To find the least natural number \( n \) such that \( f(n\pi + x) = f(x) \) for the function \( f(x) = \sin\left(\frac{x}{3}\right) + \cos\left(\frac{3x}{10}\right) \), we will analyze the periodicity of both terms in the function.
### Step 1: Analyze the sine term
The first term is \( \sin\left(\frac{x}{3}\right) \). The sine function has a period of \( 2\pi \). Therefore, we need to find \( n \) such that:
\[
\sin\left(\frac{n\pi + x}{3}\right) = \sin\left(\frac{x}{3}\right)
\]
Using the periodic property of sine, we have:
\[
\frac{n\pi + x}{3} = \frac{x}{3} + 2k\pi \quad \text{for some integer } k
\]
This simplifies to:
\[
\frac{n\pi}{3} = 2k\pi
\]
From this, we can solve for \( n \):
\[
n = 6k
\]
Thus, \( n \) must be a multiple of 6.
### Step 2: Analyze the cosine term
The second term is \( \cos\left(\frac{3x}{10}\right) \). The cosine function also has a period of \( 2\pi \). We need to find \( n \) such that:
\[
\cos\left(\frac{3(n\pi + x)}{10}\right) = \cos\left(\frac{3x}{10}\right)
\]
This can be written as:
\[
\frac{3(n\pi + x)}{10} = \frac{3x}{10} + 2m\pi \quad \text{for some integer } m
\]
This simplifies to:
\[
\frac{3n\pi}{10} = 2m\pi
\]
From this, we can solve for \( n \):
\[
n = \frac{20m}{3}
\]
Thus, \( n \) must be a multiple of \( \frac{20}{3} \).
### Step 3: Find the least natural number \( n \)
To satisfy both conditions, we need \( n \) to be a multiple of 6 and also a multiple of \( \frac{20}{3} \).
The least common multiple (LCM) of \( 6 \) and \( \frac{20}{3} \) can be found as follows:
1. The multiples of \( 6 \) are \( 6, 12, 18, 24, 30, 36, \ldots \)
2. The multiples of \( \frac{20}{3} \) can be expressed as \( 20, \frac{40}{3}, \frac{60}{3}, \ldots \)
To find the least natural number \( n \) that satisfies both conditions, we can check the multiples of \( 6 \) to see which one is also a multiple of \( \frac{20}{3} \).
Calculating:
- For \( n = 6 \): \( \frac{20 \cdot 1}{3} = \frac{20}{3} \) (not natural)
- For \( n = 12 \): \( \frac{20 \cdot 2}{3} = \frac{40}{3} \) (not natural)
- For \( n = 18 \): \( \frac{20 \cdot 3}{3} = 20 \) (natural)
- For \( n = 24 \): \( \frac{20 \cdot 4}{3} = \frac{80}{3} \) (not natural)
- For \( n = 30 \): \( \frac{20 \cdot 5}{3} = \frac{100}{3} \) (not natural)
- For \( n = 36 \): \( \frac{20 \cdot 6}{3} = 40 \) (natural)
The least natural number \( n \) that satisfies both conditions is \( 6 \).
### Final Answer
Thus, the least natural number \( n \) such that \( f(n\pi + x) = f(x) \) for all real \( x \) is:
\[
\boxed{6}
\]
