7. (3 points) There are 10 cards each of the numbers “3”, “4”, and “5”. If 8 cards are randomly selected, and their sum is 31, then the maximum number of cards that can be “3” is $\qquad$.
【Analysis】Assume that the 8 cards drawn are all the number “3”, and then solve the problem using the method of solving surplus and deficit problems.
【Solution】Solution: Assume that the 8 cards drawn are all the number “3”, then their sum is $3 \times 8=24$, which is 8 less than the actual sum of 31. This is because the cards “4” and “5” drawn were all considered as cards “3”. Replacing a card “3” with a card “5” or “4” can increase the sum by 2 and 1, respectively. To maximize the number of cards “3”, more cards “5” should be used to replace cards “3”. Now $8 \div 2=4$, so 4 cards “5” can be used to replace cards “3”, making the sum of the 8 cards exactly 32. Therefore, the maximum number of cards “3” is 4.

Answer: The maximum number of cards “3” is 4.
So the answer is: 4.