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 \[n = 1, 2, 3 . . . etc.\]

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.