$\underline{\text { Folklore }}$

What is the highest power of three that divides the product $3 \cdot 33 \cdot 333 \cdot \ldots \cdot 3333333333$?
$3 \cdot 33 \cdot 333 \cdot \ldots \cdot 3333333333=3^{10} \cdot 1 \cdot 11 \cdot 111 \cdot \ldots \cdot 1111111111$. Among the factors written only with ones, those divisible by 3 are the numbers with a sum of digits that is a multiple of 3: 111, 111111, and 111111111. 111 and 111111 are not divisible by 9, while $111111111=111 \cdot 100010001$ is divisible by 9 but not by 27 (each factor is divisible by 3 but not by 9).

Thus, the given product is divisible by $3^{14}$, but not by $3^{15}$.

## Answer

By $3^{14}$.