The greatest 4-digit number exactly divisible | vedclass

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(B) Step $1$: Find the $L.C.M.$ of $10, 15$ and $20$.$10 = 2 	imes 5$$15 = 3 	imes 5$$20 = 2^2 	imes 5$$L.C.M. = 2^2 	imes 3 	imes 5 = 60$.Step $

Vedclass · Accepted Answer

$9960$

Vedclass · Answer

(B) Step $1$: Find the $L.C.M.$ of $10, 15$ and $20$.$10 = 2 	imes 5$$15 = 3 	imes 5$$20 = 2^2 	imes 5$$L.C.M. = 2^2 	imes 3 	imes 5 = 60$.Step $2$: Identify the greatest $4-$digit number,which is $9999$.Step $3$: Divide $9999$ by $60$ to find the remainder.$9999 \div 60 = 166$ with a remainder of $39$.Step $4$: Subtract the remainder from the greatest $4-$digit number to get the largest number divisible by $60$.$9999 - 39 = 9960$.Therefore,the greatest $4-$digit number exactly divisible by $10, 15$ and $20$ is $9960$.

The greatest $4-$digit number exactly divisible by $10, 15$ and $20$ is

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