The remaider when `4^0 + 4^1 + 4^2 + 4^3 + …… + 4^10` is divided by 17 is:

Options

A. a. 0
B. b. 16
C. c. 4
D. d. none of these

Answer

To solve the problem of finding the remainder when \( 4^0 + 4^1 + 4^2 + 4^3 + \ldots + 4^{10} \) is divided by 17, we can follow these steps:

### Step 1: Identify the series
The series can be expressed as:
\[ S = 4^0 + 4^1 + 4^2 + 4^3 + \ldots + 4^{10} \]
This is a geometric series where the first term \( a = 4^0 = 1 \) and the common ratio \( r = 4 \).

### Step 2: Use the formula for the sum of a geometric series
The sum 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, \( n = 11 \) (from \( 0 \) to \( 10 \)), so:
\[ S = 1 \cdot \frac{4^{11} - 1}{4 - 1} = \frac{4^{11} - 1}{3} \]

### Step 3: Calculate \( 4^{11} \mod 17 \)
To find \( S \mod 17 \), we first need to calculate \( 4^{11} \mod 17 \). We can use Fermat's Little Theorem, which states that if \( p \) is a prime and \( a \) is not divisible by \( p \), then:
\[ a^{p-1} \equiv 1 \mod p \]
Here, \( p = 17 \) and \( a = 4 \), so:
\[ 4^{16} \equiv 1 \mod 17 \]
Thus, we can reduce \( 4^{11} \) directly:
\[ 4^{11} \mod 17 \]
To compute \( 4^{11} \mod 17 \), we can calculate lower powers of 4:
\[ 4^1 \equiv 4 \mod 17 \]
\[ 4^2 \equiv 16 \mod 17 \]
\[ 4^3 \equiv 64 \equiv 13 \mod 17 \quad (\text{since } 64 - 3 \cdot 17 = 13) \]
\[ 4^4 \equiv 4 \cdot 13 \equiv 52 \equiv 1 \mod 17 \quad (\text{since } 52 - 3 \cdot 17 = 1) \]
Since \( 4^4 \equiv 1 \mod 17 \), we can simplify \( 4^{11} \):
\[ 4^{11} = (4^4)^2 \cdot 4^3 \equiv 1^2 \cdot 13 \equiv 13 \mod 17 \]

### Step 4: Substitute back into the sum formula
Now substitute \( 4^{11} \) back into the sum formula:
\[ S \equiv \frac{4^{11} - 1}{3} \mod 17 \]
\[ S \equiv \frac{13 - 1}{3} \equiv \frac{12}{3} \equiv 4 \mod 17 \]

### Step 5: Conclusion
Thus, the remainder when \( 4^0 + 4^1 + 4^2 + \ldots + 4^{10} \) is divided by 17 is:
\[ \boxed{4} \]