Find the greatest number of five digits which | vedclass

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(B) First,find the $L.C.M.$ of $10, 12, 15$ and $18$.$10 = 2 	imes 5$$12 = 2^2 	imes 3$$15 = 3 	imes 5$$18 = 2 	imes 3^2$$L.C.M. = 2^2 	imes 3^2

Vedclass · Accepted Answer

$99911$

Vedclass · Answer

(B) First,find the $L.C.M.$ of $10, 12, 15$ and $18$.$10 = 2 	imes 5$$12 = 2^2 	imes 3$$15 = 3 	imes 5$$18 = 2 	imes 3^2$$L.C.M. = 2^2 	imes 3^2 	imes 5 = 4 	imes 9 	imes 5 = 180$.Let the required number be $x$. According to the problem,$(x + 3769)$ must be divisible by $180$.The greatest $5$-digit number is $99999$.We want the largest $x$ such that $x + 3769 = 180k$ for some integer $k$.Since $x$ is a $5$-digit number,$x \leq 99999$,so $x + 3769 \leq 99999 + 3769 = 103768$.Dividing $103768$ by $180$ gives $103768 = 180 	imes 576 + 88$.To find the largest multiple of $180$ less than or equal to $103768$,we subtract the remainder: $103768 - 88 = 103680$.Thus,$x + 3769 = 103680$.$x = 103680 - 3769 = 99911$.

Find the greatest number of five digits which becomes exactly divisible by $10, 12, 15$ and $18$ when $3769$ is added to it.

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