If `2^2006 - 2006` divided by 7, the remainder is

Options

A. a) 0
B. b) 1
C. c) 2
D. d) 4

Answer

To solve the problem of finding the remainder when \(2^{2006} - 2006\) is divided by 7, we can break it down into two parts: finding the remainder of \(2^{2006}\) when divided by 7 and finding the remainder of \(2006\) when divided by 7. Finally, we will combine these results to find the overall remainder.

### Step-by-step Solution:

1. **Finding \(2^{2006} \mod 7\)**:
- 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, \(p = 7\) and \(a = 2\). Since 2 is not divisible by 7, we can apply the theorem:
\[ 2^{6} \equiv 1 \mod 7 \]
- Now, we need to find \(2006 \mod 6\) (since \(6\) is \(7-1\)):
\[ 2006 \div 6 = 334 \quad \text{(with a remainder of 2)} \]
- Thus, \(2006 \mod 6 = 2\).
- Therefore:
\[ 2^{2006} \equiv 2^2 \mod 7 \]
- Calculating \(2^2\):
\[ 2^2 = 4 \]
- So:
\[ 2^{2006} \mod 7 = 4 \]

2. **Finding \(2006 \mod 7\)**:
- Now, we divide \(2006\) by \(7\):
\[ 2006 \div 7 = 286 \quad \text{(with a remainder of 4)} \]
- Thus:
\[ 2006 \mod 7 = 4 \]

3. **Combining the results**:
- Now we need to find the remainder of \(2^{2006} - 2006\) when divided by \(7\):
\[ 2^{2006} - 2006 \equiv 4 - 4 \mod 7 \]
- This simplifies to:
\[ 0 \mod 7 \]

### Final Answer:
The remainder when \(2^{2006} - 2006\) is divided by \(7\) is \(0\).