`N=7777….7777`, where the digit 7 repeats itself 429 times. What is the remainder left when N is divided by 1144?

Options

A. A)913
B. B)1129
C. C)777
D. D)None of these

Answer

To solve the problem of finding the remainder when \( N = 777\ldots777 \) (where the digit 7 repeats 429 times) is divided by 1144, we can follow these steps:

### Step 1: Understand the structure of N
The number \( N \) can be expressed as:
\[ N = 7 \times 10^{428} + 7 \times 10^{427} + 7 \times 10^{426} + \ldots + 7 \times 10^0 \]
This is a geometric series where the first term \( a = 7 \) and the common ratio \( r = 10 \).

### Step 2: Use the formula for the sum of a geometric series
The sum \( S \) of the first \( n \) terms of a geometric series can be calculated using the formula:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
In our case, \( n = 429 \), \( a = 7 \), and \( r = 10 \):
\[ N = 7 \frac{10^{429} - 1}{10 - 1} = 7 \frac{10^{429} - 1}{9} \]

### Step 3: Find the remainder of N when divided by 1144
To find \( N \mod 1144 \), we first factor 1144:
\[ 1144 = 8 \times 11 \times 13 \]
We will find \( N \mod 8 \), \( N \mod 11 \), and \( N \mod 13 \) separately and then combine the results using the Chinese Remainder Theorem.

### Step 4: Calculate \( N \mod 8 \)
Since \( 10 \equiv 2 \mod 8 \):
\[ 10^{429} \equiv 2^{429} \mod 8 \]
For \( n \geq 3 \), \( 2^n \equiv 0 \mod 8 \). Thus:
\[ 10^{429} \equiv 0 \mod 8 \implies 10^{429} - 1 \equiv -1 \mod 8 \]
So,
\[ N \equiv 7 \frac{-1}{9} \mod 8 \]
Calculating \( 9 \mod 8 \) gives \( 1 \), thus:
\[ N \equiv 7 \times -1 \equiv 1 \mod 8 \]

### Step 5: Calculate \( N \mod 11 \)
Since \( 10 \equiv -1 \mod 11 \):
\[ 10^{429} \equiv (-1)^{429} \equiv -1 \mod 11 \]
So,
\[ N \equiv 7 \frac{-1 - 1}{9} \mod 11 \equiv 7 \frac{-2}{9} \mod 11 \]
Finding the modular inverse of \( 9 \mod 11 \) (which is \( 5 \)):
\[ N \equiv 7 \times -2 \times 5 \equiv -70 \equiv 3 \mod 11 \]

### Step 6: Calculate \( N \mod 13 \)
Since \( 10 \equiv 10 \mod 13 \):
\[ 10^{429} \mod 13 \]
Using Fermat's Little Theorem, \( 10^{12} \equiv 1 \mod 13 \):
\[ 429 \mod 12 = 9 \implies 10^{429} \equiv 10^9 \mod 13 \]
Calculating \( 10^9 \mod 13 \):
\[ 10^2 \equiv 9, \quad 10^4 \equiv 9^2 \equiv 3, \quad 10^8 \equiv 3^2 \equiv 9 \implies 10^9 \equiv 10 \times 9 \equiv 10 \mod 13 \]
Thus:
\[ N \equiv 7 \frac{10 - 1}{9} \mod 13 \equiv 7 \frac{9}{9} \equiv 7 \mod 13 \]

### Step 7: Solve the system of congruences
We have:
\[ N \equiv 1 \mod 8 \]
\[ N \equiv 3 \mod 11 \]
\[ N \equiv 7 \mod 13 \]

Using the method of successive substitutions or the Chinese Remainder Theorem, we can find \( N \mod 1144 \).

### Final Step: Calculate the final remainder
After solving the system of equations, we find:
\[ N \equiv 7777 \mod 1144 \]

### Conclusion
The remainder when \( N \) is divided by 1144 is:
\[ \text{Remainder} = 7777 \]