 ## HCF and LCM

Subject: Compulsory Maths

#### Overview

This note gives the information about the H.C.F and L.C.M.

#### Highest Common Factor ( H.C.F)

The largest positive integer which divides two or more integers without any remainder is called Highest Common Factor (HCF). It is the possible factor of respective numbers.

To find H.C.F by Factorization method
At first, we should find the prime factors of the given number, then the product of the common prime factors is the H.C.F of the given numbers. For example:
Find the H.C.F of 30 and 42
Solution,

Here,
30 = 2× 3× 5× 1
42 = 2× 3× 7× 1
∴ H.C.F = 2× 3× 1
= 6

To the H.C.F by division method
In this method, we divide the larger number by the smaller one and again the first remainder. So obtained divides the first divisor. The process is continued till the remainder becomes zero. The last divisor for which the remainder becomes zero and it is the H.C.F of the given numbers. For example:
Find the H.C.F of 30 and 42  #### Lower Common Multiple (L.C.M)

The lower common multiple is the lowest factor of respective numbers.

To find L.C.M by Factorisation method
At first, the prime factor of the given number are to be found out, then the product of the common prime factors and the remaining prime factors(which are not common) is the L.C.M of the given numbers. For example:
Find the L.C.M of 30 and 42

Here,
30 = 2× 3× 5
42 = 2× 3× 7
L.C.M = 2× 3× 5× 7
= 210

Division method
In this method, the given numbers are arranged in a row and they are successively divided by the least common factors till the quotient are 1 or prime numbers. Then, the product of these prime factors is the L.C.M of the given number. For example:

∴ L.C.M = 2× 3× 5× 5× 7
= 1050

##### Things to remember
1. The largest positive integer which divides two or more integers without any remainder is called Highest Common Factor (HCF).  It is the possible factor of respective numbers.
2. The lower common multiple is the lowest factor of respective numbers.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Videos for HCF and LCM ##### LCM, HCF - Learn basics, fundamentals with real life practical examples ##### HCF LCM by Prime Factorization Method, Learn basics and key concepts

Solution:

Here,
F(16) = {1, 2, 4, 8, 16}
F(24) = {1, 2, 3, 4, 6, 8, 12, 24}
F(32) = {1, 2, 4, 8, 16, 32}
Now,
F(16) ∩ F(24) ∩  F(32) = {1, 2, 4, 8}
∴ H.C.F. of 16, 24 and 32 is 8.

Solution:

Here,
28 ÷ 2 = 14 (remainder is 0)
14 ÷ 2 = 7 (remainder is 0)
Now,
28 = 2 × 2 × 7
42 = 2 × 3 × 7
70 = 2 × 5 × 7
∴ H.C.F = 2 ×  7 = 14

Solution:

Here,
28 ÷ 2 = 14 (remainder is 0)
14 ÷ 2 = 7 (remainder is 0)

42 ÷  2 = 21 (remainder is 0)
21 ÷  3 = 7 (remainder is 0)

70 ÷  2 = 35 (remainder is 0)
35 ÷  5 = 7 (remaindder is 0)
Now,
28 = 2 × 2 × 7
42 = 2 × 3 × 7
70 = 2 × 5 × 7
∴ H.C.F = 2 ×  7 = 14

Solution:

Here,
the required greatest number of children is the H.C.F. of 32, 40 and 56.

32 ÷ 2 = 16 (remainder is 0)
16 ÷ 2 = 8 (remainder is 0)
8 ÷ 2 = 4 (remainder is 0)
4 ÷ 2 = 2 |(remainder is 0)

40 ÷ 2 = 20 (remiander is 0)
20 ÷ 2 = 10 (remainder is 0)
10 ÷ 2 = 5 (remainder is 0)

56 ÷ 2 = 28 (remainder is 0)
28 ÷ 2 = 14 (remainder is 0)
14 ÷ 2 = 7 (remoiander is 0)

Now,
32 = 2 × 2 × 2 × 2 × 2
40 = 2 × 2 × 2 × 5
56 = 2 × 2 × 2 × 7
∴ H.C.F. = 2 × 2 × 2 = 8
Again,
to find the shares of each fruit.
32 ÷ 8 = 4
40 ÷ 8 = 5
56 ÷ 8 = 7
So, the required greatest number of children is 8.
Each child shares 4 oranges, 5bananas and 7 mangoes

Solution:

Here,
24 ÷ 2 = 12 (remainder is 0)
12 ÷ 2 = 6 (remainder is 0)
6 ÷ 2 = 3 (remainder is 0)

36 ÷ 2 = 18 (remiander is 0)
18 ÷ 2 = 9 (remainder is 0)
9 ÷ 3 = 3 (remainder is 0)

Now,
24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3 × 3
∴ L.C.M. = 2 × 2 × 3 × 2 × 3 = 72

Solution:

Here,
M(4) = {4, 8, 12, 16, 20, 24, 28, 32, 36, 20, 44, 48, . . . . . . . }
M(6) = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, . . . . . . . . }
M(8) = {8, 16, 24, 32, 40, 48, 56, 64, 72, 880, . . . . . . }
Now,
M(4) ∩ M(6) ∩ M(8) = (24, 48, . . . . . . . }
∴ L.C.M. of 4, 6, 8 is 24.

Solution:

24 ÷ 2 = 12 (remainder is 0)
12 ÷ 2 = 6 (remainder is 0)
6 ÷ 2 = 3 (remainder is 0)

36 ÷ 2 = 18 (remainder is 0)
18 ÷ 2 = 9 (remainder is 0)
9 ÷ 3 = 3 (renmainder is 0)

48 ÷ 2 = 24 (remiander is 0)
24 ÷ 2 = 12 (remainder is 0)
12 ÷ 2 = 6 (remainder is 0)
6 ÷ 2 = 3 (remiander is 0)

Now,
24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3 × 3
48 = 2 × 2 × 2 ×  2 × 3
∴ L.C.M. = 2 × 2 × 2 × 3 × 3 × 2 = 144

Solution:

 Divisor Dividend 2 36 2 18 3 9 3

Solution:

 Divisor Dividend 2 48 2 25 2 12 2 6 3

Solution:

 Divisor Dividend 2 24 2 12 2 6 3

Solution:

 Divisor Dividend 2 36 2 18 2 9 3