Find the smallest natural number $x > 0$ such that all the following fractions are simplified:

$$\frac{3x+9}{8}, \frac{3x+10}{9}, \frac{3x+11}{10}, \ldots, \frac{3x+49}{48}$$

This means that the numerators and denominators are relatively prime.

Let's think step by step and output the final answer within \boxed{}.
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