Question

By euclid algorithm to find hcf of 1651 and 2032 and express it in the form of 1651m+2032n

Answer 1

this is the required form

Answer 2

$\huge{\underline{\bf{\red{Solution:-}}}}$

we know that 2032>1651,

So we divide 2032 by 1651 .

1651)2032(1

⠀⠀⠀1651

⠀⠀⠀⠀381)1651(4

⠀⠀⠀⠀⠀⠀⠀1524

⠀⠀⠀⠀⠀⠀⠀⠀127)381(3

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀381.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀0

${\pink{\underbrace{\sf{By\: Euclid's\: division\:lemma}}}}\:$

$:\implies\sf\:2032=1651\times1+381...... ......(i)$

$:\implies\sf\:1651=381\times4+127............(ii)$

$:\implies\sf\:381=127\times3+0...........(iii)$

$\sf\:Since\: remainder\: becomes\:zero.$

$\sf\:So,$

$:\implies\sf\:HCF\:(1651,2032)=127.$

$:\implies\sf\:Now ,$

$\sf\: From\: equation\:(ii)$

$:\implies\sf\:1651=381\times4+127$

$:\implies\sf\:127=1651-381\times4$

$:\implies\sf\:127=1651-(2032-1651\times1)\times4\:\:\:[from\: (i)]$

$:\implies\sf\:127=1651-2032\times4+1651\times4$

$:\implies\sf\:127=1651\times5+2032\times(-4)$

$\bf\:\green{Hence\:,m=5,\:and\:\:n=-4.}$

$\bf\pink{So,\:Hcf\:of \:1651\:and\:2032\:is\: }$

$\bf\pink{expressed\:in\:the\:form\:of}$

$\bf\pink{=1651m+2032n}$

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