The remainder when `9^103` is divided by 25 is equal to

Options

A. 5
B. 6
C. 4
D. none of these

Explanation

To find the remainder when \( 9^{103} \) 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 \( 9^{103} \)
We can express \( 9^{103} \) as:
\[ 9^{103} = 9 \times 9^{102} = 9 \times (9^2)^{51} \]

### Step 2: Calculate \( 9^2 \)
Next, we calculate \( 9^2 \):
\[ 9^2 = 81 \]

### Step 3: Rewrite \( 81^{51} \)
Now we can rewrite \( 9^{103} \) as:
\[ 9^{103} = 9 \times 81^{51} \]

### Step 4: Use Binomial Expansion
Using the Binomial Theorem, we can express \( 81^{51} \) as:
\[ 81^{51} = (80 + 1)^{51} \]
According to the Binomial Theorem:
\[ (80 + 1)^{51} = \sum_{k=0}^{51} \binom{51}{k} 80^k \cdot 1^{51-k} \]

### Step 5: Identify Terms Modulo 25
In this expansion, we notice that:
- The term \( \binom{51}{0} \cdot 1^{51} = 1 \)
- The term \( \binom{51}{1} \cdot 80^1 = 51 \cdot 80 \)
- All other terms have \( 80^2 \) or higher powers, which are divisible by \( 25 \).

Thus, we can simplify:
\[ (80 + 1)^{51} \equiv 1 + 51 \cdot 80 \mod 25 \]

### Step 6: Calculate \( 51 \cdot 80 \mod 25 \)
Now we compute \( 51 \cdot 80 \mod 25 \):
\[ 51 \mod 25 = 1 \quad \text{(since } 51 = 2 \times 25 + 1\text{)} \]
\[ 80 \mod 25 = 5 \quad \text{(since } 80 = 3 \times 25 + 5\text{)} \]
Thus,
\[ 51 \cdot 80 \mod 25 \equiv 1 \cdot 5 = 5 \]

### Step 7: Combine Terms
Now we can combine the terms:
\[ (80 + 1)^{51} \equiv 1 + 5 \equiv 6 \mod 25 \]

### Step 8: Final Calculation
Now substituting back into our expression for \( 9^{103} \):
\[ 9^{103} \equiv 9 \times 6 \mod 25 \]
Calculating this gives:
\[ 9 \times 6 = 54 \]
Now, find \( 54 \mod 25 \):
\[ 54 \mod 25 = 4 \quad \text{(since } 54 = 2 \times 25 + 4\text{)} \]

### Conclusion
Thus, the remainder when \( 9^{103} \) is divided by 25 is:
\[ \boxed{4} \]