Subject: Compulsory Maths

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

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

- 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.
- 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.

Write the sets of all possible factors of 16, 24 and 32. Then make a set of their common factors and find their H.C.F.

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.

Find the H.C.F. of 28, 42 and 70 by prime factorization method.

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

Find the H.C.F. of 28, 42 and 70 by prime factorization method.

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

Find the greatest number of children to whom 32 oranges, 40 bananas, and 56 mangoes can be distributed equally. Also find the shares of each fruit among them.

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

Find the L.C.M. of 24 and 36.

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

Write the sets of a few multiples of 4, 6 and 8. Make a set of their common multiples and find their L.C.M.

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.

Find the L.C.M of 24, 36 and 48 by prime factorisation method.

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

Find the HCF of 36.

Solution:

Divisor | Dividend |

2 | 36 |

2 | 18 |

3 | 9 |

3 |

Find the HCF of 48.

Solution:

Divisor | Dividend |

2 | 48 |

2 | 25 |

2 | 12 |

2 | 6 |

3 |

Find the LCM of 24.

Solution:

Divisor | Dividend |

2 | 24 |

2 | 12 |

2 | 6 |

3 |

Find the LCM of 36.

Solution:

Divisor | Dividend |

2 | 36 |

2 | 18 |

2 | 9 |

3 |

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