If the fractional part of the number `(2^(403))/(15)` is `(k)/(15)` then k is equal to

Options

A. 14
B. 6
C. 4
D. 8

Answer

To solve the problem of finding the value of \( k \) such that the fractional part of \( \frac{2^{403}}{15} \) is \( \frac{k}{15} \), we can follow these steps:

### Step-by-Step Solution

1. **Express \( 2^{403} \) in a more manageable form**:
\[ 2^{403} = 2^{400} \times 2^3 = 2^{400} \times 8 \]

2. **Rewrite \( 2^{400} \)**:
We can express \( 2^{400} \) as:
\[ 2^{400} = (2^4)^{100} = 16^{100} \]

3. **Use the Binomial Theorem**:
We can apply the Binomial Theorem to expand \( (1 + 15)^{100} \):
\[ (1 + 15)^{100} = \sum_{k=0}^{100} \binom{100}{k} 1^{100-k} (15)^k \]
This gives us:
\[ 1^{100} + \binom{100}{1} 15 + \binom{100}{2} 15^2 + \ldots + \binom{100}{100} 15^{100} \]

4. **Focus on the terms modulo 15**:
The terms \( \binom{100}{k} 15^k \) for \( k \geq 1 \) will be multiples of 15, and thus will not contribute to the fractional part when divided by 15. Therefore, we only need to consider the first term:
\[ 1 \quad \text{(the constant term)} \]

5. **Calculate \( 2^{403} \mod 15 \)**:
Since we are interested in \( 2^{403} \mod 15 \):
\[ 2^{403} = 8 \cdot 2^{400} \]
We can find \( 2^{400} \mod 15 \) using Euler's theorem. First, we find \( \phi(15) \):
\[ \phi(15) = 15 \left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{5}\right) = 15 \cdot \frac{2}{3} \cdot \frac{4}{5} = 8 \]
Since \( 2^8 \equiv 1 \mod 15 \), we can reduce the exponent:
\[ 400 \mod 8 = 0 \quad \Rightarrow \quad 2^{400} \equiv 1 \mod 15 \]
Thus,
\[ 2^{403} \equiv 8 \cdot 1 \equiv 8 \mod 15 \]

6. **Find the fractional part**:
Now, we can express \( \frac{2^{403}}{15} \):
\[ \frac{2^{403}}{15} = \frac{8 + 15m}{15} \quad \text{for some integer } m \]
The fractional part is:
\[ \frac{8}{15} \]
Therefore, we have:
\[ k = 8 \]

### Conclusion
The value of \( k \) is \( 8 \).