Find the smallest number of two digits which on being divided by 2, 3, 4 and 5 leaves 0, 1, 2 and 3 as remainder respectively.

• A. 65
• B. 55
• C. 61
• D. 58

Answer: (D)

 LCM of 2, 3, 4 and 5 is 60.The smallest number of two digits is 10.But we also see that the smallest number divisible by 2, 3, 4 and 5 is 60. Therefore, 10 has no significance here.The difference between divisor and remainder is 2 in each case. So, we subtract 2 from 60.60 - 2 = 58