Use euclid's division algorithm to find …

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Sia ? 6 years, 4 months ago
2032 = 1651 {tex} \times{/tex} 1 + 381 .
1651 = 381 {tex} \times{/tex} 4 + 127
381 = 127 {tex} \times{/tex} 3 + 0.
Since the remainder becomes 0 here, so HCF of 1651 and 2032 is 127.
{tex} \therefore{/tex} HCF (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} [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.
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