`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 expression
We start by rewriting \( 7^{103} \) in a more manageable form:
\[ 7^{103} = 7 \cdot 7^{102} = 7 \cdot (7^2)^{51} \]

### Step 2: Calculate \( 7^2 \)
Next, we calculate \( 7^2 \):
\[ 7^2 = 49 \]
We can express 49 in terms of 25:
\[ 49 = 50 - 1 \]

### Step 3: Substitute into the expression
Now substitute \( 49 \) back into our expression:
\[ 7^{103} = 7 \cdot (50 - 1)^{51} \]

### Step 4: Apply the Binomial Theorem
Using the Binomial Theorem, we expand \( (50 - 1)^{51} \):
\[ (50 - 1)^{51} = \sum_{k=0}^{51} \binom{51}{k} 50^k (-1)^{51-k} \]
This gives us:
\[ = \binom{51}{0} 50^{51} (-1)^{51} + \binom{51}{1} 50^{50} (-1)^{50} + \ldots + \binom{51}{51} 50^0 (-1)^0 \]
The first term is negative since \( (-1)^{51} = -1 \).

### Step 5: Factor out common terms
The first term \( -50^{51} \) is very large and will be divisible by 25. The second term \( \binom{51}{1} 50^{50} \) is also divisible by 25. In fact, all terms with \( k \geq 2 \) will have at least \( 50^2 \) which is also divisible by 25.

### Step 6: Focus on the first two terms
Thus, we can focus on the first two terms:
\[ (50 - 1)^{51} = -50^{51} + 51 \cdot 50^{50} + \text{(higher order terms)} \]
Since all higher order terms are divisible by 25, we can ignore them for the purpose of finding the remainder.

### Step 7: Substitute back into the expression
Now substituting this back into our expression for \( 7^{103} \):
\[ 7^{103} = 7 \cdot (-50^{51} + 51 \cdot 50^{50} + \ldots) \]
This simplifies to:
\[ 7^{103} \equiv -7 \cdot 1 \pmod{25} \]
Thus:
\[ 7^{103} \equiv -7 \pmod{25} \]

### Step 8: Find the positive remainder
To find the positive remainder, we can add 25 to -7:
\[ -7 + 25 = 18 \]

### Final Answer
Thus, the remainder when \( 7^{103} \) is divided by 25 is:
\[ \boxed{18} \]