One too many

We're searching for $\displaystyle 1+11+111+\cdots+\overbrace{1\ldots1}^{2014 \text{ ones}}\pmod{10^8}$ .

$\displaystyle 1+11+111+\cdots+\overbrace{1\ldots1}^{7 \text{ ones}}+\overbrace{1\ldots1}^{8 \text{ ones}}+\cdots+\overbrace{1\ldots1}^{2014 \text{ ones}}\equiv 1 234 567+2007\cdot \overbrace{1\ldots1}^{8 \text{ ones}}\pmod{10^8}$

Notice that $\displaystyle 9\cdot\overbrace{1\ldots 1}^{8 \text{ ones}}\equiv -1\pmod {10^8}$ and $2007=223\cdot 9$ .

$\displaystyle 1 234 567 + 223\cdot9\cdot\overbrace{1\ldots 1}^{8 \text{ ones}}\equiv 1 234 567 + 223\cdot(-1)\equiv \boxed {1 234 344}\pmod {10^8}$

If you were to write all the numbers as a column (column addition), the last digit will be made up of a sum of 2014 lots of 1s. Hence, the last digit is a 4 with 201 being carried on to the next column. This column has one less 1, so the digit for this column is 2013+201=2214 which is another 4 with 221 being carried onto the next column. Again in this column there is one less 1 than the one before, so the digit for this column is 2012+221=2233 which is a 3 with 223 being carried onto the next column. So far the last three digits are 344. This process is repeated until you have the last 8 digits which are 01234344.

digit........... place ....................... Place value

1 × 2014 × 1 = .....................................2014

1 × 2013 × 10 = ................................20130

1 × 2012 × 100= .............................201200

1 × 2011 × 1000= .........................2011000

1 × 2010 × 10000= ....................20100000

1 × 2009 × 100000= ................200900000

1 × 2008 × 1000000= ...........2008000000

1 × 2007 × 10000000= ......20070000000

Total ........................ .............. 22301234344

Last 8 digits 01234344

$\displaystyle The\quad last\quad 8\quad digits\quad of\quad the\quad sum\quad will\quad be\\ the\quad same\quad as\quad the\quad last\quad 8\quad digits\quad of\quad \\ \sum _{ k=0 }^{ 8 }{ (2014-k)\times { 10 }^{ k } } =222,901'234,344.$

$Sum~ of ~first~ 8~ terms~ is ~~~~~~~~~~~~~ 12345678$
$2014-8=2006~terms~ remain~ all ~~11111111. ~~~~~~~~$
$So ~~1111111*6 = 66666666```````1111111*2000 (mod 10^8)=22222000$
$Sum~=~12345678+66666666+22222000=1234344$

Notice that the sum of the first 2 terms is 12, and the sum of the first 3 is 123. Based on the pattern, the sum of the first 8 terms is 12345678. Now we just have to add the rest of the terms after that. But since we only need the last 8 digits, we only take the last 8 digits of the remaining terms, which are all 11,111 ,111. We've added the first 8 terms already so there must be 2014 - 8 terms remaining. Muliply 11,111,111 by 2006, take only the last eight digits, add that to 12,345,678 and you have your answer