Use Euclid’s algorithm to find the HCF of 4052 and 12576.
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12576 = 4052 × 3 + 420
4052 = 420 ×9 + 272
420 = 272 × 1 + 148
272 = 148 × 1 + 124
148 = 124 × 1 + 24
124 = 24 × 5 + 4
24 = ④ × 6 + 0.
∴HCF (12576,4052) = 4. ANS
4052 = 420 ×9 + 272
420 = 272 × 1 + 148
272 = 148 × 1 + 124
148 = 124 × 1 + 24
124 = 24 × 5 + 4
24 = ④ × 6 + 0.
∴HCF (12576,4052) = 4. ANS
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