Question

An ascending series of numbers satisfies the following conditions: i. When divided by 3, 4, 5 or 6, the numbers leave a remainder of 2.
ii. When divided by 11, the numbers leave no remainder. The 6th number in this series will be:

• A. 242
• B. 2882
• C. 3542
• D. 4202
• E. None of the above

Hint:

For remainder-type problems, reduce to modular arithmetic. First solve using LCM for simultaneous remainders, then apply extra divisibility conditions.

Answer 1

The Correct Option is C

Step 1: Apply first condition.
Number leaves remainder 2 when divided by 3, 4, 5, 6. So, number \(n \equiv 2 \pmod{3,4,5,6}\). That means \(n-2\) is divisible by LCM(3,4,5,6). \[ \text{LCM}(3,4,5,6)=60. \] Thus, numbers are of form: \[ n=60k+2. \] Step 2: Apply second condition.
\(n \equiv 0 \pmod{11}\). So, \[ 60k+2 \equiv 0 \pmod{11}. \] Since \(60 \equiv 5 \pmod{11}\), \[ 60k+2 \equiv 5k+2 \equiv 0 \pmod{11}. \] \[ 5k \equiv -2 \equiv 9 \pmod{11}. \] Multiply both sides by inverse of 5 modulo 11. Since \(5\times 9=45\equiv 1 \pmod{11}\), the inverse of 5 is 9. \[ k \equiv 9\times 9 \equiv 81 \equiv 4 \pmod{11}. \] So, \(k=11m+4\). Step 3: General form of numbers.
\[ n=60k+2=60(11m+4)+2=660m+242. \] So, series is: 242, 902, 1562, 2222, 2882, 3542, … Step 4: Find 6th term.
6th term = 3542. \[ \boxed{3542} \]