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

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.