The digit 6 is written continuously 225 times to form a natural number N. What's the remainder when N is divided by 455?

Options

A. (A) 230
B. (B) 5
C. (C) 211
D. (D) none of these

Answer

To solve the problem of finding the remainder when the number \( N \), formed by writing the digit 6 continuously 225 times, is divided by 455, we can follow these steps:

### Step 1: Understand the structure of the number \( N \)
The number \( N \) consists of the digit 6 repeated 225 times. This can be expressed as:
\[ N = 666...6 \quad (\text{225 times}) \]

### Step 2: Express \( N \) mathematically
We can express \( N \) in a more manageable form:
\[ N = 6 \times 10^{224} + 6 \times 10^{223} + 6 \times 10^{222} + \ldots + 6 \times 10^0 \]
This is a geometric series where the first term \( a = 6 \) and the common ratio \( r = 10 \).

### Step 3: Sum the geometric series
The sum \( S \) of a geometric series can be calculated using the formula:
\[ S = a \frac{r^n - 1}{r - 1} \]
where \( n \) is the number of terms. Here, \( n = 225 \), \( a = 6 \), and \( r = 10 \):
\[ S = 6 \frac{10^{225} - 1}{10 - 1} = 6 \frac{10^{225} - 1}{9} \]

### Step 4: Find the remainder when \( N \) is divided by 455
To find \( N \mod 455 \), we can simplify our calculations by breaking down \( 455 \) into its prime factors:
\[ 455 = 5 \times 91 = 5 \times 7 \times 13 \]

We will find \( N \mod 5 \), \( N \mod 7 \), and \( N \mod 13 \) separately and then combine the results using the Chinese Remainder Theorem.

#### Step 4.1: Calculate \( N \mod 5 \)
The last digit of \( N \) is 6. Therefore:
\[ N \mod 5 = 6 \mod 5 = 1 \]

#### Step 4.2: Calculate \( N \mod 7 \)
To find \( N \mod 7 \), we can look at the last three digits of \( N \), which are 666. We calculate:
\[ 666 \mod 7 \]
Calculating \( 666 \div 7 \):
\[ 666 = 7 \times 95 + 1 \]
So:
\[ N \mod 7 = 1 \]

#### Step 4.3: Calculate \( N \mod 13 \)
Again, we look at the last three digits, 666:
\[ 666 \mod 13 \]
Calculating \( 666 \div 13 \):
\[ 666 = 13 \times 51 + 3 \]
So:
\[ N \mod 13 = 3 \]

### Step 5: Use the Chinese Remainder Theorem
Now we have the following system of congruences:
1. \( N \equiv 1 \mod 5 \)
2. \( N \equiv 1 \mod 7 \)
3. \( N \equiv 3 \mod 13 \)

From the first two congruences, since both are \( N \equiv 1 \), we can combine them:
\[ N \equiv 1 \mod 35 \] (since \( 5 \times 7 = 35 \))

Now we solve:
1. \( N \equiv 1 \mod 35 \)
2. \( N \equiv 3 \mod 13 \)

Let \( N = 35k + 1 \). Substituting into the second congruence:
\[ 35k + 1 \equiv 3 \mod 13 \]
This simplifies to:
\[ 35k \equiv 2 \mod 13 \]
Since \( 35 \mod 13 = 9 \):
\[ 9k \equiv 2 \mod 13 \]

To solve for \( k \), we find the multiplicative inverse of 9 modulo 13, which is 3 (since \( 9 \times 3 = 27 \equiv 1 \mod 13 \)):
\[ k \equiv 3 \times 2 \mod 13 \]
\[ k \equiv 6 \mod 13 \]

Thus, \( k = 13m + 6 \) for some integer \( m \). Substituting back:
\[ N = 35(13m + 6) + 1 = 455m + 211 \]

### Final Step: Conclusion
Thus, the remainder when \( N \) is divided by 455 is:
\[ N \mod 455 = 211 \]

### Final Answer
The remainder when \( N \) is divided by 455 is **211**.