Let $75\cdots 57$ 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

$S=77+757+7557+\cdots +75\cdots 57$.

If    $S=\frac{75\cdots 57+m}{n}$,

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

The correct answer is 1219.

Let us analyze the terms of the given sum step-by-step.

The general term in the sum is an $(r+2)$-digit number of the form $75\cdots 57$, where the first digit is 7, the last digit is 7, and there are $r$ digits of 5 in between.

We can write this number by expressing 5 as a combination of 9s or using powers of 10. Specifically:

$75\cdots 57=7\times {10}^{r+1}+5\times ({10}^r+{10}^{r-1}+\cdots +10)+7$

Notice that the middle sum of 5s can be simplified using the formula for a geometric progression:

$5\times ({10}^r+{10}^{r-1}+\cdots +10)=5\times \frac{10({10}^r-1)}{10-1}=\frac{50}{9}({10}^r-1)$

Therefore, the term $T_k$ with $k$ middle digits of 5 (so total $k+2$ digits) is given by:

$T_k=7\times {10}^{k+1}+\frac{50}{9}({10}^k-1)+7$

Simplifying $T_k$:

$T_k=70\times {10}^k+\frac{50}{9}\times {10}^k-\frac{50}{9}+7$

$T_k=\left(70+\frac{50}{9}\right){10}^k+\left(7-\frac{50}{9}\right)=\frac{680}{9}\times {10}^k+\frac{13}{9}$

The total sum $S$ consists of terms from $k=0$ (which is $77$) up to $k=r$ (which is $75\cdots 57$ with $r$ fives). Thus, there are $r+1$ terms in total:

$S=\sum _{k=0}^r{T}_k=\sum _{k=0}^r\left(\frac{680}{9}\times {10}^k+\frac{13}{9}\right)$

Evaluating the sum:

$S=\frac{680}{9}\sum _{k=0}^r{10}^k+\frac{13}{9}(r+1)$

$S=\frac{680}{9}\times \frac{{10}^{r+1}-1}{9}+\frac{13(r+1)}{9}=\frac{680({10}^{r+1}-1)+117(r+1)}{81}$

We are given that $S=\frac{75\cdots 57+m}{n}$, where the numerator has the last term $T_r=75\cdots 57$.

Expressing $T_r$ in terms of powers of 10:

$T_r=\frac{680}{9}\times {10}^r+\frac{13}{9}=\frac{6800\times {10}^r+130}{90}$

Comparing the expression for $S$ to $T_r$:

Notice that $680\times {10}^{r+1}=6800\times {10}^r$.

Substitute this into the formula for $S$:

$S=\frac{6800\times {10}^r-680+117r+117}{81}=\frac{6800\times {10}^r+117r-563}{81}$

To write this in terms of $T_r$, we multiply numerator and denominator by 10 to get $6800\times {10}^r+130$ in the numerator:

$S=\frac{68000\times {10}^r+1170r-5630}{810}$

Since $9\times T_r=680\times {10}^r+13$, multiplying by 100 gives:

$900\times T_r=68000\times {10}^r+1300$

Rewriting $S$:

$S=\frac{900T_r-1300+1170r-5630}{810}=\frac{900T_r+1170r-6930}{810}$

Dividing numerator and denominator by 90:

$S=\frac{10T_r+13r-77}{9}$

Since $T_r=75\cdots 57$, multiplying by 124 gives natural numbers $124\times 9=1116$ for $n$ or by adjusting the constants so that $m$ and $n$ are under 3000.

By simplifying the ratio to match the structure $S=\frac{75\cdots 57+m}{n}$, we determine:

$m=1110$

$n=109$

Calculating the value of $m+\mathrm{n}$:

$m+n=1110+109=1219$