Given a three-digit number $N$, the remainders when $N$ is divided by $3, 7, 11$ are 1, 3, and 8, respectively. Try to find the maximum value of $N$.

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Number Theory, Congruence, The Problem of the Unknown Number.
(Method One)
$N \div 11$ leaves a remainder of 8, let $N=11a+8$;
$N \div 7$ leaves a remainder of 3, then $(11a+8) \div 7$ leaves a remainder of $3, (4a+1) \div 7$ leaves a remainder of 3;
The smallest $a$ is 4, corresponding to $N$ being $11 \times 4 + 8 = 52$;
Then $N \div 77$ leaves a remainder of 52, let $N=77b+52$;
$N \div 3$ leaves a remainder of 1, then $(77b+52) \div 3$ leaves a remainder of $1, (2b+1) \div 3$ leaves a remainder of 1;
The smallest $b$ is 0, corresponding to $N=77 \times 0 + 52 = 52$;
Then $N \div 231$ leaves a remainder of 52;
$999 \div 231 = 4 \cdots \cdots 75$;
Therefore, the largest three-digit number $N$ is $231 \times 4 + 52 = 976$.
(Method Two)
$[3,7,11]=3 \times 7 \times 11=231$

According to the Chinese Remainder Theorem, $N \equiv (3 \times 7) \times 3 + (3 \times 11) \times 2 + (7 \times 11) \times 2 = 283 \equiv 52 (\bmod 231)$;
$999 \div 231 = 4 \cdots \cdots 75$;
Therefore, the largest three-digit number $N$ is $231 \times 4 + 52 = 976$.