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, toget

4052 = 420 × 9 + 272

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

420 = 272 × 1 + 148

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

272 = 148 × 1 + 124

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

148 = 124 × 1 + 24

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

124 = 24 × 5 + 4

Consider the new divisor 24 and the new remainder 4, and apply the divisionlemma 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.

Also, 4 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124) =HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052)

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