If `{x}` denotes the fractiona part of x, then `{(4^(2n))/(15)}, n in N`, is ________________

Answer

To solve the problem, we need to find the fractional part of \(\frac{4^{2n}}{15}\), where \(n\) is a natural number.

### Step-by-step Solution:

1. **Rewrite \(4^{2n}\)**:
\[ 4^{2n} = (2^2)^{2n} = 2^{4n} \]
This means we can express \(4^{2n}\) in terms of powers of 2.

2. **Express \(2^{4n}\) in terms of \(15\)**:
We want to find \(\frac{2^{4n}}{15}\). To do this, we can use the fact that \(2^{4n}\) can be expressed as \(15k + r\), where \(k\) is an integer and \(r\) is the remainder when \(2^{4n}\) is divided by \(15\).

3. **Calculate \(2^{4n} \mod 15\)**:
We can find the pattern of \(2^{4n} \mod 15\) for different values of \(n\):
- For \(n=1\): \(2^{4 \cdot 1} = 16 \equiv 1 \mod 15\)
- For \(n=2\): \(2^{4 \cdot 2} = 256 \equiv 1 \mod 15\)
- For \(n=3\): \(2^{4 \cdot 3} = 4096 \equiv 1 \mod 15\)

From this, we can see that \(2^{4n} \equiv 1 \mod 15\) for all \(n \in \mathbb{N}\).

4. **Substitute back to find the fractional part**:
Since \(2^{4n} = 15k + 1\) for some integer \(k\), we can write:
\[ \frac{2^{4n}}{15} = k + \frac{1}{15} \]
The integer part of this expression is \(k\), and the fractional part is:
\[ \left\{ \frac{2^{4n}}{15} \right\} = \frac{1}{15} \]

5. **Final Answer**:
Therefore, the fractional part \(\{ \frac{4^{2n}}{15} \}\) is:
\[ \frac{1}{15} \]