The remainder when `3^0 + 3^1 + 3^2 + ….. + 3^(200)` is divided by 13 is :

Options

A. 0
B. 12
C. 3
D. none of these

Answer

To find the remainder when \(3^0 + 3^1 + 3^2 + \ldots + 3^{200}\) is divided by 13, we can follow these steps:

### Step 1: Identify the sum of the series
The expression \(3^0 + 3^1 + 3^2 + \ldots + 3^{200}\) is a geometric series. The sum \(S\) of a geometric series can be calculated using the formula:
\[ S = a \frac{(r^n - 1)}{(r - 1)} \]
where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.

Here, \(a = 3^0 = 1\), \(r = 3\), and the number of terms \(n = 200 - 0 + 1 = 201\).

### Step 2: Calculate the sum
Using the formula:
\[ S = 1 \cdot \frac{(3^{201} - 1)}{(3 - 1)} = \frac{(3^{201} - 1)}{2} \]

### Step 3: Find the remainder when divided by 13
Now, we need to find the remainder of \(S\) when divided by 13. This requires finding \(3^{201} \mod 13\).

### Step 4: Use Fermat's Little Theorem
According to Fermat's Little Theorem, since 13 is prime, we have:
\[ a^{p-1} \equiv 1 \mod p \]
for any integer \(a\) not divisible by \(p\). Here, \(p = 13\) and \(a = 3\):
\[ 3^{12} \equiv 1 \mod 13 \]

### Step 5: Reduce the exponent modulo 12
To find \(3^{201} \mod 13\), we first reduce 201 modulo 12:
\[ 201 \div 12 = 16 \quad \text{(remainder 9)} \]
Thus, \(201 \equiv 9 \mod 12\), which means:
\[ 3^{201} \equiv 3^9 \mod 13 \]

### Step 6: Calculate \(3^9 \mod 13\)
Now we need to compute \(3^9 \mod 13\):
\[ 3^1 = 3 \]
\[ 3^2 = 9 \]
\[ 3^3 = 27 \equiv 1 \mod 13 \]
\[ 3^6 = (3^3)^2 \equiv 1^2 \equiv 1 \mod 13 \]
\[ 3^9 = 3^6 \cdot 3^3 \equiv 1 \cdot 1 \equiv 1 \mod 13 \]

### Step 7: Substitute back into the sum
Now we substitute back into our sum:
\[ S = \frac{(3^{201} - 1)}{2} \equiv \frac{(1 - 1)}{2} \equiv \frac{0}{2} \equiv 0 \mod 13 \]

### Conclusion
Thus, the remainder when \(3^0 + 3^1 + 3^2 + \ldots + 3^{200}\) is divided by 13 is:
\[ \boxed{0} \]