Using Euclid's division algorithm,f find HCF of 56,88 and 404
Answers
Answer :
HCF (56, 88 and 404) = 4
Step-by-step explanation :
\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}