Let $7 \overset{r}{\overset{⏞}{5 \dots . 5} 7}$ 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 $S = 77 + 757 + 7557 + \dots + 7 \overset{98}{\overset{⏞}{5 \dots 5}} 7 .$ . If $S = \frac{7 \overset{99}{\overset{⏞}{5 \dots 5}} 7 + m}{n}$ , where m and n are natural numbers less than 3000, then the value of m+n is

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answer is 1219.

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Detailed Solution

$S = 77 + 757 + 7557 + \dots \dots . + 7 \overset{98}{\overset{⏞}{5 \dots \dots 57}} \begin{array}{l} = 7 (10 + 10^{2} + 10^{3} + \dots . + 10^{99}) + 50 (1 + 11 + 111 + \dots \dots) + 7 \times 99 \\ = \frac{7 (10^{100} - 10)}{9} + \frac{50}{9} (10 - 1 + 10^{2} - 1 + \dots + 10^{98} - 1) + 7 \times 99 \\ = \frac{7 \times 10^{100}}{9} - \frac{70}{9} + \frac{50}{9} (\frac{10^{99} - 10}{9} - 98) + 7 \times 99 \\ = \frac{7 \times 10^{100} + \frac{50}{9} (10^{99} - 1)}{9} - \frac{50}{9} (99) + 7 \times 99 - \frac{70}{9} \\ = \frac{7 + 7 \times 10^{100} + \frac{50}{9} (10^{99} - 1) + 1280 - 70}{9} \\ \begin{array}{l} \frac{7 \overset{99}{\overset{⏞}{5 \dots . 5}} 7 + 1210}{9} \Rightarrow m = 1210, n = 9 \\ m + n = 1219 \end{array} \end{array}$