Show that the cube of a positive integer is of the form $6 q+r$, where $q$ is ana integer and $r=0,1,2,3,4,5$.

Solution:

Suppose a be any arbitrary positive integer, then by Euclid's division algorithm, corresponding to the positive integers a and 6, there exists non-negative integers a and r such that

$a=6 q+r$, where $0 \leq r<6$

$\Rightarrow a^{3}=(6 q+r)^{3}=216 q^{3}+r^{3}+3 \times 6 q \times r(6 q+r)$

 

$\Rightarrow a^{3}=6\left(216 q^{3}+108 q^{2} r+18 q r^{2}\right)+r^{3} \quad \ldots \ldots(1) \quad$ where, $0 \leq r<6$

Case: 1

When r = 0.

Putting r = 0 in (1), we get

$a^{3}=216 q^{3}=6\left(36 q^{3}\right)=6 m$

Where, $m=36 q^{3}$ is an integer

Case: 2

When r = 1.

Putting r = 1 in (1), we  get

$a^{3}=\left(216 q^{3}+108 q^{2}+18 q\right)+1$

$\Rightarrow a^{3}=6\left(36 q^{3}+18 q^{2}+3 q\right)+1$

$\Rightarrow a^{3}=6 m+1$

Where, $m=\left(36 q^{3}+18 q^{2}+3 q\right)$ is an integer

Case: 3

When r = 2.

Putting r = 2 in (1), we get

$a^{3}=\left(216 q^{3}+216 q^{2}+72 q\right)+8$

$\Rightarrow a^{3}=\left(216 q^{3}+216 q^{2}+72 q+6\right)+2$

$\Rightarrow a^{3}=6\left(36 q^{3}+36 q^{2}+12 q+1\right)+2$

$\Rightarrow a^{3}=6 m+2$

Where, $m=\left(36 q^{3}+36 q^{2}+12 q+1\right)$ is an integer

Case: 4

When r = 3.

Putting r = 3 in (1), we get

$a^{3}=\left(216 q^{3}+324 q^{2}+162 q\right)+27$

$\Rightarrow a^{3}=\left(216 q^{3}+324 q^{2}+162 q+24\right)+3$

$\Rightarrow a^{3}=6\left(36 q^{2}+54 q^{2}+27 q+4\right)+3$

$\Rightarrow a^{3}=6 m+3$

Where, $m=\left(36 q^{2}+54 q^{2}+27 q+4\right)$ is an integer

Case: 5

When r = 4

Put r = 4 in (1), we get

$a^{3}=\left(216 q^{3}+432 q^{2}+288 q\right)+64$

$\Rightarrow a^{3}=\left(36 q^{3}+72 q^{2}+48 q+60\right)+4$

$\Rightarrow a^{3}=6\left(36 q^{3}+72 q^{2}+48 q+10\right)+4$

$\Rightarrow a^{3}=6 m+4$

Where, $m=\left(36 q^{3}+72 q^{2}+48 q+10\right)$ is an integer

Case: 6

When r = 5.

Putting r = 5 in (1), we get

$a^{3}=216 q^{3}+540 q^{2}+450 q+125$

$\Rightarrow a^{3}=6\left(36 q^{3}+90 q^{2}+75 q+20\right)+5$

$\Rightarrow a^{3}=6 m+5$

Where, $m=\left(36 q^{3}+90 q^{2}+75 q+20\right)$ is an integer

Hence, the cube of any positive integer of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.