Show that every positive even integer is of the form (6m+1) or (6m+3) or (6m+5)where m is some integer.

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#### Solution

Let n be any arbitrary positive odd integer.

On dividing n by 6, let m be the quotient and r be the remainder. So, by Euclid’s division lemma, we have

n = 6m + r, where 0 ≤ r ˂ 6.

As 0 ≤ r ˂ 6 and r is an integer, r can take values 0, 1, 2, 3, 4, 5.

⇒ n = 6m or n = 6m + 1 or n = 6m + 2 or n = 6m + 3 or n = 6m + 4 or n = 6m + 5

But n ≠ 6m or n ≠ 6m + 2 or n ≠ 6m + 4 ( ∵ 6m, 6m + 2, 6m + 4 are multiples of 2, so an even integer whereas n is an odd integer)

⇒ n = 6m + 1 or n = 6m + 3 or n = 6m + 5

Thus, any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.

Concept: Euclid’s Division Lemma

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