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Use euclid\'s division algorithm to find HCF of 1651 and 2032.write steps aldo |
| Answer» \xa02032 = 1651 {tex} \\times{/tex} 1 + 381 .1651 = 381 {tex} \\times{/tex} 4 + 127\xa0381 = 127 {tex} \\times{/tex} 3 + 0.\xa0Since the remainder becomes 0 here, so HCF of 1651 and 2032 is 127.{tex} \\therefore{/tex}\xa0HCF (1651, 2032) = 127.Now,{tex} 1651 = 381 \\times 4 + 127{/tex}{tex} \\Rightarrow \\quad 127 = 1651 - 381 \\times 4{/tex}{tex} \\Rightarrow \\quad 127 = 1651 - ( 2032 - 1651 \\times 1 ) \\times 4{/tex}\xa0[from 2032 = 1651 {tex} \\times{/tex} 1 + 381]{tex} \\Rightarrow \\quad 127 = 1651 - 2032 \\times 4 + 1651 \\times 4{/tex}{tex} \\Rightarrow \\quad 127 = 1651 \\times 5 + 2032 \\times ( - 4 ){/tex}Hence, m = 5, n = -4. | |