JEE Advanced 2023 Paper 1Question 1
Let $7\overbrace{5\cdots5}^{r}7$ 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 $S=77+757+7557+\cdots+7\overbrace{5\cdots5}^{98}7$. If $S=\frac{7\overbrace{5\cdots5}^{99}7+m}{n}$, where $m$ and $n$ are natural numbers less than 3000, then the value of $m+n$ is ______.
Sequence and Series · Geometric Progression
Answer$1219$
Solution
Concept. Write each number as a formula in $r$. A block of $r$ fives is $\dfrac{5(10^r-1)}{9}$, so every term becomes a power of $10$ plus a constant, and the sum is a G.P. Formulas used. (1) $\underbrace{5\cdots5}_{r}=\frac{5\left(10^r-1\right)}{9}$(2) $10+10^2+\cdots+10^{n}=\frac{10\left(10^n-1\right)}{9}$Step 1 - One term. The number $7\overbrace{5\cdots5}^{r}7$ is $7\cdot10^{r+1}+10\cdot\underbrace{5\cdots5}_{r}+7$: $$N_r=7\cdot10^{r+1}+\frac{50(10^r-1)}{9}+7=\frac{63\cdot10^{r+1}+5\cdot10^{r+1}-50+63}{9}=\frac{68\cdot10^{r+1}+13}{9}.$$ Check: $N_1=\dfrac{6800+13}{9}=757$. Step 2 - The sum ($r=0$ to $98$, i.e. $99$ terms): $$S=\frac19\left[68\left(10+10^2+\cdots+10^{99}\right)+13\cdot99\right]=\frac19\left[\frac{680\left(10^{99}-1\right)}{9}+1287\right].$$ Step 3 - Bring in $N_{99}$. $N_{99}=\dfrac{68\cdot10^{100}+13}{9}=\dfrac{680\cdot10^{99}+13}{9}$. So $$9S=\frac{680\cdot10^{99}+13}{9}-\frac{13+680}{9}+1287=N_{99}-77+1287=N_{99}+1210 .$$ Step 4 - Read off. $$S=\frac{7\overbrace{5\cdots5}^{99}7+1210}{9},$$ so $m=1210,\ n=9$ (both less than $3000$): $$m+n=1219 .$$ Answer: 1219