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# Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q +5, where q is some integer.

Asked by Topperlearning User 4th June 2014, 1:23 PM

Let a be any odd positive integer we need to prove that a is of the form, or , or , where q is some integer.

Since a is an integer. Consider b = 6 as another integer.

Applying Euclid's division lemma, we get,

a = 6q + r, for some integer q  0, and r = 0, 1, 2, 3, 4, 5 since 0 r < 6.

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

However since a is odd so a cannot take the values 6q, 6q+2 and 6q+4 (since all these are divisible by 2)

Also, 6q + 1 = 2 x 3q + 1 = 2k1 + 1, where k1 is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer. So, 6q + 1, 6q + 3, 6q + 5 are odd numbers. Thus, any odd integer can be expressed in of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer

Answered by Expert 4th June 2014, 3:23 PM
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