Question

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:

Question:-

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

Solution:-

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.

Answer 2

$\underline\mathtt\red{Question:}$

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

$\underline\mathtt\green{Solution:}$

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 + r 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.

Step-by-step explanation:

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