Find the remainder when `5^(123) ` is divided by 7.

Answer

To find the remainder when \( 5^{123} \) is divided by 7, we can use Fermat's Little Theorem, which is useful for finding remainders in modular arithmetic.

### Step-by-step Solution:

1. **Identify the modulus and the base**:
We need to find \( 5^{123} \mod 7 \).

2. **Check if the base and modulus are coprime**:
Since 5 and 7 are coprime (they have no common factors other than 1), we can apply Fermat's Little Theorem.

3. **Apply Fermat's Little Theorem**:
According to Fermat's Little Theorem, if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then:
\[ a^{p-1} \equiv 1 \mod p \]
Here, \( p = 7 \) and \( a = 5 \). Therefore:
\[ 5^{6} \equiv 1 \mod 7 \]

4. **Reduce the exponent modulo \( p-1 \)**:
We need to reduce the exponent 123 modulo 6 (since \( p-1 = 6 \)):
\[ 123 \div 6 = 20 \quad \text{remainder} \quad 3 \]
So, \( 123 \mod 6 = 3 \).

5. **Rewrite the exponent**:
We can now rewrite \( 5^{123} \) as:
\[ 5^{123} \equiv 5^{3} \mod 7 \]

6. **Calculate \( 5^{3} \)**:
\[ 5^{3} = 125 \]

7. **Find \( 125 \mod 7 \)**:
Now we need to divide 125 by 7 to find the remainder:
\[ 125 \div 7 = 17 \quad \text{remainder} \quad 6 \]
This means:
\[ 125 \equiv 6 \mod 7 \]

8. **Conclusion**:
Therefore, the remainder when \( 5^{123} \) is divided by 7 is:
\[ \boxed{6} \]