if H is the hcf of 4052 and 12576 and H=4052×A+12576×B then find the value of H+A+B
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Hcf (4052, 12576) = h 12576 = 3 * 4052 + 420, 4052 = 420 * 9 + 272. 420 = 272 * 1 + 148 272 = 148 * 1 + 12. 148 = 124 * 1 + 24. 124 = 5 * 24 + 4. 24 = 6 * 4 + 0. So 4 is the HCF.
4 = 124 - 5 * 24 = 124 - 5 * (148 - 124) = 6 * 124 - 5 * 148 = 6 * (272 - 148) - 5 * 148 = 6 * 272 - 11 * 148 = 509 * 4052 - 164 * 12576. So a = 509 and b = -164.
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