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’.

Given:

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.

To do:

We have to write the cubes of 5 natural numbers which are of the form $3n+1$ (e.g.,4, 7, 10, …………) and verify the given statement.

Solution: â€Š

$3n + 1$

Let $n = 1, 2, 3, 4, 5$

This implies,

If $n = 1$, then $3n +1= 3(1)+1= 3+1= 4$

If $n = 2$, then $3n +1=3(2)+1=6+1=7$

If $n = 3$, then $3n + 1= 3(3) + 1= 9 + 1 = 10$

If $n = 4$, then $3n + 1= 3(4)+1 = 12 + 1= 13$

If $n = 5$, then $3n +1=3(5) + 1 = 15 +1 = 16$

Therefore,

$(4)^3=4\times4\times4=64$

$64=21\times3+1$

$(7)^3=7\times7\times7=343$

$343=114\times3+1$

$(10)^3=10\times10\times10=1000$

$1000=333\times3+1$

$(13)^3=13\times13\times13=2197$

$2197=732\times3+1$

$(16)^3=16\times16\times16=4096$

$4096=1365\times3+1$

$64, 343, 1000, 2197, 4096$ all leave a remainder of 1 when divided by 3.

Hence, the given statement is true.

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Updated on: 10-Oct-2022

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