Find the least number which when doubled will be exactly divisible by 12, 16, 18, 21 and 28.
$$12=2^2×3$$
$$16=2^4$$
$$18=3^2×2$$
$$21=3×7$$
$$28=2^2×7$$
So, LCM(12,16,18,21,28)=$$(2^4×3^2×7)$$
=1008.
So, A number is exactly divisible by 12,16,18,21 and 28 ,if it is in the form of (1008k) ,where k is an integer.
So, by choices we can say that the number could be 1008 where k=1.
we can get 1008 by doubling up 504 .
D is correct choice.
Create a FREE account and get:
