What is the HCF of 1032 &408 and express it in the form 1032m-408*5

1.	What is the HCF of 1032 &408 and express it in the form 1032m-408*5
Answer» Given integers are 408 and 1032 where 408 < 1032By applying Euclid’s division lemma, we get 1032 = 408 {tex}\\times{/tex}\xa02 + 216.Since the remainder ≠ 0, so apply division lemma again on divisor 408 and remainder 216, we get the relation as408 = 216 {tex}\\times{/tex}\xa01 + 192.Since the remainder ≠ 0, so apply division lemma again on divisor 216 and remainder 192216 = 192 {tex}\\times{/tex}\xa01 + 24.Since the remainder ≠ 0, so apply division lemma again on divisor 192 and remainder 24\xa0192 = 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}\xa051032m = 24 + 408 {tex}\\times{/tex}\xa051032m = 24 + 20401032m = 2064 {tex}m = \\frac{{2064}}{{1032}}{/tex}Therefore, m = 2.

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