Question

Find the sets of two numbers whose lcm in 720 and hcf is 24

Answer 1

Solution :-

HCF×LCM=product of two numbers. Therefore,

12×720- product of two numbers

Factors of HCF and LCM are

12-2×2×3 and

720 12×5×3×2×2.

So two numbers can be

(12,720),(24,360), (36,240),

(48,180),(60,144),(72,120)

but in

(24,360),(72,120)

HCF will increase,

so rejecting these

required numbers are

(12,720), (36,240), (48,180), (60,144)

Answer 2

$\large{ \bold{\pink{GIVEN:-}}}$

$\large{HCF=24}$

$\large{LCM=720}$

$\large{ \boxed{ \bold{to \: find = sets \: of \: two \: numbers}}}$

$\huge{ \bold{ \purple{SOLUTION:-}}}$

$= > LCM \times HCF = product \: of \: the \: two \\ numbers \\ Therefore, \: 24 \times 720 = product \: of \: the \\ given \: numbers$

$\small{ \boxed{ \green{By \: factorising \: both \: of \: the \: numbers}}} \\ we \: get... \\ = > (2 \times 2 \times 2 \times 3) (2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5) \\ the \: required \: sets \: of \: combinations \: will \: be \\ different \: of \: the \: above \: product.$

Examples

$2 \: and \: 2 \times 2 \times 2 \times 2 \times 2 \times 3\times 3\times 3\times \:5$

$2\times 2 and \: 2 \times 2 \times 2 \times 2 \times 3\times 3\times 3\times \:5$

$2 \times 2 \times 2 \: and \: 2 \times 2 \times 2 \times 3\times 3\times 3\times \:5$

$And \: so \: on...$

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