2. The City Mathematics Invitational Competition sets up three types of medals: gold, silver, and bronze. The organizing committee puts these medals in five boxes, with each box containing only one type of medal. The number of medals in each box is 3, 6, 9, 14, and 18, respectively. It is known that the silver medals are in only one box, and the number of bronze medals is twice the number of gold medals. Therefore, there are gold medals $\qquad$, silver medals $\qquad$, and bronze medals $\qquad$.
2.12, 14, 24.

Based on "the number of bronze medals is twice the number of gold medals," the sum of bronze and gold medals should be a multiple of 3.

The sum of bronze and gold medals should also be the sum of four numbers from the known 3, 6, 9, 14, 18, due to the fact that "there is only one box of silver medals."

Therefore, the number of silver medals is 14, the number of gold medals is $(3+6+9+18) \div 3=12$, and the number of bronze medals is 24.