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Question

(A) To find the smallest number $x$ that must be added to $2000$ to make it divisible by $12, 16, 18$,and $21$,we first find the Least Common Multiple

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) To find the smallest number $x$ that must be added to $2000$ to make it divisible by $12, 16, 18$,and $21$,we first find the Least Common Multiple $(LCM)$ of these numbers.Prime factorization:$12 = 2^2 	imes 3$$16 = 2^4$$18 = 2 	imes 3^2$$21 = 3 	imes 7$$LCM = 2^4 	imes 3^2 	imes 7 = 16 	imes 9 	imes 7 = 1008$.Multiples of $1008$ are $1008, 2016, 3024, \dots$The smallest multiple of $1008$ greater than $2000$ is $2016$.Therefore,the number to be added is $x = 2016 - 2000 = 16$.The sum of the digits of $x$ is $1 + 6 = 7$.

Let $x$ be the smallest number,which when added to $2000$ makes the resulting number divisible by $12, 16, 18$ and $21$. The sum of the digits of $x$ is

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