In $11!(11!=11 \times 10 \times \ldots \times 1)$, the largest factor that can be expressed as $6k+1$ (where $k$ is a natural number) is
Number Theory, Prime Factorization, Congruence
Factorize 11! into prime factors: Let $11!=2^{a} \times 3^{b} \times 5^{c} \times 7^{d} \times 11^{e}$
$$
\begin{array}{ll}
11 \div 2=5 \ldots \ldots .1, & 5 \div 2=2 \ldots . .1, \quad 2 \div 2=1, \quad a=5+2+1=8 \\
11 \div 3=3 \ldots . .2, & 3 \div 3=1, b=3+1=4 \\
11 \div 5=2 \ldots \ldots .1, & c=2 \\
11 \div 7=1 \ldots \ldots .4, & d=1 \\
11 \div 11=1, e=1 &
\end{array}
$$

Thus, $11!=2^{8} \times 3^{4} \times 5^{2} \times 7^{1} \times 11^{1}$
Since $6 k+1$ is neither a multiple of 2 nor a multiple of 3,
the number we are looking for 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).
$$
\text { And } 5^{2} \times 7^{1} \times 11^{1}=5(\bmod 6)
$$

Thus, we consider $5 \times 7 \times 11, 5 \times 7 \times 11=385 \equiv 1(\bmod 6)$
Therefore, 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.