5. (3 points) There are 211 students and four different types of chocolate. Each type of chocolate has more than 633 pieces. It is stipulated that each student can take at most three chocolates, or they can take none. If the students are grouped according to the types and quantities of chocolates they take, then the largest group must have at least $\qquad$ students.
【Analysis】Each student can take: (the number in parentheses indicates how many situations this can happen). 1, not taking any (1 situation); 2, taking any one of the four candies (4 situations); 3. taking two, both of the same type (4 situations); 4. taking two different candies, randomly (6 situations); 5. taking three, all the same (4 situations); 6. taking three, two of which are the same (12 situations); 7. taking three different candies (4 situations); Therefore, the total number of different ways a student can take the candies is $1+4+4+6+4+12+4=35$; Since each type of candy has more than 633 pieces, the fifth situation can occur, $3 \times 211=633$, which is sufficient. Therefore, the other six situations can also occur. So, to minimize the number of people in the largest group: $211 \div 35=6 \cdots 1$ (remainder 1); that is, the largest group has at least $6+1=7$ people.

【Solution】According to the analysis of the problem, a student can take in $1+4+4+6+4+12+4=35$ different ways; these 35 ways can be considered as 35 drawers,
$$
211 \div 35=6 \cdots 1 \text {; }
$$

So $6+1=7$ (people),
Answer: The largest group has at least 7 people.
The answer is: 7.