Question 25

Find the least number which when doubled will be exactly divisible by 12, 16, 18, 21 and 28.

Solution

$$12=2^2×3$$

$$16=2^4$$

$$18=3^2×2$$

$$21=3×7$$

$$28=2^2×7$$

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.

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