JEE Advanced 2023 · Paper 1 · Mathematics Q10

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MathematicsInteger+4Question 10 of 51

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 $S = \dfrac{7\overbrace{5\cdots5}^{99}7 + m}{n}$ , where $m$ and $n$ are natural numbers less than 3000, then the value of $m + n$ is _____.

Your answer

Answer and worked solution

Official answer: 1219

Closed form. With $r$ fives,
$7\underbrace{5\cdots5}_r7=7\cdot10^{r+1}+\frac{50(10^r-1)}{9}+7=\frac{68\cdot10^{r+1}+13}{9}.$
Check: $r=0$ gives 77, and $r=1$ gives 757.

Sum over $r=0,\dots,98$ :
$S=\frac19\left[68\cdot\frac{10^{100}-10}{9}+13\cdot99\right].$

In terms of the last number. Let $N=7\underbrace{5\cdots5}_{99}7=\frac{68\cdot10^{100}+13}{9}$ , so $68\cdot10^{100}=9N-13$ . Then
$S=\frac19\left[\frac{9N-13-680}{9}+1287\right]=\frac{9N+10890}{81}=\frac{N+1210}{9}.$

So $m=1210$ , $n=9$ , and $m+n=1219$ .

Topics:Special series sums Geometric progression

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JEE Advanced 2023 · Paper 1, Mathematics question 10