**Factors, Prime factors (revision)**

40 ÷ 8 = 5 and 40 ÷ 5 = 8

8 and 5 divide into 40 without remainder.

8 and 5 are **factors **of 40.

A **prime number** has only two factors, itself and 1, but 1 is not a prime number.

2, 3, 5, 7, 11, 13, … are prime numbers.

**Common factors**

The number 12, 21 and 33 are all divisible by 3. We say that 3 is a **common factor** 0f 12, 21 33.

There may be more than one common factor of a set of numbers. For example, both 2 and 7 are common factors of 28, 42 and 70. Since 2 and 7 are common factors and are both prime numbers, then 14 (= 2 x 7) must also be a common factors of the set of numbers.

1 is a common factor of all numbers.

**Highest Common Factor (HCF)**

2, 7 and 14 are common factors of 28, 42 and 70; 14 is the greatest of three common factors. We say that 14 is the **highest common factor** of 28, 42 and 70.

To find the HCF of a set of numbers:

Express the number as a product of prime factors;

- Find the common prime factors
- Multiply the current prime factor together to give the HCF.

Example

Find the HCF of 18, 24 and 42.

18 = 2 x 3 x 3

24 = 2 x 2 x 2 3

42 = 2 x 3 x 7

The common prime factors are 2 and 3.

The HCF = 2 x 3 = 6.

Find the HCF of 216 and 288

2 | 216 2 | 288

2 | 108 2 | 144

2 | 54 2 | 72

3 | 27 2 | 36

3 | 9 2 | 18

3 | 3 3 | 9

…… 3 | 3

0 1 ……..

0 1

In index notation

216 = 2^{3 }x 3^{3 }

288 = 2^{5 }x 3^{2 }

2^{3 }is the lowest power of two contained in the two numbers. Thus the HCF contains 2^{3}.

3^{2} is the lowest power of 3 contained in the tow numbers. The HCF contains 3^{2}.

216 = (2^{3 }x 3^{3}) x 3

288 = (2^{2 }x 3^{3}) x 2^{2 }

The HCF = 2^{2 }x 3^{3 }= 8 x 6 = 72

**Rules of divisibility**

Table 1.2 gives some rules for divisors of whole numbers.

Any whole number is exactly divisible by … |

2 if its last digit is even or 0 |

3 if the sum of its digit is divisible by 3 |

4 if its last two digits form a number divisible by 4 |

5 if its last digit is five or zero |

6if its last digit is even and the sum of its digits is divisible by 3 |

8 if its lat three digits forma number divisible by 8 |

9if the sum of its digit is divisible by 9 |

10 if its last digit is 0 |

**Table**

There is no easy rule for division by 7.

Notice the following:

- If a number m is divisible by another number n, m is also divisible by the factors of n. For example, a number divisible by 8 is also divisible by 2 and 4.
- If a number is divisible by two or more numbers, it is also divisible by the LCM of these numbers. For example, a number divisible by both 6 and 9 is also divisible by 18, 18 is the LCM of 6 and 9.

Example

Test the following numbers to see which are exactly divisible by 9. a. 51 066 b. 9 039

Solution

- 5 + 1 + 0 + 6 + 6 = 18

18 is divisible by 9.

Thus 51 066 is divisible by 9.

- 9 + 0 + 3 + 9 = 21

21 is not divisible by 9.

Thus 9 039 is not divisible by 9.

**Squares and square roots**

Square roots

7^{2 }= 7 x 7 = 49.

In words ‘ the **square** of 7 is 49’. We can turn this statement round and say. ‘the **square root** of 49 is 7’.

In symbols, √49 = 7. The symbol √ means *the square root of* .

To find the square root of a number, first find its factors.

Example

Find √11 025.

Method: Try the prime numbers 2, 3, 5. 7, …

Working:

3 | 11 025 |

3 | 3 675 |

5 | 1 225 |

5 | 245 |

7 | 49 |

7 | 7 |

1 |

11 025 = 3^{2 }x 5^{2 }x 7^{2 }

= (3 x 5 x7) x (3 x 5 x 7)

= 105 x 105

Thus √11 025 = 105

It is not always necessary to write a number in its prime factors.

Example

√6 400

6400 = 64 x 100

= 8^{2 }x 10^{2 }

Thus √6 400 = 8 x 10 = 80

The rules for divisibility can be useful when finding square root.

**EXERCISES**

Lets see how much you’ve learnt, attach the following answers to the comment below:

Find by factors the square roots of the following:

- 225
- 194
- 342
- 484

AnonymousNovember 7, 2017 at 3:28 pmFind the factors of the following numbers?