The remainder when `6^(n) - 5n` is divided by 25, is

Options

A. 1
B. 24
C. 0
D. n

Answer

To find the remainder when \( 6^n - 5n \) is divided by 25, we can use the Binomial Theorem and properties of modular arithmetic. Here’s a step-by-step solution:

### Step 1: Express \( 6^n \) using the Binomial Theorem
We can express \( 6^n \) as \( (5 + 1)^n \). According to the Binomial Theorem:
\[ (5 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} 5^k 1^{n-k} = \sum_{k=0}^{n} \binom{n}{k} 5^k \]

### Step 2: Expand the expression
Expanding \( (5 + 1)^n \) gives us:
\[ 6^n = \binom{n}{0} 5^0 + \binom{n}{1} 5^1 + \binom{n}{2} 5^2 + \binom{n}{3} 5^3 + \ldots + \binom{n}{n} 5^n \]
This can be rewritten as:
\[ 6^n = 1 + n \cdot 5 + \binom{n}{2} 5^2 + \binom{n}{3} 5^3 + \ldots + 5^n \]

### Step 3: Consider terms modulo 25
Since we are interested in the remainder when divided by 25, we note that any term with \( 5^2 \) or higher will contribute 0 to the remainder when divided by 25. Therefore, we only need to consider the first two terms:
\[ 6^n \equiv 1 + n \cdot 5 \mod 25 \]

### Step 4: Substitute into the original expression
Now we substitute this back into the expression \( 6^n - 5n \):
\[ 6^n - 5n \equiv (1 + 5n) - 5n \mod 25 \]
This simplifies to:
\[ 6^n - 5n \equiv 1 \mod 25 \]

### Step 5: Conclusion
Thus, the remainder when \( 6^n - 5n \) is divided by 25 is:
\[ \boxed{1} \]