 # Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5, 8, 11, ...) and verify the following: `
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'.

Solution:

Five natural numbers of the form $(3 n+2)$ could be written by choosing $n=1,2,3 \ldots$ etc.

Let five such numbers be $5,8,11,14$, and 17 .

The cubes of these five numbers are:

$125=3 \times 41+2$, which is of the form $(3 n+2)$ for $n=41$

$512=3 \times 170+2$, which is of the form $(3 n+2)$ for $n=170$

$1331=3 \times 443+2$, which is of the form $(3 n+2)$ for $n=443$

$2744=3 \times 914+2$, which is of the form $(3 n+2)$ for $n=914$

$4913=3 \times 1637+2$, which is of the form $(3 n+2)$ for $n=1637$

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