Numbers too easy

13 divides 111111. So up to 24 digits 13 divides the number . From 26th digit to 50th digit 13 divides it. So 13 should divide the 25th and 26th digit hence 26th digit is 3

Powers of ten, mod $13$ , make cycle of length $6$ . the members of the cycle are $1,-3,-4,-1,3,4$ .

The weird number can be written as below.

$(1)^{(25)} \times 10^{25}+x \times 10^{24}+(1)^{(24)}$

where the notation $(1)^{(n)}$ is basically the digit $1$ , written $n$ times in a row. $x$ is the wanted single-digit number.

If $n$ in $(1)^{(n)}$ is a multiple of $6$ , then $(1)^{(n)}$ is divisible by $13$ . So, $(1)^{(24)}$ is zero, modulo $13$ and $(1)^{(25)}$ is $1$ , modulo $13$ .

For powers of ten, the exponent is taken modulo $6$ and the corresponding answer is taken from the list $1,-3,-4,-1,3,4$ .

Therefore,

$(1)^{(25)} \times 10^{25}+x \times 10^{24}+(1)^{(24)}=1 \times -3+x \times 1 + 0 \ mod \ 13$