Let 7 / brace5 ... 5^r 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+ ... +7 / brace5 ... 57^98. If S=frac7 / brace5 ... 57^99+mn, where m and n are natural numbers less than 3000 , then the value of m+n is \_\_\_\_ .

S=7(10+10^2+10^3+ * s . .+10^99)+50(1+11+111+ * s +111 ... .11)+99 x 7 =frac7 x 10^1009-(70 / 9)+(50 / 9)[(10-1)+(10^2-1)+ * s .+(10^98-1)] =frac7 x 10^1009+(50 / 9)[frac10^99-109-98]+99 x 7-(70 / 9) +99 x 7 =frac7 x 10^1009+(50 / 9)[frac10^99-19-99]+693-(70 / 9) =frac / brace755 ... 57^999+693-(5020 / 9)-(7 / 9) =frac / brace755 ... 57^99+12109 \ Therefore m+n=1210+9=1219

Let 7 {5 5}^r 7 denote the (r+2) digit number where t... | Sequence & Series (Mathematics)

$S=7\left(10+10^2+10^3+\cdots . .+10^{99}\right)+50(1+11+111+ \cdots +111 \ldots .11)+99 \times 7 \\$ = $\frac{7 \times 10^{100}}{9}-\frac{70}{9}+\frac{50}{9}\left[(10-1)+\left(10^2-1\right)+\cdots .+\left(10^{98}-1\right)\right] \\$ = $\frac{7 \times 10^{100}}{9}+\frac{50}{9}\left[\frac{10^{99}-10}{9}-98\right]+99 \times 7-\frac{70}{9} +99 \times 7 \\$ = $\frac{7 \times 10^{100}}{9}+\frac{50}{9}\left[\frac{10^{99}-1}{9}-99\right]+693-\frac{70}{9}\\$ = $\frac{\overbrace{755 \ldots 57}^{99}}{9}+693-\frac{5020}{9}-\frac{7}{9}\\$ = $\frac{\overbrace{755 \ldots 57}^{99}+1210}{9} \\ \therefore m+n=1210+9=1219$

Explanation

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