Question

Show that any positive odd integer is of the form $6\mathrm{q}+1$ or $6\mathrm{q}+2$ or $6\mathrm{q}+3$; where $\mathrm{q}$ is some integer.

Answer 1

To prove that any positive odd integer is of the form $6\mathrm{q}+1$ or $6\mathrm{q}+2$ or $6\mathrm{q}+3$; where $\mathrm{q}$ is some integer.

Step $1$: Use Euclid's Division lemma

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

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

According to Euclid's division algorithm,

$\mathrm{a}=\mathrm{bq}+\mathrm{r}$

$\mathrm{a}=6\mathrm{q}+\mathrm{r}$…………………..$\left(1\right)$

where, $(0\le \mathrm{r}<6)$

So$\mathrm{r}$can be either $0$, $1$, $2$, $3$, $4$ and $5$.

Step $2$: Check whether $\mathrm{a}$ is even or odd for different values of $\mathrm{r}$

Case 1: If$\mathrm{r}=0$, then equation$\left(1\right)$becomes

$\mathrm{a}=6\mathrm{q}+0$

$\mathrm{a}=6\mathrm{q}$

The above equation is multiple of $2$.It is an even integer.

Case 2:

If$\mathrm{r}=1$, then equation$\left(1\right)$becomes

$\mathrm{a}=6\mathrm{q}+1$

The above equation will be always an odd integer.

Case 3:

If $\mathrm{r}=2$, then equation $\left(1\right)$ becomes

$\mathrm{a}=6\mathrm{q}+2$

$=2(3\mathrm{q}+1)$

The above equation is multiple of $2$. so it is an even integer.

Case 4:

If $\mathrm{r}=3$ , then the equation$\left(1\right)$ becomes

$\mathrm{a}=6\mathrm{q}+3$

$=3(2\mathrm{q}+1)$

The above equation is multiple of $3$. It is an odd integer.

Case 5: If $\mathrm{r}=4$ , the equation$\left(1\right)$ becomes

$\mathrm{a}=6\mathrm{q}+4$

$=2(3\mathrm{q}+2)$

The above equation is multiple of $2$. It is an even integer.

Case 6: If $\mathrm{r}=5$ , then equation$\left(1\right)$ becomes

$\begin{array}{rcl}\mathrm{a}& =& 6\mathrm{q}+5\\ & =& 6q+3+2\\ & =& 3(q+1)+2\\ & =& 3k+2\left[\text{Where}k=q+1\right]\end{array}$

The above equation is not multiple of $2$. so, it is an odd integer.

On the basis of all the above cases, It is evident that except the case $1$,$3$ and $5$ all the cases are true for the given condition.

Thus, any positive odd integer is of the form $6\mathrm{q}+1$,$6\mathrm{q}+2$ and $6\mathrm{q}+3$