What is the remainder when `2^(2010)` is divided by 7 ?

Answer

To find the remainder when \( 2^{2010} \) is divided by 7, we can use Fermat's Little Theorem, which states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then:

\[ a^{p-1} \equiv 1 \mod p \]

In our case, \( a = 2 \) and \( p = 7 \). Since 2 is not divisible by 7, we can apply the theorem.

### Step 1: Apply Fermat's Little Theorem

According to Fermat's Little Theorem:

\[ 2^{7-1} \equiv 1 \mod 7 \]

This simplifies to:

\[ 2^6 \equiv 1 \mod 7 \]

### Step 2: Reduce the exponent modulo 6

Next, we need to reduce the exponent 2010 modulo 6, since \( 2^6 \equiv 1 \mod 7 \).

\[ 2010 \mod 6 \]

Calculating \( 2010 \div 6 \):

\[ 2010 = 6 \times 335 + 0 \]

This means:

\[ 2010 \mod 6 = 0 \]

### Step 3: Substitute back into the expression

Now we can substitute back into our expression using the result from Step 1:

\[ 2^{2010} = 2^{6 \times 335} = (2^6)^{335} \]

Since \( 2^6 \equiv 1 \mod 7 \):

\[ (2^6)^{335} \equiv 1^{335} \equiv 1 \mod 7 \]

### Conclusion

Thus, the remainder when \( 2^{2010} \) is divided by 7 is:

\[ \boxed{1} \]