Question

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

Answer 1

Hint: Here, we have to apply the Euclid’s division lemma. According to Euclid’s division lemma, if we have two positive integers a and b, then there exists unique integers q and r which satisfy the condition $a=bq+r$ where $0\le r\le b$. Here, let a be any positive integer and $b=3$. Therefore, we can write $a=3q+r,q\ge 0,\text{ }0\le r<3$. We have to solve the problem in 3 cases, for r = 0, r = 1 and r = 2

Complete step-by-step answer:

Here, by applying Euclid’s division lemma, we have to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Hence, according to Euclid’s division lemma, if we have two positive integers a and b, then there exists unique integers q and r which satisfy the condition $a=bq+r$ where $0\le rHere, let $a$ be any positive integer and b = 3. Therefore we can write, $a=3q+r,q\ge 0,\text{ }0\le r<3$.

Now, we have to solve the problem in 3 cases:

Case 1: When r = 0, we get:

 $a=3q$

Now, by taking cube on both the sides we get:

${{a}^{3}}={{(3q)}^{3}}$

We know that ${{(xy)}^{n}}={{x}^{n}}{{y}^{n}}$

 Therefore we will get:

$\begin{align}

  & {{a}^{3}}={{3}^{3}}{{q}^{3}} \\

 & {{a}^{3}}=27{{q}^{3}} \\

 & {{a}^{3}}=9(3{{q}^{3}}) \\

\end{align}$$a=9m$, where $m=3{{q}^{3}}$.

Case 2: When r = 1

We have $a=3q+1$

Now by taking cube on both the sides we get:

${{a}^{3}}={{(3q+1)}^{3}}$

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

Here a = 3q and b = 1, therefore our equation becomes:

$\begin{align}

  & {{a}^{3}}={{(3q)}^{3}}+3\times {{(3q)}^{2}}\times 1+3\times 3q\times {{1}^{2}}+{{1}^{3}} \\

 & {{a}^{3}}=27{{q}^{3}}+3\times {{3}^{2}}{{q}^{2}}+9q+1 \\

 & {{a}^{3}}=27{{q}^{3}}+27{{q}^{2}}+9q+1 \\

\end{align}$

From the first three terms, by taking 9 outside we get:

${{a}^{3}}=9\left( 3{{q}^{3}}+3{{q}^{2}}+q \right)+1$

${{a}^{3}}=9m+1$, where $m=3{{q}^{3}}+{{q}^{2}}+q$

Next consider,

Case 3: When r = 2, we get:

$a=3q+2$

Now, by taking cube on both sides we obtain:

$\begin{align}

  & {{a}^{3}}={{(3q+2)}^{3}} \\

 & {{a}^{3}}={{(3q)}^{3}}+3\times {{(3q)}^{2}}\times 2+3\times 3q\times {{2}^{2}}+{{2}^{3}} \\

 & {{a}^{3}}=27{{q}^{3}}+3\times {{3}^{2}}{{q}^{2}}\times 2+9q\times 4+8 \\

 & {{a}^{3}}=27{{q}^{3}}+54{{q}^{2}}+36q+8 \\

\end{align}$

Next, by taking 9 outside from the first 3 terms we get:

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

${{a}^{3}}=9m+8$, where $m=3{{q}^{3}}+6{{q}^{2}}+4q$

Therefore, we can say that a cube of any integer is of the form 9m, 9m + 1 or 9m + 2.


Note: Here, a = 3q + r, the value we can take for is r = 0, r = 1 and r = 3 because r is the remainder and remainder is always less than the divisor. Here the divisor is 3. Therefore r < 3.