If a number $x$ can be written in terms of its prime factor $2$ , $3$ and $5$ as $x = 2^{6} 3^{3} 5^{a}$ , and $x$ is a perfect cube , then find the minimum value of $a$ if $a$ is positive number.

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Solution

The correct option is C $3$
If $a = 3$
$x = 2^{6} 3^{3} 5^{a}$
$2, 3$ form triplets, for $5$ to form a triplet minimum value of $a$ is $3.$
$∴ x$ is a perfect cube if $a = 3.$

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Taking the set of natural numbers as the universal set, write down the complement of the following sets :

(i) {x : x is an even natural number}

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(iii) {x : x is a positive multiple of 3}

(iv) {x : x is a prime number}

(v) {x : x is a natural number divisible by 3 and 5}

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(x) {x : x $\geq$ 7}

(xi) {x : x $\in$ N and 2x + 1 > 10}

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