Let x be the least number,which when divided | vedclass

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(B) The number $x$ is of the form $x = \operatorname{LCM}(5, 6, 7, 8) 	imes k + 3$,where $k$ is a positive integer.First,find the $\operatorname{LCM}

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) The number $x$ is of the form $x = \operatorname{LCM}(5, 6, 7, 8) 	imes k + 3$,where $k$ is a positive integer.First,find the $\operatorname{LCM}$ of $5, 6, 7, 8$:$5 = 5$$6 = 2 	imes 3$$7 = 7$$8 = 2^3$$\operatorname{LCM}(5, 6, 7, 8) = 5 	imes 3 	imes 7 	imes 8 = 840$.So,$x = 840k + 3$.We are given that $x$ is divisible by $9$. Thus,$840k + 3 \equiv 0 \pmod{9}$.Since $840 = 9 	imes 93 + 3$,we have $840 \equiv 3 \pmod{9}$.So,$3k + 3 \equiv 0 \pmod{9}$,which implies $3(k + 1) = 9n$ for some integer $n$,or $k + 1$ must be a multiple of $3$.The smallest positive integer $k$ is $2$ (since for $k=1$,$843$ is not divisible by $9$).For $k = 2$,$x = 840(2) + 3 = 1680 + 3 = 1683$.Check divisibility by $9$: $1 + 6 + 8 + 3 = 18$,which is divisible by $9$.The sum of the digits of $x = 1683$ is $1 + 6 + 8 + 3 = 18$.

Let $x$ be the least number,which when divided by $5, 6, 7$ and $8$ leaves a remainder $3$ in each case,but when divided by $9$ leaves no remainder. The sum of the digits of $x$ is:

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