Question

Show that the cube of a positive integer of the form $6\mathrm{q}+\mathrm{r}$, where $\mathrm{q}$ is an integer and$\mathrm{r}=0,1,2,3,4,5$ is also of the form $6\mathrm{m}+\mathrm{r}$ for some integer $\mathrm{m}$.

Answer 1

Given that

A positive integer of the form $6\mathrm{q}+\mathrm{r}$ ,$\mathrm{q}$ is an integer

To Prove

The cube of a positive integer of the form $6\mathrm{q}+\mathrm{r}$, $\mathrm{q}$ is an integer and$\mathrm{r}=0,1,2,3,4,5$ is also of the form $6\mathrm{m}+\mathrm{r}$

Proof

Step $1$: Use Eucild's division lemma to write a positive integer.

$6\mathrm{m}+\mathrm{r}$ is a positive integer, where $\mathrm{q}$ is an integer and $\mathrm{r}=0,1,2,3,4,5$

Then, the positive integers are of form$6\mathrm{q}$, $6\mathrm{q}+1$, $6\mathrm{q}+2$, $6\mathrm{q}+3$, $6\mathrm{q}+4$ and $6\mathrm{q}+5$.

Step $2$: Take a cube of every positive integer of step $1$, we obtain

Case 1: For$6\mathrm{q}$ ,

$$\begin{gathered} \left(6\mathrm{q}\right)³=216\mathrm{q}³ \\ =6\left(6\mathrm{q}\right)³+0 \end{gathered}$$

$=6\mathrm{m}+0$, where $\mathrm{m}$ is an integer $=\left(6\mathrm{q}\right)³$

Case 2: For $6\mathrm{q}+1$,

$(6\mathrm{q}+1)³=\left(6\mathrm{q}\right)³+{\left(1\right)}^3+3\times {\left(6\mathrm{q}\right)}^2\times 1+3\times 6\mathrm{q}\times {\left(1\right)}^2$ $[\because (\mathrm{a}+\mathrm{b})³=(\mathrm{a})³+{\left(\mathrm{b}\right)}^3+3\times {\left(\mathrm{a}\right)}^2\times \mathrm{b}+3\times \mathrm{a}\times {\left(\mathrm{b}\right)}^2]$

$=216\mathrm{q}³+1+3\times 36{\mathrm{q}}^2+18\mathrm{q}$

$=216\mathrm{q}³+1+108{\mathrm{q}}^2+18\mathrm{q}$

$=6(36\mathrm{q}³+18{\mathrm{q}}^2+3\mathrm{q})+1$

$=6\mathrm{m}+1$ , here$\mathrm{m}$ is an integer and $\mathrm{m}$$=(36\mathrm{q}³+18{\mathrm{q}}^2+3\mathrm{q})$

Case 3: For $6\mathrm{q}+2$,

$(6\mathrm{q}+2)³=\left(6\mathrm{q}\right)³+{\left(2\right)}^3+3\times {\left(6\mathrm{q}\right)}^2\times 2+3\times 6\mathrm{q}\times {\left(2\right)}^2$ $[\because (\mathrm{a}+\mathrm{b})³=(\mathrm{a})³+{\left(\mathrm{b}\right)}^3+3\times {\left(\mathrm{a}\right)}^2\times \mathrm{b}+3\times \mathrm{a}\times {\left(\mathrm{b}\right)}^2]$

$=216\mathrm{q}³+8+6\times 36{\mathrm{q}}^2+18\mathrm{q}\times 4$

$=216\mathrm{q}³+72\mathrm{q}+216{\mathrm{q}}^2+6+2$ here $8$ is split into $6+2$ in order to obtain the multiple of $6$

$=6(36\mathrm{q}³+12{\mathrm{q}}^2+36\mathrm{q}+1)+2$

$=6\mathrm{m}+2$ , here$\mathrm{m}$ is an integer and $\mathrm{m}=(36\mathrm{q}³+12{\mathrm{q}}^2+36\mathrm{q}+1)$

