The smallest positive number greater than 5 s | vedclass

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Question

(D) To find the smallest number that leaves a remainder of $5$ when divided by $20, 30,$ and $40,$ we first find the Least Common Multiple $(LCM)$ of

Vedclass · Accepted Answer

$125$

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(D) To find the smallest number that leaves a remainder of $5$ when divided by $20, 30,$ and $40,$ we first find the Least Common Multiple $(LCM)$ of the divisors.$LCM(20, 30, 40)$:$20 = 2^2 	imes 5$$30 = 2 	imes 3 	imes 5$$40 = 2^3 	imes 5$$LCM = 2^3 	imes 3 	imes 5 = 8 	imes 3 	imes 5 = 120$.The number must be of the form $(LCM 	imes k) + 	ext{remainder}$.Here,$LCM = 120$ and $	ext{remainder} = 5$.So,the number $= 120k + 5$.For the smallest positive number greater than $5$,we take $k = 1$.Number $= 120(1) + 5 = 125$.

Part $I$	Part $II$
$1. \text{l.c.m.}(8, 16, 24)$	$a. 36$
$2. \text{g.c.d.}(8, 16, 24)$	$b. 48$
$3. \text{l.c.m.}(6, 12, 18)$	$c. 8$
$4. \text{g.c.d.}(6, 12, 18)$	$d. 12$
	$e. 6$

The smallest positive number greater than $5$ such that when it is divided by $20, 30,$ or $40,$ it leaves the remainder $5$ is:

Similar Questions

Find the $HCF(156, 455)$.

Find the $g.c.d.$ of the following numbers using Euclid's division algorithm: $117$ and $65$.

Find $g.c.d.$ of $210$ and $55$ by Euclid's algorithm.

Which of the following groups correctly matches the data of Part $I$ with the data of Part $II$?

Part $I$ Part $II$

$1. \text{l.c.m.}(8, 16, 24)$ $a. 36$

$2. \text{g.c.d.}(8, 16, 24)$ $b. 48$

$3. \text{l.c.m.}(6, 12, 18)$ $c. 8$

$4. \text{g.c.d.}(6, 12, 18)$ $d. 12$

$e. 6$

Express the number $140$ as a product of its prime factors.

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