To solve this problem, we are looking for the smallest number of students in the class such that when divided by 4, 6, or 11, there are always 3 students left over. This means we seek a number \( N \) that satisfies the following conditions:
1. \( N \equiv 3 \pmod{4} \)
2. \( N \equiv 3 \pmod{6} \)
3. \( N \equiv 3 \pmod{11} \)
We can rewrite these congruences for clarity:
- When \( N \) is divided by 4, the remainder is 3. \( N \equiv 3 \pmod{4} \)
- When \( N \) is divided by 6, the remainder is 3. \( N \equiv 3 \pmod{6} \)
- When \( N \) is divided by 11, the remainder is 3. \( N \equiv 3 \pmod{11} \)
This implies that \( N-3 \) must be divisible by 4, 6, and 11. Let \( M = N - 3 \). Then:
- \( M \) is divisible by 4.
- \( M \) is divisible by 6.
- \( M \) is divisible by 11.
To find the smallest \( N \), we need the least common multiple (LCM) of 4, 6, and 11 because \( M \) should be a common multiple of these numbers.
First, calculate the LCM:
- The prime factorization of 4 is \( 2^2 \).
- The prime factorization of 6 is \( 2 \times 3 \).
- The prime factorization of 11 is \( 11 \).
The LCM must include each prime factor the maximum number of times it appears in any single factorization:
- \( 2 \) appears at most twice (from 4).
- \( 3 \) appears once (from 6).
- \( 11 \) appears once (from 11).
Thus, the LCM is:
[tex]\[ 2^2 \times 3 \times 11 = 4 \times 3 \times 11 = 12 \times 11 = 132\][/tex]
Hence, the smallest \( M \) is 132. Adding back the remainder:
[tex]\[ N = M + 3 = 132 + 3 = 135\][/tex]
Therefore, the smallest possible number of students in the class is:
[tex]\[ \boxed{135} \][/tex]
