Find the smallest number which when increased by 5 is divisible by 9, 21, 25 and 30.?

• A. 3145
• B. 1882
• C. 2378
• D. 2567

Answer: (A)

LCM of the given numbers.3 |9, 21, 25, 305 |3, 7, 25, 10   | 3, 7, 5, 2LCM = 3 × 5 × 3 × 7 × 5 × 2 = 3150By definition, LCM of a given set of numbers is their smallest common multiple. In other words, it is the smallest numbers which is divisible by all the given numbers.So, to get the answer , we have to  deduct 5 from the LCM.3150 – 5 = 3145