Question

3) What is the smallest number
which when dismished by 7, is
divisible by 12, 16,21 and 28.​

Answer 1

Answer: 1015

Step-by-step explanation:

LCM of the numbers = 1008 thus the required no=1008+7=1015

Answer 2

Answer:

first find the smallest number divisible by 12,16,21 and28. it is the LCM of these number. so the number will be 1008+7 =1015 . so the required number is 1015.