QJEEAdv23Ma3c0e: JEE Advanced 2023 Mathematics - Class 11 JEEAdv23Ma3c0e Solution: Let denote the digit number where the first and the last digits …

Question

Verified Solution • 2025–26 Session moderate

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

Accepted Answer

S=77+757+7557+\cdots+ 7\overbrace{5 \cdots 5}^{98}7

\begin{aligned}&amp; =7\left(10+10^2+\ldots+10^{99}\right)+50(1+11+\ldots+\overbrace{111 \ldots 1}^{98})+7 \times 99 \\\\&amp; =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}&amp; =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 \\\\&amp; =\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}&amp; =\frac{7 \times 10^{100}}{9}-\frac{70}{9}+\frac{50}{9}[\overbrace{111 \ldots 1}^{99}-99]+7 \times 99 \\\\&amp; =\frac{7 \times 10^{100}-70+\overbrace{555 \ldots 50}^{99}}{9}-550+693\end{aligned}

\begin{aligned}&amp; =\frac{7 \overbrace{555 \ldots 5}^{99}-70+143 \times 9}{9} \\\\&amp; =\frac{7 \overbrace{55 \ldots 5}^{99}+1210}{9}\end{aligned}

\therefore

m+n=1219

Step-by-Step Verified Solution