Question

Write the cubes of 5 natural numbers which are of the form 3n + 1 (e.g. 4, 7, 10, ...) and verify the following:
'The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1'.

Answer 1

Five natural numbers of the form (3n + 1) could be written by choosing $n=1,2,3...\text{etc.}$
Let five such numbers be $4,7,10,13,\text{and}16.$
The cubes of these five numbers are:
$4^3=64,7^3=343,{10}^3=1000,{13}^3=2197\text{and}{16}^3=4096$
The cubes of the numbers $4,7,10,13,\text{and}16$ could expressed as:

$64=3\times 21+1$, which is of the form (3n + 1) for n = 21
$343=3\times 114+1$, which is of the form (3n + 1) for n = 114
$1000=3\times 333+1,$ which is of the form (3n + 1) for n = 333
$2197=3\times 732+1,$which is of the form (3n + 1) for n = 732
$4096=3\times 1365+1,$ which is of the form (3n + 1) for n = 1365

The cubes of the numbers $4,7,10,13,\text{and}16$ could be expressed as the natural numbers of the form (3n + 1) for some natural number n; therefore, the statement is verified.