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 determine the 26th digit of the 50-digit number \( N \) which consists of all digits being 1 except for the 26th digit. The number \( N \) must be divisible by 13.

Let's denote the 26th digit as \( A \). The number \( N \) can be expressed as:

\[ N = 111...1A111...1 \]

where there are 25 ones before \( A \) and 24 ones after \( A \).

### Step 1: Express \( N \) in terms of powers of 10
The number \( N \) can be represented in expanded form as:

\[ N = 1 \times 10^{49} + 1 \times 10^{48} + \ldots + 1 \times 10^{25} + A \times 10^{24} + 1 \times 10^{23} + \ldots + 1 \times 10^0 \]

### Step 2: Simplify the expression
The sum of the powers of 10 can be simplified. The first part (the sum of the first 25 ones) is a geometric series:

\[ 1 \times (10^{49} + 10^{48} + \ldots + 10^{25}) + A \times 10^{24} + 1 \times (10^{23} + \ldots + 10^0) \]

The sum of the first 25 terms is:

\[ \frac{10^{50} - 10^{25}}{9} \]

And the sum of the last 24 terms is:

\[ \frac{10^{24} - 1}{9} \]

Thus, we can write \( N \) as:

\[ N = \frac{10^{50} - 10^{25}}{9} + A \times 10^{24} + \frac{10^{24} - 1}{9} \]

Combining these gives:

\[ N = \frac{10^{50} - 10^{25} + A \times 10^{24} + 10^{24} - 1}{9} \]

### Step 3: Find \( N \mod 13 \)
To check if \( N \) is divisible by 13, we need to evaluate \( N \mod 13 \).

Calculating powers of 10 modulo 13, we find:

- \( 10^0 \equiv 1 \)
- \( 10^1 \equiv 10 \)
- \( 10^2 \equiv 9 \)
- \( 10^3 \equiv 12 \)
- \( 10^4 \equiv 3 \)
- \( 10^5 \equiv 4 \)
- \( 10^6 \equiv 1 \) (and it repeats every 6)

Using this periodicity, we can find \( 10^{24} \mod 13 \):

- \( 10^{24} \equiv 1 \) (since \( 24 \mod 6 = 0 \))

Now, we can substitute back into the expression for \( N \):

\[ N \equiv \frac{10^{50} - 10^{25} + A \times 1 + 1 - 1}{9} \mod 13 \]

Calculating \( 10^{50} \mod 13 \) and \( 10^{25} \mod 13 \):

- \( 10^{50} \equiv 1 \)
- \( 10^{25} \equiv 10 \)

Thus,

\[ N \equiv \frac{1 - 10 + A \times 1}{9} \mod 13 \]
\[ N \equiv \frac{A - 9}{9} \mod 13 \]

### Step 4: Set up the divisibility condition
For \( N \equiv 0 \mod 13 \):

\[ A - 9 \equiv 0 \mod 13 \]

This implies:

\[ A \equiv 9 \mod 13 \]

### Step 5: Determine possible values for \( A \)
The possible values for \( A \) (which is a digit) that satisfy \( A \equiv 9 \mod 13 \) are:

- \( A = 9 \)

### Conclusion
Thus, the 26th digit \( A \) must be 9 for \( N \) to be divisible by 13.

**Final Answer: The 26th digit is 9.**