Question

2. Show that any positive odd integer is of the form
6q + 1, or 6q + 3, or 64 + 5, where is
some integer


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Answer 1

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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

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Solution:

To solve this question, let us use Euclid’s division algorithm.

Let's assume any positive integer ‘a’ of the form 6q + r, where q is some integer.

This means that 0 ≤ r < 6, i.e, r = 0 or 1 or 2 or 3 or 4 or 5 but it can’t be 6 because r is smaller than 6.

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

Therefore, possible values of a are 6q + 0 or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5.

Now, 6q + 1 = 2 × 3 q + 1 = 2k1 + 1, where k1 is a positive integer

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

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

Clearly, 6q + 1, 6q + 3 and 6q + 5 are of the form 2k + 1, where k is an integer. Therefore, 6q + 1, 6q + 3 and 6q + 5 are not exactly divisible by 2.

Hence, these expressions of numbers are odd numbers and therefore any odd integers can be expressed in the form 6q + 1 or 6q + 3 or 6q + 5.