as a product of its prime factors

$36 = 2 \times 2 \times 3 \times 3$

We can find prime factorisation by dividing the numbers with 2, 3, 5, 7 etc. in this order repeatedly so long as the quotient is divisible by that number

Steps to find HCF or GCD

a. Find the prime factorisation of the numbers

b. Choose the common factors in them

c. Multiply those common factors to obtain HCF

(a) Find the HCF of 8 and 12

Prime Factorisation of the numbers

$8 = 2 \times 2 \times 2$

$12 = 2 \times 2 \times 3$

Common factors are 2,2

So $HCF = 2 \times 2 =4$

(b) Find the HCF of 20, 28 and 36

Prime Factorisation of the numbers

$20 = 2 \times 2 \times 5$

$28 = 2 \times 2 \times 7$

$36=2 \times 2 \times 3 \times 3$

Common factors are 2,2

So $HCF = 2 \times 2 =4$

(b) The Lowest Common Multiple (LCM) of two or more given numbers is the lowest of their common multiples.

Steps to find LCM

(a) Find the prime factorisation of the numbers

(b) look for the maximum occurrence of all the prime factors in these numbers

(c) The LCM of the numbers will be the product of the prime factors counted the maximum number of times they occur in any of the numbers

(a) Find the LCM of 8 and 12

Prime Factorisation of the numbers

$8 = 2 \times 2 \times 2$

$12 = 2 \times 2 \times 3$

So, $LCM = (2 \times 2 \times 2) \times (3) = 24$

(b) Find the LCM of 20, 28 and 36

Prime Factorisation of the numbers

$20 = 2 \times 2 \times 5$

$28 = 2 \times 2 \times 7$

$36=2 \times 2 \times 3 \times 3$

$LCM = (2 \times 2) \times (3) \times (5) \times (7) =540$

Here we divide the given numbers by common prime number until the remainder is a prime number or one. LCM will be the product obtained by multiplying all divisors and remaining prime numbers.

Steps are

(1) We place number in the line

(2) We start dividing the number by least prime number which is common among all of them or group of them

(3) Keep dividing by least until we have 1's in the remainder

(4) LCM is the product of the divisors

Example

(1) Find the LCM of 14 and 20

$LCM = 2 \times 2 \times 7 \times 5=140$

(2) Find the LCM of 12 ,18 and 44

$LCM = 2 \times 2 \times 3 \times 3 \times 11=396$

**Notes****Assignments**

Class 6 Maths Class 6 Science