Find the remainder when `10^(1) + 10^2 + 10^3 + 10^4 + 10^5 + …….. + 10^(99)` is divided by 6.

Answer

To find the remainder when \( 10^1 + 10^2 + 10^3 + \ldots + 10^{99} \) is divided by 6, we can follow these steps:

### Step 1: Find the remainder of \( 10^n \) when divided by 6
First, we will calculate the remainder of \( 10^n \) for various values of \( n \) when divided by 6.

- For \( n = 1 \):
\[ 10^1 = 10 \quad \Rightarrow \quad 10 \div 6 = 1 \quad \text{(remainder 4)} \]

- For \( n = 2 \):
\[ 10^2 = 100 \quad \Rightarrow \quad 100 \div 6 = 16 \quad \text{(remainder 4)} \]

- For \( n = 3 \):
\[ 10^3 = 1000 \quad \Rightarrow \quad 1000 \div 6 = 166 \quad \text{(remainder 4)} \]

From these calculations, we observe that the remainder of \( 10^n \) when divided by 6 is consistently 4 for \( n = 1, 2, 3 \).

### Step 2: Generalize the result
Since \( 10 \equiv 4 \mod 6 \), we can conclude that:
\[ 10^n \equiv 4 \mod 6 \quad \text{for all } n \geq 1 \]

### Step 3: Sum the series
Now, we need to sum the series:
\[ 10^1 + 10^2 + 10^3 + \ldots + 10^{99} \]
Since each term \( 10^n \) contributes a remainder of 4, we can write:
\[ \text{Total remainder} = 4 + 4 + 4 + \ldots + 4 \quad \text{(99 times)} \]
This can be expressed as:
\[ \text{Total remainder} = 4 \times 99 = 396 \]

### Step 4: Find the remainder of the total sum when divided by 6
Now, we need to find the remainder of 396 when divided by 6:
\[ 396 \div 6 = 66 \quad \text{(remainder 0)} \]

### Conclusion
Thus, the remainder when \( 10^1 + 10^2 + 10^3 + \ldots + 10^{99} \) is divided by 6 is:
\[ \boxed{0} \]