If the HCF of 408 and …
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Sia ? 6 years, 6 months ago
Given integers are 408 and 1032 where 408 < 1032
By applying Euclid’s division lemma, we get 1032 = 408 {tex}\times{/tex} 2 + 216.
Since the remainder ≠ 0, so apply division lemma again on divisor 408 and remainder 216, we get the relation as
408 = 216 {tex}\times{/tex} 1 + 192.
Since the remainder ≠ 0, so apply division lemma again on divisor 216 and remainder 192
216 = 192 {tex}\times{/tex} 1 + 24.
Since the remainder ≠ 0, so apply division lemma again on divisor 192 and remainder 24
192 = 24 × 8 + 0.
Now the remainder has become 0. Therefore, the H.C.F of 408 and 1032 = 24.
Therefore,
24 = 1032m - 408 {tex}\times{/tex} 5
1032m = 24 + 408 {tex}\times{/tex} 5
1032m = 24 + 2040
1032m = 2064
{tex}m = \frac{{2064}}{{1032}}{/tex}
Therefore, m = 2.
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