Let $\overline{75\cdots57}$ denote the $(r+2)$ digit number where the first and the last digits are $7$ and the remaining $r$ digits are $5$. Consider the sum $S = 77 + 757 + 7557 + \cdots + \overbrace{75\cdots57}^{98}$. If $S = \frac{\overbrace{75\cdots57}^{99} + m}{n}$, where $m$ and $n$ are natural numbers less than $3000$, then the value of $m + n$ is

Let 7̅5̅·̅s̅5̅7̅ denote the (r+2) digit number where the first and the last digits are 7 and the remaining r digits are 5. Consider the sum S = 77 + 757 + 7557 + ·s + overbrace75·s57⁹⁸. If S = (overbrace75·s57⁹⁹ + m)/n, where m and n are natural numbers less than 3000, then the value of m + n is $\text{Given: } \overline{75\cdots57} = (r+2)\text{-digit number}$ $\text{first \& last digit } = 7,\ \text{the } r \text{ middle digits} = 5$ $S = 77 + 757 + 7557 + \cdots + \overbrace{75\cdots57}^{98}$ $\text{To find: } S = \dfrac{\overbrace{75\cdots57}^{99} + m}{n}$ $m,\, n \in \mathbb{N},\quad m,\, n < 3000$ $\text{Find } m + n$ $\text{Plan:}\quad 1.\ N_r \text{ in terms of } r$ $2.\ \text{Sum } r = 0 \to 98$ $3.\ \text{Write } S = \dfrac{N_{99} + m}{n}\text{ and read } m,\, n$ $r=0:\ 77,\qquad r=1:\ 757,\qquad r=2:\ 7557$ $N_r = \underbrace{7\,5\,5\cdots5\,7}_{r+2\ \text{digits}}$ $N_r = 7\cdot 10^{r+1} + 5\left(10^{r} + 10^{r-1} + \cdots + 10\right) + 7$ $= 7\cdot 10^{r+1} + 5\cdot\dfrac{10^{r+1}-10}{9} + 7$ $N_r = \dfrac{63\cdot 10^{r+1} + 5\cdot 10^{r+1} - 50 + 63}{9}$ $N_r = \dfrac{68\cdot 10^{r+1} + 13}{9}$ $r=1:\ N_1 = \dfrac{68\cdot 100 + 13}{9} = \dfrac{6813}{9} = 757$ $S = \sum_{r=0}^{98} N_r = \dfrac{1}{9}\left(68\sum_{r=0}^{98}10^{r+1} + 13\cdot 99\right)$ $\sum_{r=0}^{98} 10^{r+1} = 10 + 10^2 + \cdots + 10^{99}$ $= \dfrac{10\left(10^{99}-1\right)}{9}$ $68\sum_{r=0}^{98} 10^{r+1} = \dfrac{680\left(10^{99}-1\right)}{9}$ $S = \dfrac{1}{9}\left[\dfrac{680\left(10^{99}-1\right)}{9} + 13\cdot 99\right] = \dfrac{680\left(10^{99}-1\right)}{81} + 143$ $S = \dfrac{680\cdot 10^{99} - 680 + 11583}{81}$ $S = \dfrac{680\cdot 10^{99} + 10903}{81}$ $N_{99} = \dfrac{68\cdot 10^{100} + 13}{9} = \dfrac{680\cdot 10^{99} + 13}{9}$ $680\cdot 10^{99} = 9N_{99} - 13$ $S = \dfrac{(9N_{99} - 13) + 10903}{81}$ $= \dfrac{9N_{99} + 10890}{81} = \dfrac{N_{99} + 1210}{9}$ $\text{Compare: } S = \dfrac{N_{99} + m}{n} \Rightarrow m = 1210,\ n = 9$ $m,\, n < 3000\ \text{(satisfied)}$ $m + n = 1210 + 9 = 1219$