# 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 $6m + r$.

Given:

"Cube of a positive integer is of the form $6q\ +\ r$, where $q$ is an integer and $r\ =\ 0,\ 1,\ 2,\ 3,\ 4,\ 5$ is also of the form $6m\ +\ r$".

To do:

We have to prove the given statement.

Solution:

According to Euclid's division lemma;

If $a$ and $b$ are two positive integers;

$a\ =\ bq\ +\ r$, where $0\ \underline{< }\ r\ <\ b$.

If $b\ =\ 6$, then;

$a\ =\ 6q\ +\ r$, where $0\ \underline{< }\ r\ <\ 6$.

So, $r\ =\ 0,\ 1,\ 2,\ 3,\ 4,\ 5$

When $r\ =\ 0$:

$a\ =\ 6q$

Cubing both sides, we get:

$a^3\ =\ (6q)^3$

$a^3\ =\ 216q^3$

$a^3\ =\ 6(36q^3)$

$a^3\ =\ 6m$, where $m\ =\ 36q^3$

When $r\ =\ 1$:

$a\ =\ 6q\ +\ 1$

Cubing both sides, we get:

$a^3\ =\ (6q\ +\ 1)^3$

$a^3\ =\ 216q^3\ +\ 108q^2\ +\ 18q\ +\ 1$

$a^3\ =\ 6(36q^3\ +\ 18q^2\ +\ 3q)\ +\ 1$

$a^3\ =\ 6m\ +\ 1$, where $m\ =\ 36q^3\ +\ 18q^2\ +\ 3q$

When $r\ =\ 2$:

$a\ =\ 6q\ +\ 2$

Cubing both sides, we get:

$a^3\ =\ (6q\ +\ 2)^3$

$a^3\ =\ 216q^3\ +\ 216q^2\ +\ 72q\ +\ 8$

$a^3\ =\ 216q^3\ +\ 216q^2\ +\ 72q\ +\ 6\ +\ 2$

$a^3\ =\ 6(36q^3\ +\ 36q^2\ +\ 12q\ +\ 1)\ +\ 2$

$a^3\ =\ 6m\ +\ 2$, where $m\ =\ 36q^3\ +\ 54q^2\ +\ 12q\ +\ 1$

When $r\ =\ 3$:

$a\ =\ 6q\ +\ 3$

Cubing both sides, we get:

$a^3\ =\ (6q\ +\ 3)^3$

$a^3\ =\ 216q^3\ +\ 324q^2\ +\ 162q\ +\ 27$

$a^3\ =\ 216q^3\ +\ 324q^2\ +\ 162q\ +\ 24\ +\ 3$

$a^3\ =\ 6(36q^3\ +\ 54q^2\ +\ 27q\ +\ 4)\ +\ 3$

$a^3\ =\ 6m\ +\ 3$, where $m\ =\ 36q^3\ +\ 54q^2\ +\ 27q\ +\ 4$

When $r\ =\ 4$:

$a\ =\ 6q\ +\ 4$

Cubing both sides, we get:

$a^3\ =\ (6q\ +\ 4)^3$

$a^3\ =\ 216q^3\ +\ 432q^2\ +\ 288q\ +\ 64$

$a^3\ =\ 216q^3\ +\ 432q^2\ +\ 288q\ +\ 60\ +\ 4$

$a^3\ =\ 6(36q^3\ +\ 72q^2\ +\ 48q\ +\ 10)\ +\ 4$

$a^3\ =\ 6m\ +\ 4$, where $m\ =\ 36q^3\ +\ 72q^2\ +\ 48q\ +\ 10$

When $r\ =\ 5$:

$a\ =\ 6q\ +\ 5$

Cubing both sides, we get:

$a^3\ =\ (6n\ +\ 5)^3$

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

$a^3\ =\ 216q^3\ +\ 540q^2\ +\ 450q\ +\ 120\ +\ 5$

$a^3\ =\ 6(36q^3\ +\ 90q^2\ +\ 75q\ +\ 20)\ +\ 5$

$a^3\ =\ 6m\ +\ 5$, where $m\ =\ 36q^3\ +\ 72q^2\ +\ 48q\ +\ 10$

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

Updated on: 10-Oct-2022

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