【Question 6】
Among all the factors of $11!(11!=11 \times 10 \times \cdots \times 1)$, the largest factor that can be expressed as $6k+1$ (where $k$ is a natural number) is $\qquad$.
【Analysis and Solution】
Number Theory, Prime Factorization, Congruence.
Factorize 11! into prime factors: Let $11!=2^{a} \times 3^{b} \times 5^{c} \times 7^{d} \times 11^{e}$;
$$
\begin{array}{l}
11 \div 2=5 \cdots \cdots 1, \quad 5 \div 2=2 \cdots \cdots 1, \quad 2 \div 2=1, \quad a=5+2+1=8 ; \\
11 \div 3=3 \cdots \cdots 2, \quad 3 \div 3=1, \quad b=3+1=4 ; \\
11 \div 5=2 \cdots \cdots 1, \quad c=2 ; \\
11 \div 7=1 \cdots \cdots 4, \quad d=1 ; \\
11 \div 11=1, \quad e=1 ;
\end{array}
$$

Therefore, $11!=2^{8} \times 3^{4} \times 5^{2} \times 7^{1} \times 11^{1}$;
Since $6 k+1$ is neither a multiple of 2 nor a multiple of 3;
Thus, the number sought does not contain the prime factors 2 or 3;
This is equivalent to finding the largest factor of $5^{2} \times 7^{1} \times 11^{1}$ that can be expressed as $6 k+1$ (where $k$ is a natural number). And $5^{2} \times 7^{1} \times 11^{1} \equiv 5(\bmod 6)$;

Therefore, we consider $5 \times 7 \times 11, 5 \times 7 \times 11=385 \equiv 1(\bmod 6)$;
Hence, among all the factors of $11!$, the largest factor that can be expressed as $6 k+1$ (where $k$ is a natural number) is 385.