A number, when divided by \( 3,4,5 \), and 6 , has a remainder of 1 . A 21 B 41 C 61 D 81

To determine which number among the options \( 21, 41, 61, \) and \( 81 \) leaves a remainder of 1 when divided by \( 3, 4, 5, \) and \( 6 \), we can use the concept of modular arithmetic and the Least Common Multiple (LCM). 1. **Find the LCM of the divisors:** - The divisors are \( 3, 4, 5, \) and \( 6 \). - LCM of \( 3 \) and \( 6 \) is \( 6 \). - LCM of \( 6 \) and \( 4 \) is \( 12 \). - LCM of \( 12 \) and \( 5 \) is \( 60 \). So, the LCM of \( 3, 4, 5, \) and \( 6 \) is \( 60 \). 2. **Formulate the condition:** - We need to find numbers \( x \) such that: \[ x \equiv 1 \pmod{3}, \quad x \equiv 1 \pmod{4}, \quad x \equiv 1 \pmod{5}, \quad x \equiv 1 \pmod{6} \] - This means \( x \equiv 1 \pmod{60} \), since \( 60 \) is the LCM. 3. **Check the options:** - **A. 21** \[ 21 \div 60 = 0 \text{ remainder } 21 \quad (\text{Not } \equiv 1) \] - **B. 41** \[ 41 \div 60 = 0 \text{ remainder } 41 \quad (\text{Not } \equiv 1) \] - **C. 61** \[ 61 = 60 \times 1 + 1 \quad (\equiv 1 \pmod{60}) \] - **D. 81** \[ 81 = 60 \times 1 + 21 \quad (\text{Not } \equiv 1) \] Only **61** satisfies the condition of leaving a remainder of 1 when divided by \( 3, 4, 5, \) and \( 6 \). **Answer:** C) 61