Question

Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5, 8, 11, ...) and verify the following:
'The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2'.

Answer 1

Five natural numbers of the form (3n + 2) could be written by choosing $n=1,2,3...\text{etc.}$
Let five such numbers be $5,8,11,14,\text{and}17.$
The cubes of these five numbers are:
$5^3=125,8^3=512,{11}^3=1331,{14}^3=2744,\text{and}{17}^3=4913.$
The cubes of the numbers $5,8,11,14\text{and}17$ could expressed as:

$125=3\times 41+2$, which is of the form (3n + 2) for n = 41
$512=3\times 170+2$, which is of the form (3n + 2) for n = 170
$1331=3\times 443+2,$ which is of the form (3n + 2) for n = 443
$2744=3\times 914+2,$which is of the form (3n + 2) for n = 914
$4913=3\times 1637+2,$ which is of the form (3n + 2) for n = 1637

The cubes of the numbers $5,8,11,14,\text{and}17$ can be expressed as the natural numbers of the form (3n + 2) for some natural number n. Hence, the statement is verified.