Question

Write true (T)I or false (f) for the followiong statements:
(i) 392 is a perfect cube.
(ii) 8640 is not a perfect cube.
(iii) No cube can end with exactly two zeros.
(iv) There is no perfect cube which ends in 4.
(v) For an integer $a,a^3$ is always greater than $a^2$
(vi) if a and b are integers such that $a^2>b^2,then{a}^3>b^3$
(vii) If a adivides b, then $a^3$ divides $b^3$
(viii) If $a^2$ ends in 9 then $a^3$ ends in 7.
(ix) If $a^2$ ends in 5, then $a^3$ ends in 25.
(x) If $a^2$ ends in an even number of zeros. than $a^3$ ends in an odd number of zeros.

Answer 1

(i) False $\because 392=\underset{–}{2\times 2\times 2}\times 7\times 7$
$\because$ Its all factors are not in triplets of equal factors.
(ii) True
$\because 8640=\underset{–}{2\times 2\times 2}\times \underset{–}{2\times 2\times 2}\times \underset{–}{3\times 3\times 3}\times 5$
$\because$ 5 is left.
(iii) True: A number ending three zeros can be a perfect cube.
(iv) False : $\because (4{)}^3=4\times 4\times 4=64$ Which ends with 4.
(v) False: If n is a proper fraction It is not possible
(vi) False: it is not true if a and b are proper fraction.
(vii) rue.
(viii) False, as $a^2$ ends in 9 then $a^3$ does not necessarily ends in 7 it ends in 3 also.
(ix) False, it is not necessarily that $a^3$ ends in 25 it can end also in 75.
(x) False : If $a^2$ ends with even zeros then $a^3$ will ends with odd zeros but of multiple of 3.