When `10^1 + 10^2 + 10^3 + ....... + 10^32` is divided by 6, the remainder obtained is ?

Options

A. 4
B. 5
C. 2
D. 1

Answer

To solve the problem of finding the remainder when \( 10^1 + 10^2 + 10^3 + \ldots + 10^{32} \) is divided by 6, we can follow these steps:

### Step 1: Find the pattern of \( 10^n \mod 6 \)

First, we need to calculate the values of \( 10^n \) modulo 6 for the first few powers of \( n \):

- \( 10^1 \mod 6 = 4 \)
- \( 10^2 \mod 6 = 4 \)
- \( 10^3 \mod 6 = 4 \)
- \( 10^4 \mod 6 = 4 \)

From the calculations, we can observe that \( 10^n \mod 6 = 4 \) for all \( n \geq 1 \).

### Step 2: Sum the results

Since \( 10^n \mod 6 = 4 \) for each \( n \) from 1 to 32, we can sum these results:

\[ 10^1 + 10^2 + 10^3 + \ldots + 10^{32} \mod 6 = 4 + 4 + 4 + \ldots + 4 \quad (\text{32 times}) \]

This can be expressed as:

\[ 4 \times 32 = 128 \]

### Step 3: Find the remainder of the total sum when divided by 6

Now, we need to find \( 128 \mod 6 \):

\[ 128 \div 6 = 21 \quad \text{(whole number part)} \]
\[ 128 - (21 \times 6) = 128 - 126 = 2 \]

Thus, the remainder when \( 128 \) is divided by \( 6 \) is \( 2 \).

### Final Answer

Therefore, the remainder obtained when \( 10^1 + 10^2 + 10^3 + \ldots + 10^{32} \) is divided by \( 6 \) is \( \boxed{2} \).

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