`7^(103)` when divided by 25 leaves the remainder .

Options

A. 20
B. 16
C. 18
D. 15

Explanation

To find the remainder when \( 7^{103} \) is divided by 25, we can use the Binomial Theorem and properties of modular arithmetic. Here’s a step-by-step solution:

### Step 1: Rewrite the base
We can express \( 7 \) as \( 25 - 18 \) (since \( 7 \equiv -18 \mod 25 \)). This helps in simplifying calculations.

### Step 2: Apply the Binomial Theorem
Using the Binomial Theorem, we can expand \( (25 - 18)^{103} \):
\[ (25 - 18)^{103} = \sum_{k=0}^{103} \binom{103}{k} 25^k (-18)^{103-k} \]

### Step 3: Identify relevant terms
When we divide by 25, only the terms where \( k = 0 \) and \( k = 1 \) will contribute to the remainder, as all higher powers of 25 will be divisible by 25.

- For \( k = 0 \):
\[ \binom{103}{0} 25^0 (-18)^{103} = (-18)^{103} \]

- For \( k = 1 \):
\[ \binom{103}{1} 25^1 (-18)^{102} = 103 \cdot 25 \cdot (-18)^{102} \]

### Step 4: Calculate \( (-18)^{103} \mod 25 \)
Now we need to find \( (-18)^{103} \mod 25 \). We can simplify \( -18 \mod 25 \) to \( 7 \) (since \( -18 + 25 = 7 \)):
\[ (-18)^{103} \equiv 7^{103} \mod 25 \]

### Step 5: Use Euler's theorem
To simplify \( 7^{103} \mod 25 \), we can use Euler's theorem. First, we calculate \( \phi(25) \):
\[ \phi(25) = 25 \left(1 - \frac{1}{5}\right) = 25 \cdot \frac{4}{5} = 20 \]
According to Euler's theorem, since \( \gcd(7, 25) = 1 \):
\[ 7^{20} \equiv 1 \mod 25 \]

### Step 6: Reduce the exponent
Now we reduce \( 103 \mod 20 \):
\[ 103 \div 20 = 5 \quad \text{(remainder 3)} \]
Thus, \( 103 \equiv 3 \mod 20 \).

### Step 7: Calculate \( 7^3 \mod 25 \)
Now we compute \( 7^3 \):
\[ 7^3 = 343 \]
Now, find \( 343 \mod 25 \):
\[ 343 \div 25 = 13 \quad \text{(remainder 18)} \]
Thus, \( 343 \equiv 18 \mod 25 \).

### Final Answer
The remainder when \( 7^{103} \) is divided by 25 is \( \boxed{18} \).

---