Question

find the HCF of 44,58and88 by using eculid division algorithm​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

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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