JEE Advance - Mathematics (2023 - Paper 1 Online - No. 10)
Let $7 \overbrace{5 \cdots 5}^r 7$ 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 $S=77+757+7557+\cdots+7 \overbrace{5 \cdots 5}^{98}7$. If $S=\frac{7 \overbrace{5 \cdots 5}^{99}7+m}{n}$, where $m$ and $n$ are natural numbers less than 3000 , then the value of $m+n$ is
Answer
1219
Explanation
$S=77+757+7557+\cdots+ 7\overbrace{5 \cdots 5}^{98}7$
$$ \begin{aligned} & =7\left(10+10^2+\ldots+10^{99}\right)+50(1+11+\ldots+\overbrace{111 \ldots 1}^{98})+7 \times 99 \\\\ & =70\left(\frac{10^{99}-1}{9}\right)+\frac{50}{9}\left[(10-1)+\left(10^2-1\right)+\ldots+\left(10^{98}-1\right)\right]+7 \times 99 \end{aligned} $$
$$ \begin{aligned} & =70\left(\frac{10^{99}-1}{9}\right)+\frac{50}{9}\left[10\left(\frac{10^{98}-1}{9}\right)-98\right]+7 \times 99 \\\\ & =\frac{7 \times 10^{100}}{9}-\frac{70}{9}+\frac{50}{9}\left[\frac{10^{99}-1-9}{9}-98\right]+7 \times 99 \end{aligned} $$
$$ \begin{aligned} & =\frac{7 \times 10^{100}}{9}-\frac{70}{9}+\frac{50}{9}[\overbrace{111 \ldots 1}^{99}-99]+7 \times 99 \\\\ & =\frac{7 \times 10^{100}-70+\overbrace{555 \ldots 50}^{99}}{9}-550+693 \end{aligned} $$
$$ \begin{aligned} & =\frac{7 \overbrace{555 \ldots 5}^{99}-70+143 \times 9}{9} \\\\ & =\frac{7 \overbrace{55 \ldots 5}^{99}+1210}{9} \end{aligned} $$
$$ \therefore $$ $$ m+n=1219 $$
$$ \begin{aligned} & =7\left(10+10^2+\ldots+10^{99}\right)+50(1+11+\ldots+\overbrace{111 \ldots 1}^{98})+7 \times 99 \\\\ & =70\left(\frac{10^{99}-1}{9}\right)+\frac{50}{9}\left[(10-1)+\left(10^2-1\right)+\ldots+\left(10^{98}-1\right)\right]+7 \times 99 \end{aligned} $$
$$ \begin{aligned} & =70\left(\frac{10^{99}-1}{9}\right)+\frac{50}{9}\left[10\left(\frac{10^{98}-1}{9}\right)-98\right]+7 \times 99 \\\\ & =\frac{7 \times 10^{100}}{9}-\frac{70}{9}+\frac{50}{9}\left[\frac{10^{99}-1-9}{9}-98\right]+7 \times 99 \end{aligned} $$
$$ \begin{aligned} & =\frac{7 \times 10^{100}}{9}-\frac{70}{9}+\frac{50}{9}[\overbrace{111 \ldots 1}^{99}-99]+7 \times 99 \\\\ & =\frac{7 \times 10^{100}-70+\overbrace{555 \ldots 50}^{99}}{9}-550+693 \end{aligned} $$
$$ \begin{aligned} & =\frac{7 \overbrace{555 \ldots 5}^{99}-70+143 \times 9}{9} \\\\ & =\frac{7 \overbrace{55 \ldots 5}^{99}+1210}{9} \end{aligned} $$
$$ \therefore $$ $$ m+n=1219 $$
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