To determine which of the given numbers are solutions to the system of simultaneous linear congruences, we need to check each number against the following conditions:

1. $2n \equiv 1 \,(\text{mod}\, 3)$
2. $3n \equiv 1 \,(\text{mod}\, 5)$
3. $5n \equiv 1 \,(\text{mod}\, 7)$
4. $7n \equiv 1 \,(\text{mod}\, 11)$

We will check each number one by one:

1. For $n = 1592$:

1. $2(1592) = 3184 \equiv 1 \,(\text{mod}\, 3)$ (because $3184 \div 3$ gives a remainder of 1)
2. $3(1592) = 4776 \equiv 1 \,(\text{mod}\, 5)$ (because $4776 \div 5$ gives a remainder of 1)
3. $5(1592) = 7960 \equiv 1 \,(\text{mod}\, 7)$ (because $7960 \div 7$ gives a remainder of 5)
4. $7(1592) = 11144 \equiv 1 \,(\text{mod}\, 11)$ (because $11144 \div 11$ gives a remainder of 1)

$n = 1592$ does not satisfy all the conditions.

2. For $n = 2747$:

1. $2(2747) = 5494 \equiv 1 \,(\text{mod}\, 3)$
2. $3(2747) = 8241 \equiv 1 \,(\text{mod}\, 5)$
3. $5(2747) = 13735 \equiv 1 \,(\text{mod}\, 7)$
4. $7(2747) = 19229 \equiv 1 \,(\text{mod}\, 11)$

$n = 2747$ satisfies all the conditions.

3. For $n = 3002$:

1. $2(3002) = 6004 \equiv 1 \,(\text{mod}\, 3)$
2. $3(3002) = 9006 \equiv 1 \,(\text{mod}\, 5)$
3. $5(3002) = 15010 \equiv 1 \,(\text{mod}\, 7)$
4. $7(3002) = 21014 \equiv 1 \,(\text{mod}\, 11)$

$n = 3002$ does not satisfy all the conditions.

4. For $n = 3782$:

1. $2(3782) = 7564 \equiv 1 \,(\text{mod}\, 3)$
2. $3(3782) = 11346 \equiv 1 \,(\text{mod}\, 5)$
3. $5(3782) = 18910 \equiv 1 \,(\text{mod}\, 7)$
4. $7(3782) = 26474 \equiv 1 \,(\text{mod}\, 11)$

$n = 3782$ does not satisfy all the conditions.

5. For $n = 4562$:

1. $2(4562) = 9124 \equiv 1 \,(\text{mod}\, 3)$
2. $3(4562) = 13686 \equiv 1 \,(\text{mod}\, 5)$
3. $5(4562) = 22810 \equiv 1 \,(\text{mod}\, 7)$
4. $7(4562) = 31934 \equiv 1 \,(\text{mod}\, 11)$

$n = 4562$ does not satisfy all the conditions.

6. For $n = 5057$:

1. $2(5057) = 10114 \equiv 1 \,(\text{mod}\, 3)$
2. $3(5057) = 15171 \equiv 1 \,(\text{mod}\, 5)$
3. $5(5057) = 25285 \equiv 1 \,(\text{mod}\, 7)$
4. $7(5057) = 35399 \equiv 1 \,(\text{mod}\, 11)$

$n = 5057$ does not satisfy all the conditions.

Conclusion:

The number $\boxed{2747}$ is the only number that satisfies all the simultaneous linear congruences.

Would you like more details, or do you have any questions?

Here are 5 follow-up questions:

1. How can the Chinese Remainder Theorem help solve simultaneous linear congruences?
2. What are the steps to check congruences manually for larger numbers?
3. Why do some numbers satisfy only a subset of the congruences?
4. How can modular arithmetic be applied in cryptography?
5. What is the significance of the moduli being pairwise coprime in solving congruences?

Tip: When dealing with multiple congruences, ensure the moduli are pairwise coprime to use the Chinese Remainder Theorem effectively.