Question

Use Euclid’s Division Algorithm to show that the cube of following numbers is either of form 9m, 9m + 1 or 9m + 8 for some integer m.

89

Answer 1

Euclid’s Division Algorithm: if there are any two integers a and b, there exists q and r such that it satisfies the given condition, $a=bq+r,0\le r<b$

Assume:

Let $a$ be any positive integer and $b=3$

$a=3q+r$, where $q\ge 0and0\le r<3$

$a=3q\mathrm{or}3q+1\mathrm{or}3q+2$

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When $a=$$3q,$

$\begin{array}{rcl}{a}^3& =& {\left(3q\right)}^3\\ & =& 27q^3\\ & =& 9\times 3q^3\\ & =& 9m,\mathrm{where}m=3q^3\end{array}$

Case 2: When $a=3q+1,$

$\begin{array}{rcl}{a}^3& =& {\left(3q+1\right)}^3\\ & =& 27q^3+27q^2+9q+1\\ & =& 9\times \left(3q^3+3q^2+q\right)+1\\ & =& 9m+1,\mathrm{where}m=3q^3+3q^2+q\end{array}$

Case 3: When$a=3q+2$,

$\begin{array}{rcl}{a}^3& =& {\left(3q+2\right)}^3\\ & =& 27q^3+54q^2+36q+8\\ & =& 9\times \left(3q^3+6q^2+4q\right)+8\\ & =& 9m+8,\mathrm{where}m=3q^3+6q^2+4q\end{array}$

Therefore, the cube of any positive integer is of the form $9m,9m+1,or9m+8$.
Hence, the cube of 89 is either of the form $9m,9m+1,or9m+8$.