The least multiple of 7, which leaves a remai | vedclass

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(D) First,find the $LCM$ of $6, 9, 15,$ and $18$.$\begin{array}{c|cccc} 2 & 6, & 9, & 15, & 18 \ \hline 3 & 3, & 9, & 15, & 9 \ \hline 3 & 1, & 3, &

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$364$

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(D) First,find the $LCM$ of $6, 9, 15,$ and $18$.$\begin{array}{c|cccc} 2 & 6, & 9, & 15, & 18 \ \hline 3 & 3, & 9, & 15, & 9 \ \hline 3 & 1, & 3, & 5, & 3 \ \hline & 1, & 1, & 5, & 1 \end{array}$$LCM = 2 	imes 3 	imes 3 	imes 5 = 90$.Let the number be of the form $90k + 4$,where $k$ is a positive integer.We are given that the number is divisible by $7$. Therefore,$90k + 4 \equiv 0 \pmod{7}$.Simplifying $90 \pmod{7}$: $90 = 7 	imes 12 + 6$,so $90 \equiv 6 \equiv -1 \pmod{7}$.Thus,$-k + 4 \equiv 0 \pmod{7}$,which means $k \equiv 4 \pmod{7}$.The least positive integer $k$ is $4$.Substituting $k = 4$ into the expression $90k + 4$:Required number $= 90(4) + 4 = 360 + 4 = 364$.

The least multiple of $7,$ which leaves a remainder of $4,$ when divided by $6, 9, 15,$ and $18$ is

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