The smallest positive number greater than 5 s | vedclass

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(D) To find the smallest number that leaves a remainder of $5$ when divided by $20, 30,$ and $40,$ we first find the Least Common Multiple $(LCM)$ of

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$125$

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(D) To find the smallest number that leaves a remainder of $5$ when divided by $20, 30,$ and $40,$ we first find the Least Common Multiple $(LCM)$ of the divisors.$LCM(20, 30, 40)$:$20 = 2^2 	imes 5$$30 = 2 	imes 3 	imes 5$$40 = 2^3 	imes 5$$LCM = 2^3 	imes 3 	imes 5 = 8 	imes 3 	imes 5 = 120$.The number must be of the form $(LCM 	imes k) + 	ext{remainder}$.Here,$LCM = 120$ and $	ext{remainder} = 5$.So,the number $= 120k + 5$.For the smallest positive number greater than $5$,we take $k = 1$.Number $= 120(1) + 5 = 125$.

The smallest positive number greater than $5$ such that when it is divided by $20, 30,$ or $40,$ it leaves the remainder $5$ is:

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