Find hcf of 81and237 and express …
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Sia ? 6 years, 6 months ago
Since, 237 > 81
On applying Euclid's division algorithm, we get
237 = 81 {tex}\times{/tex} 2 + 75 ..........(i)
81 = 75 {tex}\times{/tex} 1 + 6 ..........(ii)
75 = 6 {tex}\times{/tex} 12 + 3 .........(iii)
6 = 3 {tex}\times{/tex} 2 + 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
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