Find the remainder when `2^(2) + 22^(2) + 222^(2) + …… + (222…… 49 times)^(2)` is divided by 9.

Options

A. 0
B. 2
C. 4
D. 6

Answer

To find the remainder when \(2^2 + 22^2 + 222^2 + \ldots + (222\ldots \text{ (49 times)})^2\) is divided by 9, we can follow these steps:

### Step 1: Identify the pattern in the terms
The terms in the series are \(2^2\), \(22^2\), \(222^2\), and so on, up to \(222\ldots\) (49 times) squared. We can express each term in a more manageable form.

### Step 2: Express each term
Each term can be expressed as:
- \(2 = 2\)
- \(22 = 2 \times 11\)
- \(222 = 2 \times 111\)
- \(2222 = 2 \times 1111\)
- ...
- \(222\ldots\) (49 times) = \(2 \times (10^{48} + 10^{47} + \ldots + 10^0)\)

The general term can be written as:
\[ (2 \times (10^k + 10^{k-1} + \ldots + 10^0))^2 \]
where \(k\) goes from 0 to 48.

### Step 3: Calculate the square of each term
The square of each term is:
\[ (2 \times (10^k + 10^{k-1} + \ldots + 10^0))^2 = 4 \times (10^k + 10^{k-1} + \ldots + 10^0)^2 \]

### Step 4: Find the remainder of each term when divided by 9
We can find the remainder of each term when divided by 9.

Using the property of congruences:
- \(10 \equiv 1 \mod 9\)

Thus, \(10^k \equiv 1 \mod 9\) for any \(k\). Therefore:
\[ 10^k + 10^{k-1} + \ldots + 10^0 \equiv (k + 1) \mod 9 \]

### Step 5: Calculate the contribution of each term
For each \(k\) from 0 to 48:
- The \(k\)-th term contributes:
\[ 4 \times (k + 1)^2 \mod 9 \]

### Step 6: Sum the contributions
Now we need to sum up the contributions for \(k = 0\) to \(48\):
\[ \sum_{k=0}^{48} 4 \times (k + 1)^2 \]

### Step 7: Use the formula for the sum of squares
The sum of squares formula is:
\[ \sum_{n=1}^{N} n^2 = \frac{N(N + 1)(2N + 1)}{6} \]
For \(N = 49\):
\[ \sum_{n=1}^{49} n^2 = \frac{49 \times 50 \times 99}{6} \]

### Step 8: Calculate the sum
Calculating:
\[ \frac{49 \times 50 \times 99}{6} = 49 \times 50 \times 16.5 = 40425 \]

### Step 9: Multiply by 4
Now multiply by 4:
\[ 4 \times 40425 = 161700 \]

### Step 10: Find the remainder when divided by 9
Now we need to find \(161700 \mod 9\):
Calculating the sum of the digits:
\[ 1 + 6 + 1 + 7 + 0 + 0 = 15 \quad \text{and} \quad 1 + 5 = 6 \]
Thus, \(161700 \equiv 6 \mod 9\).

### Final Answer
The remainder when \(2^2 + 22^2 + 222^2 + \ldots + (222\ldots \text{ (49 times)})^2\) is divided by 9 is **6**.