7. (10 points) There are 10 cards each of the numbers “3”, “4”, and “5”. From these, any 8 cards are selected so that their sum is 27. Then, the maximum number of cards that can be “3” is $\qquad$.

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【Analysis】Assume that the 8 cards drawn are all the number “3”, then their sum is $3 \times 8=24$, which is 3 less than the actual sum of 27. This is because the cards “4” and “5” drawn were counted 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”, followed by more cards “4” to replace cards “3”. Since $3=2+1$, one card “5” and one card “4” can be used to replace cards “3”. This way, the sum of the numbers on the 8 cards will exactly equal 27. Therefore, the maximum number of cards “3” can be 6.

【Solution】Assume that the 8 cards drawn are all the number “3”, then their sum is $3 \times 8=24$, $27-24=3$, which is because the cards “4” and “5” drawn were counted as cards “3”.
$$
5-3=2,4-3=1 \text {, }
$$

Therefore, to maximize the number of cards “3”, one card “5” and one card “4” should be used to replace cards “3”,

This way, the sum of the numbers on the 8 cards will exactly equal 27.
So the maximum number of cards “3” can be 6.
The answer is: 6.