Question

Find the hcf of 12576 &4052 &express it in the form of 12576x + 4052y

Answer 1

Answer:

ANSWERSince 12576 > 405212576 = 4052 × 3 + 420Since the remainder 420 ≠ 04052 = 420 × 9 + 272Consider the new divisor 420 and the new remainder 272420 = 272 × 1 + 148Consider the new divisor 272 and the new remainder 148272 = 148 × 1 + 124Consider the new divisor 148 and the new remainder 124148 = 124 × 1 + 24Consider the new divisor 124 and the new remainder 24124 = 24 × 5 + 4Consider the new divisor 24 and the new remainder 424 = 4 × 6 + 0The remainder has now become zero, so procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.