Question

Use Euclid’s division lemma to show that the cube of any positive integer is of

the form 9m, 9m + 1 or 9m + 8.​

Answer 1

Question:

Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8

Solution:

Let $\alpha$ be any positive integer and $b$ equals to $3$, then $\alpha = 3q + r$ where $q \geqslant 0$ and

$0 \leqslant r < 3, ∴ = 3q$ or $3q + 1$ or $3q + 2.$

Therefore,

Every number can be represented in these three forms. There are three cases.

Case 1:

When $\alpha = 3q (r=0)$

${\alpha}^{3} = {(3q)}^{3} = {27q}^{3} = 9{(3q}^{3)} = 9 m$

Where m is an integer such that $m = {3q}^{3}$

Case 2:

When $\alpha = 3q + 1 (r=1), {\alpha}^{3} = {(3q + 1)}^{3}$

${\alpha}^{3} = {27q}^{3} + {27q}^{2} + 9q + 1$

${\alpha}^{3} = 9({3q}^{3} + {3q}^{2} + q) + 1$

${\alpha}^{3} = 9m + 1$

Where m is an integer such that $m = ({3q}^{3} + {3q}^{2} + q)$

Case 3:

When $\alpha = 3q + 2 (r=2)$

${\alpha}^{3} = {(3q + 2)}^{3}$

${\alpha}^{3} = {27q}^{3} + {54q}^{2} + 36q + 8$

${\alpha}^{3} = 9({3q}^{3} + {6q}^{2} + 4q) + 8$

${\alpha}^{3} = 9m + 8$

Where m is an integer such that $m = ({3q}^{3} + {6q}^{2} + 4q)$

Therefore,

The cube of any positive integer is of the form $9m, 9m + 1$ or $9m + 8.$