N is 50 digit number in decimal form). All digits except the `26^("th")` digit (from left) are 1. If N is divisible by 13, find the `26^("th")` digit.

Answer

To solve the problem, we need to find the 26th digit of a 50-digit number \( N \) that consists of 49 digits as '1' and one digit as \( a \) (the 26th digit). The number \( N \) must be divisible by 13.

### Step-by-Step Solution:

1. **Understanding the Structure of \( N \)**:
The number \( N \) can be represented as:
\[ N = 111...111a111...111 \]
where there are 25 '1's before \( a \) and 24 '1's after \( a \).

2. **Expressing \( N \) Mathematically**:
We can express \( N \) in terms of powers of 10:
\[ N = 10^{49} + 10^{48} + \ldots + 10^{25} + a \cdot 10^{24} + 10^{23} + \ldots + 10^0 \]
This can be simplified to:
\[ N = \sum_{i=0}^{49} 10^i - 10^{24} + a \cdot 10^{24} \]
\[ N = \frac{10^{50} - 1}{9} - 10^{24} + a \cdot 10^{24} \]
\[ N = \frac{10^{50} - 1}{9} + (a - 1) \cdot 10^{24} \]

3. **Finding \( N \mod 13 \)**:
To find the value of \( a \) such that \( N \) is divisible by 13, we need to compute \( N \mod 13 \):
\[ N \equiv \left(\frac{10^{50} - 1}{9} + (a - 1) \cdot 10^{24}\right) \mod 13 \]

4. **Calculating \( 10^{50} \mod 13 \)**:
Using Fermat's Little Theorem, since \( 10^{12} \equiv 1 \mod 13 \):
\[ 10^{50} \equiv 10^{50 \mod 12} \equiv 10^2 \equiv 9 \mod 13 \]
Thus:
\[ 10^{50} - 1 \equiv 9 - 1 \equiv 8 \mod 13 \]
Therefore:
\[ \frac{10^{50} - 1}{9} \equiv \frac{8}{9} \mod 13 \]
To find the modular inverse of 9 modulo 13, we can use the Extended Euclidean Algorithm or trial:
\[ 9 \cdot 3 \equiv 27 \equiv 1 \mod 13 \quad \Rightarrow \quad 9^{-1} \equiv 3 \mod 13 \]
Thus:
\[ \frac{8}{9} \equiv 8 \cdot 3 \equiv 24 \equiv 11 \mod 13 \]

5. **Calculating \( (a - 1) \cdot 10^{24} \mod 13 \)**:
Similarly, we compute \( 10^{24} \mod 13 \):
\[ 10^{24} \equiv 10^{24 \mod 12} \equiv 10^0 \equiv 1 \mod 13 \]
Therefore:
\[ (a - 1) \cdot 10^{24} \equiv (a - 1) \mod 13 \]

6. **Setting Up the Equation**:
Now we have:
\[ N \equiv 11 + (a - 1) \mod 13 \]
For \( N \) to be divisible by 13:
\[ 11 + (a - 1) \equiv 0 \mod 13 \]
Simplifying gives:
\[ a - 1 \equiv -11 \equiv 2 \mod 13 \]
Thus:
\[ a \equiv 3 \mod 13 \]

7. **Finding Possible Values of \( a \)**:
The digit \( a \) must be a single digit (0-9). The only value satisfying \( a \equiv 3 \mod 13 \) within this range is:
\[ a = 3 \]

### Conclusion:
The 26th digit \( a \) of the number \( N \) is \( 3 \).