# Introduction to Numbers

**Important Formulas**

- (a + b)(a – b) = (a
^{2}– b^{2}) - (a + b)
^{2}= (a^{2}+ b^{2}+ 2ab) - (a – b)
^{2}= (a^{2}+ b^{2}– 2ab) - (a + b + c)
^{2}= a^{2}+ b^{2}+ c^{2}+ 2(ab + bc + ca) - (a
^{3}+ b^{3}) = (a + b)(a^{2}– ab + b^{2}) - (a
^{3}– b^{3}) = (a – b)(a^{2}+ ab + b^{2}) - (a
^{3}+ b^{3}+ c^{3}– 3abc) = (a + b + c)(a^{2}+ b^{2}+ c^{2}– ab – bc – ac) - When a + b + c = 0, then a
^{3}+ b^{3}+ c^{3}= 3abc

**Number Series**

- 1 + 2 + 3 + 4 + 5 + … + n = n(n + 1)/2
- (1
^{2}+ 2^{2}+ 3^{2}+ ….. + n^{2}) = n ( n + 1 ) (2n + 1) / 6 - (1
^{3}+ 2^{3}+ 3^{3}+ ….. + n^{3}) = (n(n + 1)/ 2)^{2} - Sum of first n odd numbers = n
^{2} - Sum of first n even numbers = n (n + 1)

**Divisibility Rules**

Number | Divisibility Rule | Example |
---|---|---|

2 | Number must be even | 124 is even, hence it is divisible by 2. 421 is odd, hence it is not divisible by 2 |

3 | Sum of the digits must be divisible by 3 | Consider 123. The sum of its digits is 1+2+3 = 6. 6 is divisible by 3. Hence 123 is divisible by 3 |

4 | The last two digits must be divisible by 4 | In 123496, the last two digits 96 is divisible by 4, hence 123496 is divisible by 4. |

5 | The last digit is either 5 or 0, then the number is divisible by 5 | 947245. Here the last digit is 5. Hence 947245 is divisible by 5. |

6 | The number must be divisible by both 2 and 3 | 96312. The number is even hence its divisible by 2 and the sum of its digits is 21, which is divisible by 3. Hence 96312 is divisible by 6 |

7 | Multiply the last digit by 2, then subtract it from the rest of the number. If the result is divisible by 7, then the number is divisible by 7 | 672. 1. Multiplying the last digit 2 by 2 gives 4. 2. Subtract 4 from the rest of the number i.e 67 - 4 = 63. Since 63 is divisible by 7. 672 is divisible by 7 |

8 | The last three digits must be divisible by 8 | 26032. Since the last three digits 032 is divisible by 8. The number 26032 is divisible by 8 |

9 | Sum of the digits must be divisible by 9 | 11106. The sum of the digits is 1+1+1+0+6 = 9, which is divisible by 9. Hence 11106 is divisible by 9. |

10 | If the last digit is 0, the number is divisible by 10 | 123450. Since the number ends in 0. It is divisible by 10 |

11 | Subtract the sum of the digits in even numbered places with the sum of the digits in odd numbered places. The result obtained must be divisible by 11 | 107415. Sum of the digits in even places is 1+7+1=9 Sum of the digits in odd places is 0+4+5 = 9 Difference in the even and odd places sum is 9-9=0, which is divisible by 11. Hence 107415 is divisible by 11 |

12 | Number must be divisible by both 3 and 4 | 132. Sum of the digits is 1+3+2 = 6, which is divisible by 3. Hence the number is divisible by 3. Last two digits 32 is divisible by 4. Hence the number is divisible by 4. Since the number is divisible by 3 and 4. The number is divisible by 12. |

13 | Multiply the last digit by 9, then subtract it from the rest of the number. If the result obtained is divisible by 13, then the number is divisible by 13 | 247. Multiply 7 by 9, which gives 63. Subtract 63 from 24, which gives -39. -39 is divisible by 13. Hence 247 is divisible by 13. |