+0  
 
0
148
2
avatar

A repunit is a positive integer whose base-ten representation consists entirely of $1$s.  For example, the first five repunits are $1, 11, 111, 1111,$ and $11111$.  If $S$ is the sum of the first $100$ repunits, what is the sum of the digits of $S$?

 Oct 31, 2019
 #1
avatar+7764 
0

Denote \(r_n\) the nth repunit.

\(\begin{array}{rcl} S &=& \displaystyle \sum^{100}_{n = 1} r_n\\ &=& \displaystyle \sum^{100}_{n = 1} \sum^{n-1}_{k = 0} 10^{k}\\ &=& \displaystyle \frac{1}{10}\sum_{n = 1}^{100} \sum_{k = 1}^{n} 10^{k}\\ &=& \displaystyle \dfrac{1}{10} \sum_{1 \leq k \leq n \leq 100} 10^{k}\\ &=& \displaystyle \dfrac{1}{10}\sum_{k = 1}^{100} \sum_{n=k}^{100} 10^{k}\\ &=& \displaystyle \dfrac{1}{10}\sum_{k = 1}^{100} \left(10^{k} (101 - k)\right)\\ &=& \displaystyle \dfrac{1}{10}\left(101\left(\dfrac{10\left(10^{100} - 1\right)}{10 - 1}\right)-\sum_{k= 1}^{100}k\cdot 10^{k}\right) \end{array}\\ \text{Let }S_{AG} = \displaystyle \sum_{k = 1}^{100} k\cdot 10^{k}\\ \begin{array}{rcl} S &=& \dfrac{101}{9}\left(10^{100} - 1\right) - \dfrac{1}{10}S_{AG}\\ S_{AG} &=& 1\cdot 10^1 + 2\cdot 10^2 + \cdots + 100\cdot 10^{100}---(1)\\ \dfrac{1}{10}S_{AG} &=& 1 + 2\cdot 10^1+\cdots +100\cdot 10^{99}---(2)\\ (1)-(2):\dfrac{9}{10}S_{AG} &=& 100\cdot 10^{100} - (1 + 10^1 + 10^2 +\cdots + 10^{99})\\ &=&10^{102} - \dfrac{10^{100} - 1}{9}\\ S_{AG} &=& \dfrac{10^{103}}{9} - \dfrac{10^{101} - 10}{81}\\ S &=& \dfrac{101}{9} (10^{100} - 1) - \left(\dfrac{10^{102}}{9}-\dfrac{10^{100} - 1}{81}\right)\\ &=& \dfrac{10^{100}}{9} - \dfrac{101}{9}+\dfrac{10^{100} - 1}{81}\\ &=& \dfrac{10^{100} - 1}{9}+\dfrac{10^{100} - 901}{81} \end{array}\\\)

.
 Oct 31, 2019
 #2
avatar
0

Isn't it an arithmetic series with terms =1, 2, 3, 4...........100 repunits.

 

The sum(S) = [100  x   101] / 2 =5050

 

Sum of digits of S =5 + 0 + 5 + 0 =10

 Oct 31, 2019

12 Online Users

avatar