find hcf 52,117 and express in form of 52x+117y
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a = bq + r
117 > 52
let a= 117, b=52
117 = 52 x 2 + 13 ( 52 is divisor )
52 = 13 x 4 + 0 ,the division process stops here, as remainder becomes 0 ,so HCF is 13 ( divisor )
13 can be also expressed as 52( -2 ) + 117 ( 1)
117 > 52
let a= 117, b=52
117 = 52 x 2 + 13 ( 52 is divisor )
52 = 13 x 4 + 0 ,the division process stops here, as remainder becomes 0 ,so HCF is 13 ( divisor )
13 can be also expressed as 52( -2 ) + 117 ( 1)
Esaiamudhu:
good do it by reverse lemma
it is by reverse lemma
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ok
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Hcf is 13. And I can't understand the second question
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