Question

Use Euclid division Lemma to show that the cube of any positive integer is either of the form $9m,9m + 1$ or $9m + 8$ for some integer m.

Answer 1

Hint: Any number can be written in the form of $3q$or of $3q + 1$or $3q + 2$. Find the cube of all of them making different cases for each.

Let $x$be any positive integer. Then $x$will be either of the form of $3q$or of $3q + 1$or $3q + 2$. So, we have the following cases:
Case 1: When $x = 3q$
In this case, we know:
$\Rightarrow {x^3} = {\left( {3q} \right)^3} = 27{q^3}, \\\ \Rightarrow {x^3} = 9\left( {3{q^3}} \right) = 9m,{\text{ where }}m = 3{q^3}{\text{ }} \\\$
Case 2: When $x = 3q + 1$
In this case we have:

$$\Rightarrow {x^3} = {\left( {3q + 1} \right)^3}, \\\ \Rightarrow {x^3} = 27{q^3} + 27{q^2} + 9q + 1, \\\ \Rightarrow {x^3} = 9q\left( {3{q^2} + 3q + 1} \right) + 1, \\\ \Rightarrow {x^3} = 9m + 1,{\text{ where }}m = q\left( {3{q^2} + 3q + 1} \right) \\\$$

Case 3: When $x = 3q + 2$
In this case we have:

$$\Rightarrow {x^3} = {\left( {3q + 2} \right)^3}, \\\ \Rightarrow {x^3} = 27{q^3} + 54{q^2} + 36q + 8, \\\ \Rightarrow {x^3} = 9q\left( {3{q^2} + 6q + 4} \right) + 8, \\\ \Rightarrow {x^3} = 9m + 8,{\text{ where }}m = q\left( {3{q^2} + 6q + 4} \right) \\\$$

Thus, ${x^3}$can be either of the form of $9m,9m + 1$ or $9m + 8$.
Note: From the above solution, we can say that the cube of any natural number can be written in the form of either $9m,9m + 1$ or $9m + 8$. From this we can conclude that when a cube of any natural number is divided by 9, it gives remainder 0, 1 or 8.