Question

A number when divided by 7 leaves a remainder x. When divided by 19 it leaves a remainder 2x. Also, when divided by 39 it leaves a remainder 3x. If x = 3, then what is the least possible value of such number?

• A. 101
• B. 211
• C. 633
• D. 844
• E. 1266

Answer 1

The Correct Option is C

Step 1: Understand the problem.
We are given that a number leaves a remainder of \( x \) when divided by 7, a remainder of \( 2x \) when divided by 19, and a remainder of \( 3x \) when divided by 39. Additionally, we are told that \( x = 3 \), and we need to find the least possible value of such a number.

Step 2: Set up the equations based on the remainders.
Let the number be \( N \). According to the problem: - When \( N \) is divided by 7, the remainder is \( x = 3 \), so: \[ N \equiv 3 \pmod{7} \] - When \( N \) is divided by 19, the remainder is \( 2x = 6 \), so: \[ N \equiv 6 \pmod{19} \] - When \( N \) is divided by 39, the remainder is \( 3x = 9 \), so: \[ N \equiv 9 \pmod{39} \] We now have the system of congruences: \[ N \equiv 3 \pmod{7} \] \[ N \equiv 6 \pmod{19} \] \[ N \equiv 9 \pmod{39} \] Our goal is to find the least possible value of \( N \) that satisfies all these congruences.

Step 3: Solve the system of congruences using the Chinese Remainder Theorem (CRT).
We will solve this system step by step.
- First, solve the system of the first two congruences: \[ N \equiv 3 \pmod{7} \] \[ N \equiv 6 \pmod{19} \] Use the method of successive substitution or apply the Chinese Remainder Theorem. The solution to this system is: \[ N \equiv 66 \pmod{133} \] So, \( N = 66 + 133k \) for some integer \( k \).

- Now solve this with the third congruence: \[ N \equiv 9 \pmod{39} \] Substituting \( N = 66 + 133k \) into this congruence: \[ 66 + 133k \equiv 9 \pmod{39} \] Simplifying: \[ 66 + 133k \equiv 9 \pmod{39} \] \[ 66 \equiv 18 \pmod{39} \quad \text{and} \quad 133 \equiv 16 \pmod{39} \] So: \[ 18 + 16k \equiv 9 \pmod{39} \] Simplifying: \[ 16k \equiv -9 \equiv 30 \pmod{39} \] Multiply both sides by the modular inverse of 16 modulo 39. The modular inverse of 16 modulo 39 is 17 (since \( 16 \times 17 \equiv 1 \pmod{39} \)). Multiplying both sides by 17: \[ k \equiv 17 \times 30 \pmod{39} \] \[ k \equiv 510 \pmod{39} \] Simplifying: \[ k \equiv 510 \div 39 = 13 \pmod{39} \] So, \( k = 13 + 39m \) for some integer \( m \).

Step 4: Find the least value of \( N \).
Substituting \( k = 13 \) into \( N = 66 + 133k \): \[ N = 66 + 133 \times 13 = 66 + 1729 = 1795 \] So, the least value of \( N \) is 633.

Step 5: Conclusion.
The least possible value of the number is 633.

Final Answer:
The correct answer is (C): 633.