7. (10 points) There are 10 cards each of the numbers “3”, “4”, and “5”. Select any 8 cards so that their sum is 33, then the maximum number of cards that can be “3” is $\qquad$.

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7. (10 points) There are 10 cards each of the numbers “3”, “4”, and “5”. Select 8 cards such that their sum is 33. The maximum number of cards that can be “3” is $\qquad$.

【Analysis】This problem requires determining the maximum number of cards that can be “3”. We can use the assumption method and consider different scenarios:
(1) All 8 cards are 3, (2) 7 cards are 3, (3) 6 cards are 3, ..., until the requirement is met.

【Solution】Solution: If all 8 cards are 3, then $8 \times 3=24$, which is not possible.
If 5 cards are 3, then $5 \times 3=15$, and the sum of the remaining 3 cards is $33-15=18, 18 \div 3=6>5$, which is not possible.
If 4 cards are 3, then $4 \times 3=12$, and the sum of the remaining 4 cards is $33-12=21, 21 \div 4>5$, which is not possible.
If 3 cards are 3, then $3 \times 3=9$, and the sum of the remaining 5 cards is $33-9=24=5+5+5+5+4$, which means taking 4 cards of 5 and 1 card of 4.

In summary, the maximum number of cards that can be 3 is 3.
Therefore, the answer is: 3.