What is the least number, which when divided by 5,6,8,10 and 12 leaves the remainder 2 in each case , but when divided by 13 leaves no remainder ?

• A. 664
• B. 962
• C. 864
• D. 762

Answer: (B)

 LCM of 5,6,8,10 and 12 = 120But in each case , we have a remainder 2  .So the required number is 120n+2 which is exactly divisible by 13.              120 n + 2 = 13 x 9 n + 2       So clearly 3n+2 must be divisible by 13.  For n=8,  3n +2 is exactly divisible by 13.     Therefore , the required number = 120n + 2                                                 = 120 x 8 + 2   = 962