Tournament Integer!

First, after all, this is an awesome problem, thanks for compiling it , Mursalin!

Suppose $m$ is a positive integer with $n$ digits.

$\bar{m}\bar{m}$ can be written as $m(10^{n} + 1)$

Thus, $\frac{m(10^{n} + 1)}{m^{2}} = k$

$\frac{10^{n} + 1}{m} = k$ --- Equation (1)

For $n = 1$ , we can clearly see that $m = \{1, 11\}$ , but the only acceptable $m$ is $1$ because $m$ must be a one-digit integer in this case. So, if $m = 1$ , that leads to $\boxed{k = 11}$

For $n = 2$ , we can clearly see that $m = \{1, 101\}$ , since $101$ is a prime number. But there are no acceptable $m$ s, because $m$ must be a two-digit integer.

For $n = 3$ , trial and error shows that $1001 = 7 \cdot 11 \cdot 13$ . This shows that $m=\{1,7,11,13,77,91,143,1001\}$ . But the only acceptable $m$ is $143$ because $m$ must be a three-digit integer in this case. So if $m = 143$ , that leads to $\boxed{k = 7}$

For $n \geq 4$ , we want to find the lowest upper bound of $k$ to limit the boundaries of $k$ , by doing that we must maximize $k$ by minimizing $m$ . The minimum value of an $n$ -digit integer is 1000...0 ( $n$ digits) = $10^{n-1}$ . So, the maximum value of $k$ is $\frac{10^{n} + 1}{10^{n-1}} = 10 + \frac{1}{10^{n-1}}$ . And we know that $10^{n-1} < 1$ , for all $n \geq 4$ . So,the boundaries of $k$ is $[0,10]$ , when $k$ is an integer.

Under the same circumstances, for $n \geq 4$ , we will rewrite the Equation (1) as $\frac{10^{n} + 1}{k} = m$ ---Equation (1)'. We will now then use the properties of parity to continue.

$10^{n} + 1$ is always an odd integer. Thus $k$ must also be an odd integer for $m$ to be an integer. So, we cancel out the even integers and our used value $7$ , and that leaves $\{1, 3, 5, 9\}$ . We cancel out $7$ because even though it will work, it will just be a repeating value and that doesn't affect our answer.

For $k = 1$ , $\forall n[k | 10^{n} + 1]$ , substituting into the equation (1)', we will get $m = 10^{n} + 1$ , which has $n+1$ digits. This contradicts the statement that $m$ has $n$ digits. So, goodbye $1$

For $10^{n} + 1$ to be divisible by $3$ , its digits' sum must be a multiple of $3$ . However, the digits' sum is always $2$ . So, goodbye $3$ .

For $10^{n} + 1$ to be divisible by $5$ , its last digit must be either $0$ or $5$ . However its last digit here is 1. So, goodbye $5$ .

For $10^{n} + 1$ to be divisible by $9$ , its digits' sum must be a multiple of $9$ . However, the digits' sum is always $2$ . So, goodbye $9$ .

Thus, $\forall n \geq 4 [m \nmid 10^{n} + 1]$

Thus, $k = \{7, 11\}$ (And this makes me think of 7-11 shops)

$k_{sum} = \boxed{18}!$

If there are any flaws, please let me know in the comments.

PS.: I'm stuck with this question for an hour, and I decided to turn on some music. Thanks to "Let it go" for cheering me up. XD

For m=1 we have 11:1=k=11

For m>1 we have m has a digits, so 10^(a-1)<= m <10^a (1)

We also have mm=(10^a+1)m, so from the problem we have (10^a+1) is divided by m (2)

From (1) and (2) we have 1<k<10

Because (10^a+1) is not divided by 2,3,5, so k=7. This happens when a=3(mod6)

Sorry for the inconvenience but I don't know how to use Latex