5. Each of the three Greeks brought the same number of wreaths. Meeting nine Muses, they divided the wreaths in such a way that each Greek and each Muse had the same number of wreaths. How many wreaths did each Greek have at first?
5. The least number of wreaths each Greek and Muse could receive after division is one. Then the Muses had 9 wreaths, and the Greeks brought a total of 12 wreaths $(3+9=12)$, with each Greek having 4 wreaths $(12: 3=4)$. If everyone received 2 wreaths, then the Muses would have a total of 18 wreaths $(2 \times 9=18)$. In this case, the Greeks would have brought 24 wreaths $(18+2 \times 3=24)$, and each Greek would have brought 8 wreaths $(24: 3=8)$. If everyone received $m$ wreaths, then the Muses would have a total of $9 m$ wreaths, and the Greeks would have brought a total of $12 m$ wreaths $(3 m+9 m=12 m)$, with each Greek bringing $4 m$ wreaths $(12 m: 3=4 m)$.

Answer: $4 m$ wreaths, where $m=1,2,3 \ldots$