If S is the sum of all the digits of the natural number N = 1+11+111+1111+ ....to 2011 terms, then the sum of the digits of S is ? thank you.Here is my response:

- 17
- 18
- 19
- 21

First I had to use a little bit of brute force to get me started thinking and establish a pattern. I decided to look at the same problem, but with a smaller number of terms. After doing that it was easy to establish that in general for n terms, you would have , since you are just counting the number of 1's that you are adding in each place value position. If you use a bit of brute force for some small values of *n* it won't take long to notice a pattern. Note, that I decided to focus on numbers of terms that are multiples of 3 plus 1, because 2011 = 3(670)+1

number of terms | N | S | Sum of digits of S |
---|---|---|---|

4 | 1234 | 10 | 1 |

7 | 1234567 | 28 | 10 |

10 | 1,234,567,900 | 37 | 10 |

13 | 1,234,567,901,233 | 46 | 10 |

16 | 1,234,567,901,234,566 | 64 | 10 |

19 | 1,234,567,901,234,567,900 | 74 | 11 |

22 | 1,234,567,901,234,567,901,233 | 83 | 11 |

25 | 1,234,567,901,234,567,901,234,566 | 101 | 2 |

28 | 1,234,567,901,234,567,901,234,567,900 | 111 | 3 |

31 | 1,234,567,901,234,567,901,234,567,901,233 | 120 | 3 |

Okay, I admit it, after 16 terms, I was no longer using brute force, but I was using the pattern that I saw emerging. Once you see that pattern, then you can see that for the part that repeats: 234,567,901 the digits add to 37. Add the leading 1 and the leftover digits on the end and you get S. So for 2011, there are 670/3= 223 with remainder 1. That means for 2011 terms,