Question

Use division algorithm to find the hcf of 4052 and 12576

Answer 1

Answer:

Step-by-step explanation:

Step 1: Since 12576 > 4052, apply the division lemma to 12576 and 4052, to get

12576 = 4052 × 3 + 420

Step 2: Since the remainder 420 ≠ 0, apply the division lemma to 4052 and 420, to get

4052 = 420 × 9 + 272

Step 3: Consider the new divisor 420 and the new remainder 272, and apply the division lemma to get

420 = 272 × 1 + 148

Consider the new divisor 272 and the new remainder 148, and apply the division lemma to get

272 = 148 × 1 + 124

Consider the new divisor 148 and the new remainder 124, and apply the division lemma to get

148 = 124 × 1 + 24

Consider the new divisor 124 and the new remainder 24, and apply the division lemma to get

124 = 24 × 5 + 4

Consider the new divisor 24 and the new remainder 4, and apply the division lemma to get

24 = 4 × 6 + 0

The remainder has now become zero, so procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.

Hope it helps you....

Answer 2

$\huge\sf\pink{Answer}$

☞ HCF of (4052,12576) = 4

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$\huge\sf\blue{Given}$

➝ 4052 and 12576

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$\huge\sf\gray{To \:Find}$

✭ Their HCF

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$\huge\sf\purple{Steps}$

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

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

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

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