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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 $6m\ +\ r$.
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