Let $75 . . .57$ denote the $(\mathrm{r}+2)$ digit number where the first and the last digits are 7 and the remaining r digits are 5. Consider the sum $\mathrm{S}=77+757+7557+\ldots+75 \ldots 57$. If $\mathrm{S}=\frac{75 \ldots 57+\mathrm{m}}{\mathrm{n}}$, where m and n are natural numbers less than 3000, then the value of m + n is

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Sequences and Series

Let

75...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 + 75 \dots 57

. If

S = \frac{75 \dots 57 + m}{n}

, where m and n are natural numbers less than 3000, then the value of m + n is

Answer: 1219

Let

S = 77 + 757 + 7557 + \dots + 75 \dots 57

(where the last term has 100 digits).

Multiplying by 10:

10 S = 770 + 7570 + 75570 + \dots + 75 \dots 570

Subtracting:

9 S = - 77 + 98 terms (770 - 757) + (7570 - 7557) + \dots + 75 \dots 570

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

9 S = - 77 + 1274 + 75 \dots 57 + 13

9 S = 75 \dots 57 + 1210

Therefore:

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

But we need the form

\frac{75 \dots 57 + m}{n}

where

g cd (m, n)

might not be 1.

From the solution:

9 S = 75 \dots 57 + 1210

, so

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

.

Simplifying:

\frac{1210}{9}

is not an integer, but

1210 = 9 \times 134 + 4

.

Actually, re-examining:

m = 1210, n = 9

gives

m + n = 1219

.

Answer: 1219

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Integer answer