2024/022) N is a 50 digit no. all digit except the 26^th (from left) are 1.if n is divisible by 13,find the 26^th digit?

Let the $26^{th}$ digit be k

so $N = 1111\cdots( 50\, times) + (k-1) * 10^{24}$

or $N = \frac{10^{50}-1}{9} + (k-1) * 10^{24}$

or     $9N = 10^{50}-1 + 9 *  ((k-1) * 10^{24})$

because $GCD(9,13)=1$ so if n is divisible by 13 9N is divisible by 13.

as 13 ia prime we have

$10^{12} \equiv 1 \pmod {13}$

so  

$10{24} \equiv 1 \pmod {13}\cdots(1)$

 and

$10{48} \equiv 1 \pmod {13}$

hence

 $10{50} \equiv 100 \pmod {13}$

 or

 $10{50} \equiv 9 \pmod {13}$

$9N = 10^{50}-1 + 9 *  ((k-1) * 10^{24}) \pmod {13}$

$= 9-1  + 9 *  ((k-1) * 1) \pmod {13}$

 $= 9k -1 \pmod {13}$

this is zero and because k is a digit trying from 0 to 9 we get k=3 satisfies the condition

so the $26^{th}$ digit is 3