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

Given:

Statement "Any positive odd integer is of the form $6q\ +\ 1$ or, $6q\ +\ 3$ or, $6q\ +\ 5$, where q is some integer".

To prove:

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

Solution:

According to Euclid's Division Lemma,

If $a$ and $b$ are two positive integers, then

$a\ =\ bq\ +\ r$ where, $0\ \underline{< }\ r\ <\ b$.

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

$a\ =\ 6q\ +\ r$ where, $0\ \underline{< }\ r\ <\ 6$.

So, $r\ =\ 0,\ 1,\ 2,\ 3,\ 4,\ 5$

Now,

When, $\mathbf{r\ =\ 0}$:

$a\ =\ 6q\ +\ 0\ =\ 6q$, which is divisible by 2, so it is an even number.

When, $\mathbf{r\ =\ 1}$:

$a\ =\ 6q\ +\ 1$, which is not divisible by 2, so it is an odd number.

When, $\mathbf{r\ =\ 2}$:

$a\ =\ 6q\ +\ 2\ =\ 2(3q\ +\ 1)$, which is divisible by 2, so it is an even number.

When, $\mathbf{r\ =\ 3}$:

$a\ =\ 6q\ +\ 3$, which is not divisible by 2, so it is an odd number.

When, $\mathbf{r\ =\ 4}$:

$a\ =\ 6q\ +\ 4\ =\ 2(3q\ +\ 2)$, which is divisible by 2, so it is an even number.

When, $\mathbf{r\ =\ 5}$:

$a\ =\ 6q\ +\ 5$, which is not divisible by 2, so it is an odd number.

Therefore, any odd integer is of the form $6q\ +\ 1$ or $6q\ +\ 3$ or $6q\ +\ 5$.

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Updated on: 10-Oct-2022

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