Express the hcf of 48 and …
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Posted by Devashish K 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let us find HCF of 48 and 18
{tex}48 = 18 \times 2 + 12{/tex}
{tex}18 = 12 \times 1 + 6{/tex}
{tex}12 = 6 \times 2 + 0{/tex}
Hence HCF (48, 18) = 6
Now, 6 = 18 - 12 {tex}\times{/tex} 1
6 = 18 - (48 - 18 {tex}\times{/tex} 2)
6 = 18 - 48 {tex}\times{/tex} 1 + 18{tex}\times{/tex}2
6 = 18 {tex}\times{/tex} 3 - 48 {tex}\times{/tex} 1
6 = 18 {tex}\times{/tex} 3 + 48 {tex}\times{/tex}(-1)
i.e., 6 = 18x + 48y ........ (1)
where x= 3, y = -1
{tex}\therefore{/tex} 6 = 18 {tex}\times{/tex} 3 + 48 {tex}\times{/tex} (-1)
= 18 {tex}\times{/tex} 3 + 48 {tex}\times{/tex} (-1) + 18 {tex}\times{/tex} 48 - 18 {tex}\times{/tex} 48
= 18(3 + 48) + 48(-1 - 18)
= 18 {tex}\times{/tex} 51 + 48 {tex}\times{/tex} (-19)
6 = 18x + 48y ...... (2)
where x = 51, y = -19
Hence, x and y are not unique.
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