Question

Question 1
Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Answer 1

Let a be an arbitrary positive integer. Then, by Euclid's division algorithm, corresponding to the positive integers a and 6, there exist non - negative integers q and r such that;
$a=6q+r,where,0\le r<6$
$\Rightarrow a^3=(61+r{)}^3=216q^3+r^3+3.6qr(6q+r)$
$[\therefore (a+b{)}^3=a^3+b^3+3ab(a+b)]$
$\Rightarrow a^3=(216q^3+108q^{2}r+18q{r}^2+r^3)$ ...(i)
Where, $0\le r<6$.
Case I
When r = 0, putting r = 0 in eq (i), we get,
$a^3=216q^3=6\left(36q^3\right)=6m$
Where, $m=36q^3$ is an integer.
Case II
When r = 2, putting r = 2 in eq. (i), we get,
$a^3=(216q^3+216q^2+72)+8$
$a^3=(216q^3+216q^2+72q+6)+2$
$\Rightarrow a^3=6(36q^3+36q^2+12q+1)+2=6m+2$
Where $m=(36q^3+36q^2+12q+1)$ is an integer.
Class III
When r = 3, putting r = 3 in eq. (i), we get,
$a^3=(216q^3+432q^2+288q)+64$
$=6(36q^3+72q^2+48q)+60+4$
$=6(36q^3+72q^2+48q+10)+4=6m+4$
Where, $m=(36q^3+72q^2+48q+10)$ is an integer.
Hence, the cube of a positive integer of the form 6q + r, where q is an integer and r = 0 1, 2, 3, 4, 5 is also of the forms 6m , 6m + 1 6m + 2, 6m + 3, 6m + 4 and 6m + 5 i.e., 6m + r.