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Find L.C.M & H.C.F of 8/9 …

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Find L.C.M & H.C.F of 8/9 ;10/27 &16/81 .
  • 1 answers

Sia ? 6 years, 4 months ago

{tex}\mathrm { HCF } \left( \frac { 8 } { 9 } , \frac { 10 } { 27 } , \frac { 16 } { 81 } \right){/tex} {tex}= \frac { \mathrm { HCF } \text { of the numerators } } { \mathrm { LCM } \text { of the denominators } }{/tex} {tex}= \frac { H C F ( 8,10,16 ) } { L C M ( 9,27,81 ) }{/tex} 
and {tex}\operatorname { LCM } \left( \frac { 8 } { 9 } , \frac { 10 } { 27 } , \frac { 16 } { 81 } \right){/tex} {tex}= \frac { L C M \text { of the numerators } } { \text { HCF of the denominators } }{/tex} {tex}= \frac { L C M ( 8,10,16 ) } { H C F ( 9,27,81 ) }{/tex}
Consider, 
8 = 2 {tex}\times{/tex} 2 {tex}\times{/tex} 2 = 23
10 = 2 {tex}\times{/tex}5
16   = 2 {tex}\times{/tex} 2 {tex}\times{/tex} 2{tex}\times{/tex} 2 = 24
So, the HCF{tex} (8, 10, 16){/tex} = 2 and LCM {tex}(8, 10, 16){/tex} {tex}= 2 ^ { 4 } \times 5 = 80{/tex}
9 = 3 {tex}\times{/tex}3 = 32
27 =3 {tex}\times{/tex}{tex}\times{/tex} 3 =  33
81 = 3 {tex}\times{/tex} 3 {tex}\times{/tex}  3 {tex}\times{/tex}  3 = 34
So, the HCF (9, 27, 81) = 32 = 9 and LCM (9, 27, 81) = 34 = 81
{tex}\Rightarrow \text { HCF } \left( \frac { 8 } { 9 } , \frac { 10 } { 27 } , \frac { 16 } { 81 } \right) = \frac { 2 } { 81 }{/tex} and {tex}\operatorname { LCM } \left( \frac { 8 } { 9 } , \frac { 10 } { 27 } , \frac { 16 } { 81 } \right) = \frac { 80 } { 9 }{/tex}

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