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Question

If the HCF of 408 and 1032 is expressible in the form 1032 m408×5, find m.


Solution

We need to find m if the H.C.F of 408 and 1032 is expressible in the form 1032 m — 408 x 5

Given integers are 408 and 1032 where 408 < 1032

By applying Euclid’s division lemma, we get 1032 = 408x 2 + 216.

Since the remainder ≠ 0, so apply division lemma on divisor 408 and remainder 216

408 = 216 x 1 + 192.

Since the remainder ≠ 0, so apply division lemma on divisor 216 and remainder 192

216 = 192 x 1 + 24.

Since the remainder ≠ 0, so apply division lemma on divisor 192 and remainder 24

192 = 24 x 8 + 0.

We observe that remainder is 0. So the last divisor is the H.C.F of 408 and 1032.

Therefore,

24 = 1032m — 408 x 5

1032m = 24 + 408 x 5

1032m = 24 + 2040

1032m = 2064

m=  20641032

m = 2

Therefore, m = 2.

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