The least number which when divided by 5, 6, | vedclass

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(B) First,find the $LCM$ of $5, 6, 7$ and $8$.$\begin{array}{c|cccc} 2 & 5, & 6, & 7, & 8 \ \hline & 5, & 3, & 7, & 4 \end{array}$$LCM = 2 	imes 5 \

Vedclass · Accepted Answer

$1683$

Vedclass · Answer

(B) First,find the $LCM$ of $5, 6, 7$ and $8$.$\begin{array}{c|cccc} 2 & 5, & 6, & 7, & 8 \ \hline & 5, & 3, & 7, & 4 \end{array}$$LCM = 2 	imes 5 	imes 3 	imes 7 	imes 4 = 840$.The number must be of the form $840k + 3$,where $k$ is a positive integer.We need to find the smallest $k$ such that $(840k + 3)$ is divisible by $9$.$840k + 3 = (837k + 3k) + 3 = 9(93k) + 3k + 3$.For $(840k + 3)$ to be divisible by $9$,$(3k + 3)$ must be divisible by $9$.If $k = 1$,$3(1) + 3 = 6$ (not divisible by $9$).If $k = 2$,$3(2) + 3 = 9$ (divisible by $9$).Therefore,the required number is $840(2) + 3 = 1680 + 3 = 1683$.

The least number which when divided by $5, 6, 7$ and $8$ leaves $3$ as a remainder,but when divided by $9$ leaves no remainder,is:

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