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Find H. C. F at 592 and 252. Express it in linear combination of 592 and 252 |
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Answer» 592=252×2+88252=88×2+7688=76×1+1276=12×6+412=4×3+0Here,r=0So,HCF OF 592 & 252 is 4 We need to find the H.C.F. of 592 and 252 and express it as a linear combination of 592 and 252.By applying Euclid’s division lemma592 = 252 x 2 + 88Since remainder ≠ 0, apply division lemma on divisor 252 and remainder 88252 = 88 x 2 + 76Since remainder ≠ 0, apply division lemma on divisor 88 and remainder 7688 = 76 x 1 + 12Since remainder ≠ 0, apply division lemma on divisor 76 and remainder 1276 = 12 x 6 + 4Since remainder ≠ 0, apply division lemma on divisor 12 and remainder 412 = 4 x 3 + 0.Therefore, H.C.F. = 4.Now, 4 = 76 – 12 x 6= 76 – 88 – 76 x 1 x 6= 76 – 88 x 6 + 76 x 6= 76 x 7 – 88 x 6= 252 – 88 x 2 x 7 – 88 x 6= 252 x 7- 88 x 14- 88 x 6= 252 x 7- 88 x 20= 252 x 7 – 592 – 252 x 2 x 20= 252 x 7 – 592 x 20 + 252 x 40= 252 x 47 – 592 x 20= 252 x 47 + 592 x (-20)Hence obtained. |
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