 # Write the cubes of 5 natural numbers which are of the form 3n + 1 (e.g. 4, 7, 10, ...) and verify the following: `
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'.

Solution:

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

Let five such numbers be $4,7,10,13$, and 16 .

The cubes of these five numbers are:

$4^{3}=64,7^{3}=343,10^{3}=1000,13^{3}=2197$ and $16^{3}=4096$

The cubes of the numbers $4,7,10,13$, and 16 could expressed as:

$64=3 \times 21+1$, which is of the form $(3 n+1)$ for $n=21$

$343=3 \times 114+1$, which is of the form $(3 n+1)$ for $n=114$

$1000=3 \times 333+1$, which is of the form $(3 n+1)$ for $n=333$

$2197=3 \times 732+1$, which is of the form $(3 n+1)$ for $n=732$

$4096=3 \times 1365+1$, which is of the form $(3 n+1)$ for $n=1365$

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