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

Let $$N_r$$ denote the number whose first and last digits are 7 and whose $$r$$ middle digits are all 5. Thus the total number of digits is $$(r+2)$$ and

$$N_r = 7\cdot 10^{\,r+1} + 5\Bigl(10^{\,r-1}+10^{\,r-2}+\dots+10^1\Bigr) + 7.$$

The parenthesis is a geometric series of $$r$$ terms with first term $$10^{\,r-1}$$ and ratio 1/10, whose sum equals $$\dfrac{10^{\,r}-1}{9} \times 10.$$ Therefore

$$N_r \;=\; 7\cdot 10^{r+1} \;+\; \dfrac{50}{9}\,(10^{\,r}-1) \;+\; 7.$$

The required sum is

$$S \;=\; \sum_{r=0}^{98} N_r.$$ We add $$S$$ term-wise by separating coefficient-wise parts.

First part (leading 7): $$S_1 = 7\sum_{r=0}^{98}10^{\,r+1}=70\sum_{r=0}^{98}10^{\,r} = 70\cdot\dfrac{10^{99}-1}{9}.$$

Second part (middle 5’s): $$S_2 = \dfrac{50}{9}\sum_{r=0}^{98}(10^{\,r}-1) = \dfrac{50}{9}\Bigl(\dfrac{10^{99}-1}{9}-99\Bigr) = \dfrac{50}{81}\bigl(10^{99}-892\bigr).$$

Third part (trailing 7): $$S_3 = 7\times 99 = 693.$$

Putting $$A = 10^{99},$$ we get

$$S = S_1+S_2+S_3 = \dfrac{70A}{9}-\dfrac{70}{9}+\dfrac{50A}{81}-\dfrac{50\!\cdot\!892}{81}+693 = \dfrac{680A + 10\,903}{81}.$$

Next write the term with 99 middle 5’s, i.e. $$N_{99}$$ (the number $$\overset{99}{75\ldots57}$$).

Since $$10^{100}=10A,$$ we have

$$N_{99}=7\cdot10^{100}+\dfrac{50}{9}(10^{99}-1)+7 = 70A+\dfrac{50A-50}{9}+7 = \dfrac{680A+13}{9}.$$

We are told that

$$S = \dfrac{\,N_{99}+m\,}{n}, \qquad m,n\in\mathbb{N},\;m,n\lt3000.$$ Substituting the obtained expressions:

$$\dfrac{680A+10\,903}{81} = \dfrac{\dfrac{680A+13}{9}+m}{n}.$$

Cross-multiplying,

$$n(680A + 10\,903) = 9(680A + 13) + 81m.$$

Equating coefficients of $$A$$ gives $$680n = 9\!\times\!680 \;\Longrightarrow\; n = 9.$$

Equating the constant terms then yields

$$10\,903n = 117 + 81m \;\Longrightarrow\; 10\,903\!\times\!9 = 117 + 81m \;\Longrightarrow\; 98\,127 = 117 + 81m \;\Longrightarrow\; 81m = 98\,010 \;\Longrightarrow\; m = 1\,210.$$

Both $$m$$ and $$n$$ are below 3000, so they satisfy the given restriction.

Finally, $$m + n = 1\,210 + 9 = 1\,219.$$

Hence the required value is 1219.