Which of the following numbers are solutions to the system of
simultaneous linear congruences: 2n≡1(mod3)
, 3n≡1(mod5)
, 5n≡1(mod7)
, 7n≡1(mod11)
?

1592

2747

3002

3782

4562

5057

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.

Explore solutions to a system of simultaneous linear congruences involving modular arithmetic. Check which numbers among 1592, 2747, 3002, 3782, 4562, 5057 satisfy the conditions for 2n ≡ 1 (mod 3), 3n ≡ 1 (mod 5), 5n ≡ 1 (mod 7), 7n ≡ 1 (mod 11). Detailed analysis and verification included.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation