Find the smallest natural number n which has last digit 6 & if this last is moved to the front of the number, the number becomes 4 times larger.

Answer

To solve the problem, we need to find the smallest natural number \( n \) that ends with the digit 6, and when we move this last digit to the front, the new number becomes 4 times larger than the original number. Let's denote the original number as \( n \).

### Step-by-Step Solution:

1. **Represent the Number**:
Let \( n \) be represented as \( 10k + 6 \), where \( k \) is the integer part of \( n \) without the last digit (which is 6).

2. **Moving the Last Digit**:
When we move the last digit (6) to the front, the new number becomes \( 6 \times 10^d + k \), where \( d \) is the number of digits in \( k \).

3. **Setting Up the Equation**:
According to the problem, moving the last digit makes the number 4 times larger:
\[ 6 \times 10^d + k = 4(10k + 6) \]

4. **Expanding the Equation**:
Expanding the right side gives:
\[ 6 \times 10^d + k = 40k + 24 \]

5. **Rearranging the Equation**:
Rearranging the equation to isolate terms gives:
\[ 6 \times 10^d - 24 = 40k - k \]
\[ 6 \times 10^d - 24 = 39k \]

6. **Solving for \( k \)**:
Rearranging for \( k \) gives:
\[ k = \frac{6 \times 10^d - 24}{39} \]

7. **Finding Integer Solutions**:
For \( k \) to be a natural number, \( 6 \times 10^d - 24 \) must be divisible by 39. We can simplify this condition:
\[ 6 \times 10^d \equiv 24 \mod 39 \]
Simplifying further:
\[ 10^d \equiv 4 \mod 13 \]
(since \( 39 = 3 \times 13 \) and \( 6 \) is invertible modulo \( 39 \)).

8. **Finding the Smallest \( d \)**:
We can check powers of \( 10 \) modulo \( 13 \):
- \( 10^1 \equiv 10 \)
- \( 10^2 \equiv 9 \)
- \( 10^3 \equiv 12 \)
- \( 10^4 \equiv 3 \)
- \( 10^5 \equiv 4 \)

We find that \( d = 5 \) satisfies the condition.

9. **Calculating \( k \)**:
Substitute \( d = 5 \) back into the equation for \( k \):
\[ k = \frac{6 \times 10^5 - 24}{39} = \frac{600000 - 24}{39} = \frac{599976}{39} = 15384 \]

10. **Finding \( n \)**:
Now, substitute \( k \) back to find \( n \):
\[ n = 10k + 6 = 10 \times 15384 + 6 = 153840 + 6 = 153846 \]

### Conclusion:
The smallest natural number \( n \) that satisfies the conditions is \( \boxed{153846} \).