111111.......

As usual my solution is the worst and involves extreme bashing.I wont even bother to write it up nicely.

Given $N = 1111...a..1$ with $1$ occurring $25$ times before $a$ and $24$ times after it.Expanding it out we get

$N = (10^{49} + 10^{48} + \cdots + 10^{25}) + (a \times 10^{24}) + (10^{23} + \cdots + 1) = \dfrac{10^{25} ( 10^{25} - 1 ) }{9} + a \times 10^{24} + \dfrac{10^{24}-1}{9} = \dfrac{10^{25}( 10^{25}-1) + 9\times a\times 10^{24} + 10^{24} - 1}{9}$ If we use the Geometric Progression Sum formula.

Now clearly $9$ $\mid$ $\text{numerator}$ .Because of Fermat 's Little theorem.We must also have that $13$ $\mid$ $\text{numerator}$ .

Again using FLT we have $10^{12} \equiv 1 \pmod {13}$ . Now u have to use this congruence on the numerator and deduce that $9 \times a \times 10^{24} \equiv 1 \pmod {13}$ . Since we already have $10^{24} \equiv 1 \pmod {13}$ from FLT $\implies 9 \times a \equiv 1 \pmod {13}$ . Since $a$ is a digit thus $0 \leq a \leq 9$ .

Make $10$ cases for $a$ put each one of them and see which one leaves remainder $1$ on division by $13$ .

Clearly $9 \times 3 = 27 \equiv 1 (\mod 13)$ thus answer is $3$ .

The number is 1111 (24 times) 1 x (1111 24 times) ------ 26 th digit is x

Test of divisibility for 7, 11 and 13

step 1 make groups of 3 digits from the end

step 2 add groups at odd places and add groups at even piaces

step 3 find the difference in above two additions

If difference is 0 or divisible by 7, 11 or 13 then the given no is divisible by 7, 11 or 13

in above case first 24 digits (8 groups) and last 24 digits divisible divisible by 13

so remaining (1 x) must be divisible by 13 so x = 3

if it is asked 26 digit from right then x = 9 to make 91 divisible by 13

Divisibility test of 13 can be used here.

Last 24 digits will be cancelled off by using the method.

Now if I observe every 6 1's from right gets divided by 13 completely.

So now we are left with 25th and 26th digits.

So it is clear that with unit digit 1 only 13 is the multiple so 3 is the answer.

We know that the cyclicity of 13 is 6. Therefore, any six digit number with the same digits will always be divisible by 13. Thus the set of first 24 and last 24 digits must always be divisible by 13. Lets consider 25th and 26th digit. Let the 26th digit be X. ∴ The remainder obtained when (25th & 26th digit is divided by 13 is 10 + x. ∴ x = 3 so that the remainder will become 13 ≡ 0.