When 666666 ........... 134 times, is divided by 13, the remainder obtained is?

Options

A. 1
B. 3
C. 11
D. 9

Answer

To solve the problem of finding the remainder when the number formed by repeating '666' a total of 134 times is divided by 13, we can follow these steps:

### Step 1: Understand the number formation
The number can be represented as:
\[ N = 666...6 \text{ (134 times)} \]
This can also be expressed as:
\[ N = 6 \times (10^{0} + 10^{1} + 10^{2} + ... + 10^{133}) \]

### Step 2: Use the formula for the sum of a geometric series
The sum of the series \( 10^{0} + 10^{1} + 10^{2} + ... + 10^{133} \) is a geometric series. The formula for the sum of the first \( n \) terms of a geometric series is:
\[ S_n = a \frac{(r^n - 1)}{(r - 1)} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

Here, \( a = 1 \), \( r = 10 \), and \( n = 134 \):
\[ S = \frac{10^{134} - 1}{10 - 1} = \frac{10^{134} - 1}{9} \]

### Step 3: Substitute back into the equation for N
Now substituting back into our equation for \( N \):
\[ N = 6 \times \frac{10^{134} - 1}{9} \]

### Step 4: Find the remainder when N is divided by 13
We need to find \( N \mod 13 \):
\[ N \mod 13 = \left( 6 \times \frac{10^{134} - 1}{9} \right) \mod 13 \]

### Step 5: Calculate \( 10^{134} \mod 13 \)
To find \( 10^{134} \mod 13 \), we can use Fermat's Little Theorem, which states that if \( p \) is a prime and \( a \) is not divisible by \( p \), then:
\[ a^{p-1} \equiv 1 \mod p \]
Here, \( p = 13 \) and \( a = 10 \):
\[ 10^{12} \equiv 1 \mod 13 \]
Now, we need \( 134 \mod 12 \):
\[ 134 \div 12 = 11 \text{ remainder } 2 \]
So, \( 134 \mod 12 = 2 \):
\[ 10^{134} \equiv 10^{2} \mod 13 \]
Calculating \( 10^{2} \):
\[ 10^{2} = 100 \]
Now, find \( 100 \mod 13 \):
\[ 100 \div 13 = 7 \text{ remainder } 9 \]
So, \( 10^{134} \equiv 9 \mod 13 \).

### Step 6: Substitute back to find N mod 13
Now substitute back:
\[ N \mod 13 = \left( 6 \times \frac{9 - 1}{9} \right) \mod 13 \]
Calculating \( 9 - 1 = 8 \):
\[ N \mod 13 = \left( 6 \times \frac{8}{9} \right) \mod 13 \]

### Step 7: Find the modular inverse of 9 mod 13
We need to find the modular inverse of 9 mod 13. We can find it by checking:
\[ 9x \equiv 1 \mod 13 \]
Testing values, we find:
\[ 9 \times 3 = 27 \equiv 1 \mod 13 \]
So, the inverse is 3.

### Step 8: Calculate N mod 13
Now substitute the inverse back:
\[ N \mod 13 = (6 \times 8 \times 3) \mod 13 \]
Calculating \( 6 \times 8 = 48 \):
\[ 48 \times 3 = 144 \]
Now find \( 144 \mod 13 \):
\[ 144 \div 13 = 11 \text{ remainder } 1 \]
Thus, the remainder when \( N \) is divided by 13 is:
\[ \boxed{1} \]