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If "d" is the HCF of 56 and 72 . find x and y . satisfying d= 56x+72y Asked by arindeep.singh | 21st Apr, 2020,  06:50: PM Expert Answer:   Here, 56 < 72 Applyig Euclid's Divisio Lemma,   72 = 56 × 1 + 16 (r ≠ 0)  ... (i) 56 = 16 × 3 + 8   (r ≠ 0)  ... (ii) 16 = 8 × 2 + 0   Since, r = 0 →HCF(56, 72) = 8 = d   8 = 56 - 16 × 3        .... from (ii)    = 56 - [72 - (56 × 1)] × 3  ... from (i)    = 56 × 4 + 72 × (-3)    = 56 × x + 72 × y   → x = 4 and y = -3 Q) In this question how 56* 4 + 72 * (-3) has come in the last 2nd step.
Asked by arindeep.singh | 23 Apr, 2020, 05:55: PM

8 = 56 - 16 × 3        .... from (ii)

= 56 - [72 - (56 × 1)] × 3  ... from (i)

= 56 - 72× 3  + (56 × 1)× 3

= 56 × 4  - 72× 3

= 56 × 4 + 72 × (-3)

= 56 × x + 72 × y

Answered by Arun | 23 Apr, 2020, 07:03: PM

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