Question

using the Euclid division Lemma find the HCF of 56, 88 and 404

Answer 1

the HCF is 4

I hope it helps u

Answer 2

$\underline {\blue { Euclid's \:Division \:Lemma}}$

Given positive integers a and b , there exist unique pair of integers q and r satisfying

a = bq + r , 0 ≤ r < b .

$HCF \: of \: 56 \:and \: 88 \: applying \\ Euclid's\: division \: lemma .$

$88 = 56 \times 1 + 32$

$Consider \: the \: division \: of \: 56 \: with \\the \: remainder \: 32 \: applying \: the \: division \\lemma \: to \:get$

$56 = 32 \times 1 + 24$

$Remainder = 24 \: \neq 0 \: and \: applying \\ the \: division \\lemma \: to \:get$

$32 = 24 \times 1 + 8$

$\implies 24 = 8 \times 3 + 0$

$Notice \:that \: the \: Remainder \: has \\become \:zero .\\The \: HCF \:of \: 56 \:and \: 88 \\: is \: the \: divisor \: at \:this \:stage , i.e., \:8 .$

$ii) Similarly ,\: find \: HCF \:of \: 8 \:and \: 404$

$404 = 8\times 50 + 4$

$8 = 4\times 2 + 0$

$Remainder = 0$

$\therefore HCF( 404,8 ) = 4$

$Now, HCF(56,88 \:and \: 404 ) = 4$

Therefore.,

$\red { HCF(56,88 \:and \: 404 )}\green { = 4}$

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