Find the hcf of 81 and 237 and express it as a linear combination of 8

1.	Find the hcf of 81 and 237 and express it as a linear combination of 81 237
Answer» Since, 237 > 81On applying Euclid\'s division algorithm, we get237 = 81 {tex}\\times{/tex}\xa02 + 75 ..........(i)81 = 75 {tex}\\times{/tex}\xa01 + 6 ..........(ii)75 = 6 {tex}\\times{/tex}\xa012 + 3 .........(iii)6 = 3 {tex}\\times{/tex}\xa02 + 0 .............(iv)Hence, HCF (81,237) = 3.In order to write 3 in the form of 81x + 237y,Now,{tex}\\style{font-family:Arial}{\\begin{array}{l}3=75-6\\times12(\\;from(iii))\\\\=75-(81-75\\times1)\\times12\\\\=75-(81\\times12-75\\times12)\\\\=75-81\\times12+75\\times12\\\\=75(1+12)-81\\times12\\\\=75\\times13-81\\times12\\\\=13\\times(237-81\\times2)-81\\times12\\;(\\;Substuting\\;75\\;from(i))\\\\=13\\times237-13\\times81\\times2-81\\times12\\\\=237\\times13-81\\times(26+12)\\\\=81(-38)+237\\times13\\\\=81x+237y\\end{array}}{/tex}Hence, x = -38 and y = 13\xa0

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