Case 4: For $6\mathrm{q}+3$

$(6\mathrm{q}+3)³=\left(6\mathrm{q}\right)³+{\left(3\right)}^3+3\times {\left(6\mathrm{q}\right)}^2\times 3+3\times 6\mathrm{q}\times {\left(3\right)}^2$ $[\because (\mathrm{a}+\mathrm{b})³=(\mathrm{a})³+{\left(\mathrm{b}\right)}^3+3\times {\left(\mathrm{a}\right)}^2\times \mathrm{b}+3\times \mathrm{a}\times {\left(\mathrm{b}\right)}^2]$

$=216\mathrm{q}³+27+9\times 36{\mathrm{q}}^2+18\mathrm{q}\times 9$

$=216\mathrm{q}³+27+324{\mathrm{q}}^2+162\mathrm{q}$

$=216\mathrm{q}³+324{\mathrm{q}}^2+162\mathrm{q}+24+3$ here $27$ is split into $24+3$ in order to obtain the multiple of $6$

$=6(36\mathrm{q}³+18{\mathrm{q}}^2+3\mathrm{q}+4)+3$

$=6\mathrm{m}+3$ , here$\mathrm{m}=6(36\mathrm{q}³+18{\mathrm{q}}^2+3\mathrm{q}+4)$

Case 5: For $6\mathrm{q}+4$

$(6\mathrm{q}+4)³=\left(6\mathrm{q}\right)³+{\left(4\right)}^3+3\times {\left(6\mathrm{q}\right)}^2\times 4+3\times 6\mathrm{q}\times {\left(4\right)}^2$ $[\because (\mathrm{a}+\mathrm{b})³=(\mathrm{a})³+{\left(\mathrm{b}\right)}^3+3\times {\left(\mathrm{a}\right)}^2\times \mathrm{b}+3\times \mathrm{a}\times {\left(\mathrm{b}\right)}^2]$

$=216\mathrm{q}³+64+12\times 36{\mathrm{q}}^2+18\mathrm{q}\times 16$

$=216\mathrm{q}³+432{\mathrm{q}}^2+288\mathrm{q}+60+4$

$=6(36\mathrm{q}³+72{\mathrm{q}}^2+48\mathrm{q}+10)+4$ here $27$ is split into $24+3$ in order to obtain the multiple of $6$

$=6\mathrm{m}+4$ , here $\mathrm{m}=6(36\mathrm{q}³+72{\mathrm{q}}^2+48\mathrm{q}+10)$

Case 6: For $6\mathrm{q}+5$

$(6\mathrm{q}+5)³=\left(6\mathrm{q}\right)³+{\left(5\right)}^3+3\times {\left(6\mathrm{q}\right)}^2\times 5+3\times 6\mathrm{q}\times {\left(5\right)}^2$ $[\because (\mathrm{a}+\mathrm{b})³=(\mathrm{a})³+{\left(\mathrm{b}\right)}^3+3\times {\left(\mathrm{a}\right)}^2\times \mathrm{b}+3\times \mathrm{a}\times {\left(\mathrm{b}\right)}^2]$

$=216\mathrm{q}³+125+15\times 36{\mathrm{q}}^2+18\mathrm{q}\times 25$

$=216\mathrm{q}³+540{\mathrm{q}}^2+450\mathrm{q}+120+5$

$=6(36\mathrm{q}³+90{\mathrm{q}}^2+75\mathrm{q}+20)+5$ here $125$ is split into $120+5$ in order to obtain the multiple of $6$

$=6\mathrm{m}+5$ , here $\mathrm{m}=6(36\mathrm{q}³+90{\mathrm{q}}^2+75\mathrm{q}+20)$

Hence, the cube of a positive integer is of the form $6\mathrm{q}+\mathrm{r}$ where $\mathrm{q}$ is an integer and$\mathrm{r}=0,1,2,3,4,5$is also of the form $6\mathrm{m}+\mathrm{r}$.

Thus proved.