Question

find the hcf 4052 and 12576 by using eculids division alorthim​

Answer 1

AnswEr:

$\normalsize\sf\ According\: to \: Euclid \: Division \: algorithm, \\ \normalsize\sf\ a = bq + r \: where, 0 \leq r \textless \: b$

$\rule{170}2$

$\underline{\bigstar\:\sf{H.C.F \: of \: 4052 \: \& \: 12576 :}}$

$\normalsize\sf\ Since, \: 12576 \textgreater 4052 \\ \\ \\ \normalsize\ : \implies\sf\ 12576 = 4052 \times\ 3 + 420 \\ \:\:\:\qquad\footnotesize\qquad\sf{(r \neq 0)} \\ \\ \\ \normalsize\ : \implies\sf\ 4052 = 420 \times\ 9 + 272 \\\qquad\footnotesize\qquad\:\:\:\:\:\sf{(r \neq 0)} \\ \\ \\ \normalsize\ : \implies\sf\ 420 = 272 \times\ 1 + 148 \\ \qquad\footnotesize\qquad\:\:\:\:\:\sf{(r \neq 0)} \\ \\ \\ \normalsize\ : \implies\sf\ 272 = 148 \times\ 1 + 124 \\ \qquad\footnotesize\qquad\:\:\:\:\:\sf{(r \neq 0)} \\ \\ \\ \normalsize\ : \implies\sf\ 148 = 124 \times\ 1 + 24 \\ \qquad\footnotesize\qquad\:\:\:\:\:\sf{(r \neq 0)} \\ \\ \\ \normalsize\ : \implies\sf\ 124 = 24 \times\ 5 + 4 \\ \qquad\footnotesize\qquad\:\:\:\:\:\sf{(r \neq 0)} \\ \\ \\ \normalsize\ : \implies\sf\ 24 = 4 \times\ 6 + 0 \\ \qquad\footnotesize\qquad\:\:\:\:\:\sf{(r = 0)} \\ \\ \\ \qquad\begin{aligned}\bf{\maltese}\:\:\sf HCF(4052,12576)= 4 \:\:\quad\end{aligned}$

Answer 2

Answer:

1. 12576=4052×3+420. Step 2: Since the remainder 420 and new divisor is 4052 apply the division lemma to 4052 and 420, to get.
2. 4052=420×9+272. ...
3. 420=272×1+148. ...
4. 272=148×1+124. ...
5. 148=124×1+24. ...
6. 124=24×5+4. ...
7. 24=4×6+0.