## Task B-4.2.

Determine the smallest natural number $m$ for which the product $3^{\frac{1}{55}} \cdot 3^{\frac{4}{55}} \cdot 3^{\frac{7}{55}} \cdot \ldots \cdot 3^{\frac{3 m-2}{55}}$ is greater than 729?
## Solution.

The given expression $3^{\frac{1}{55}} \cdot 3^{\frac{4}{55}} \cdot 3^{\frac{7}{55}} \ldots \cdot 3^{\frac{3 m-2}{55}}$ can be written as $3^{\frac{1}{55}+\frac{4}{55}+\frac{7}{55}+\ldots+\frac{3 m-2}{55}}>729=3^{6}$. \quad 1 point

Since the base is greater than 1, for the exponents we have

$\frac{1}{55}+\frac{4}{55}+\frac{7}{55}+\ldots+\frac{3 m-2}{55}>6$, or

1 point

$1+4+7+\ldots+3 m-2>330$.

1 point

On the left side, we have the sum of $m$ terms of an arithmetic sequence, which is

$$
S_{m}=\frac{m(1+3 m-2)}{2}=\frac{m(3 m-1)}{2}
$$

so the inequality becomes

$\frac{m(3 m-1)}{2}>330$ or $3 m^{2}-m-660>0$.

1 point

The solution to this inequality is the set $\left\langle-\infty,-\frac{44}{3}\right\rangle \cup\langle 15, \infty\rangle$.

1 point

The smallest natural number in this set is 16.

1 point