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’.

Given:

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

To do:

We have to write the cubes of 5 natural numbers which are of the form $3n+2$ (i.e. 5, 8, 11,……… ) and verify the given statement.

Solution: â€Š

$3n + 2$

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

This implies,

If $n = 1$, then $3n +2= 3(1)+2= 3+2= 5$

If $n = 2$, then $3n +2=3(2)+2=6+2=8$

If $n = 3$, then $3n + 2= 3(3) + 2= 9 + 2 = 11$

If $n = 4$, then $3n + 2= 3(4)+2 = 12 + 2= 14$

If $n = 5$, then $3n +2=3(5) + 2 = 15 +2 = 17$

Therefore,

$(5)^3=5\times5\times5=125$

$125=41\times3+2$

$(8)^3=8\times8\times8=512$

$512=170\times3+2$

$(11)^3=11\times11\times11=1331$

$1331=443\times3+2$

$(14)^3=14\times14\times14=2744$

$2744=914\times3+2$

$(17)^3=17\times17\times17=4913$

$4913=1637\times3+2$

$125, 512, 1331, 2744, 4913$ all leave a remainder of 2 when divided by 3.

Hence, the given statement is true.

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Updated on: 10-Oct-2022

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