Question

3) Find the smallest number which when diminished by 7 is divisible by 12, 16, 18, 21 and 28.​

Answer 1

Explanation:

First find the smallest number divisible by 12,16,18,21 and 28. It is the LCM of these numbers.

12 = 2*2*3 = 2²*3

16 = 2*2*2*2 = 2⁴

18 = 2*3*3 = 2*3²

21 = 3*7

28 = 2*2*7 = 2²*7

LCM = 2⁴ * 3² * 7 = 1008

So the number will be 1008+7 = 1015.

So the required number is 1015. If you diminish by 7, it will be divisible by all the given numbers