By taking three different values of n verify the truth of the following statements:
(i) If $n$ is even, then $n^3$ is also even.
(ii) If $n$ is odd, then $n^3$ is also odd.
(iii) If $n$ leaves remainder 1 when divided by 3, then it also leaves 1 as remainder when divided by $n^3.$
(iv) If a natural number $n$ is of the form $3p + 2$ then $n^3$ also a number of the same type.

To do:

We have to verify the truth of the given statements by taking three different values of $n$.

Solution: 

(i) $n$ is an even number.

Let $n = 2, 4, 6$ then,

$n^3 = (2)^3$

$= 8$, which is also an even number.

$n^3= (4)^3$

$= 64$, which is also an even number.

$n^3 = (6)^3$

$= 216$, which is also an even number.

Therefore,

The given statement is true.

(ii) $n$ is an odd number.

Let $n = 3, 5, 7$ then,

$n^3 = (3)^3$

$= 27$, which is also an odd number.

$n^3= (5)^3$

$= 125$, which is also an odd number.

$n^3 = (7)^3$

$= 343$, which is also an odd number.

Therefore,

The given statement is true. 

(iii) Let $n = 4, 7, 10$

If $n = 4$, then,

$n^3= 4^3$

$= 64$

$64 = 21\times3+1$

This implies,

Remainder of $64\div3$ is 1.

If $n = 7$, then,

$n^3= 7^3$

$= 343$

$343 = 114\times3+1$

This implies,

Remainder of $343\div3$ is 1.

If $n = 10$, then,

$n^3= 10^3$

$= 1000$

$1000 = 333\times3+1$

This implies,

Remainder of $1000\div3$ is 1.

Therefore,

The given statement is true. 

(iv) Let $p =1, 2, 3$.

If $p = 1$, then,

$n = 3p + 2$

$= 3 \times 1+2$

$=3+2$

$=5$

This implies,

$n^3 = (5)^3$

$= 125$

$= 3 \times 41 + 2$

$= 3p' +2$

If $p = 2$, then,

$n = 3p + 2$

$= 3 \times 2 + 2$

$= 6 + 2$

$= 8$

This implies,

$n^3= (8)^3$

$= 512$

$= 3 \times 170 + 2$

$= 3p' + 2$

If $p = 3$, then

$n = 3p + 2$

$= 3 \times 3 + 2$

$= 9 + 2$

$= 11$

This implies,

$n^3 = (11)^3$

$=1331$

$=3 \times 443 + 2$

$= 3p' + 2$

Therefore,

The given statement is true. 

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Updated on: 10-Oct-2022

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