JEE Advanced 2023 — Paper 1 — MATH-III solution

JEE Advanced2023Paper 1MATH-III

Let $r 75 \dots 57$ 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 $S = 77 + 757 + 7557 + \dots + 98 75 \dots 57$ . If $S = \frac{99 75 \dots 57 + m}{n}$ , where $m$ and $n$ are natural numbers less than $3000$ , then the value of $m + n$ is

Your answer

Solution

$S = 77 + 757 + 7557 + \dots + 98 digits 75 \dots 57$

$10 S = 770 + 7570 + \dots + 99 digits 75 \dots 570$

$9 S = - 77 + 98 times 13 + 13 + \dots + 13 + 99 digits 75 \dots 570$

$= - 77 + 13 \times 98 + 99 digits 75 \dots 57 + 13$

$= 99 digits 75 \dots 57 + 1210$

$S = \frac{99 digits 75 \dots 57 + 1210}{9}$

$m = 1210$

$n = 9$

$m + n = 1219$

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