## Sequence Divisibility

#### Problem

Prove that every term in the infinite sequence 18, 108, 1008, 10008, 100008, ... , is divisible by 18.

#### Solution

The `n` th term, `u _{n}` , of the sequence 18, 108, 1008, 10008, ... is given by

`u`= 10

_{n}^{n}+ 8.

u_{n+1} | = 10^{n+1} + 8 |

= 10.10^{n} + 8 | |

= (9 + 1).10^{n} + 8 | |

= 9.10^{n} + 10^{n} + 8 | |

= 9.10^{n} + 10^{n} + 8 |

If `u _{n}` = 10

^{n}+ 8 is divisible by 18, then so too will be

`u`

_{n+1},

as 9 [even] will be divisble by 18.

We can see that `u`_{1} = 18, hence 10^{n} + 8 must be divisible by 18 for all `n`.

Problem ID: 75 (Apr 2002) Difficulty: 3 Star