Question

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

Answer 1

Answer:

According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + where 0 ≤ r < b.

Let a be the positive odd integer which when divided by 6 gives q as quotient and r as remainder.

According to Euclid’s division lemma

a = bq + r

a = 6q + r………………….(1)

where, (0 ≤ r < 6)

So r can be either 0, 1, 2, 3, 4 and 5.

Case 1:

If r = 1, then equation (1) becomes

a = 6q + 1

The Above equation will be always as an odd integer.

Case 2:

If r = 3, then equation (1) becomes

a = 6q + 3

The Above equation will be always as an odd integer.

Case 3:

If r = 5, then equation (1) becomes

a = 6q + 5

The above equation will be always as an odd integer.

∴ Any odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.

Hence proved.

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Answer 2

Answer:

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Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

Now substituting the value of r, we get,

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.