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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.