Question

Use Euclid's division lemma to show that the cube of any positive integer is of form $7m$ or $7m+1$ or $7m+6$.

Answer 1

Let us take $a$ as any positive integer and $b=7$

according to Euclid's Division Lemma,

we get $a=7q+r$ , where $r$ is remainder, value of $q$ is more than or equal to $0$ and $r=0,1,2,3,4,5,6$ because $0\le r<b$ and the value of $b=7$

So possible forms will be $7q,7q+1,7q+2,7q+3,7q+4,7q+5,7q+6$

Now, to get cube of these values use the formula

${(a+b)}^3=a^3+3a^{2}b+3a{b}^2+b^3$

In this formula value of $a$ is always $7q$

so, we get

${(7q+b)}^3=343q^3+{147q}^{2}b+21q{b}^2+b^3$

Now divide by $7$ , we get quotient = $49q^3+{21q}^{2}r+3q{r}^2$ and remainder is $b^3$

$⟹(7q+b{)}^3=7(49q^3+21q^{2}r+3q{r}^2)+b^3=7m+b^3$, where $m=49q^3+21q^{2}r+3q{r}^2$

so we have to consider the value of $b^3$

$b=0$ we get $7m+0=7m$

$b=1$ then $1^3=1$ so we get $7m+1$

$b=2$ then $2^3=8$ divided by $7$ we get $1$ as remainder so we get $7m+1$

$b=3$ then $3^3=27$ divided by $7$ we get $6$ as remainder so we get $7m+6$

$b=4$ then $4^3=64$ divided by $7$ we get $1$ as remainder so we get $7m+1$

$b=5$ then $5^3=125$ divided by $7$ we get $6$ as remainder so we get $7m+6$

$b=6$ then $6^3=216$ divided by $7$ we get $6$ as remainder so we get $7m+6$

so all values are in the form of $7m,7m+1$ and $7m+6$