Is the fifth power of a number divisible by the cube of that number? Is the product of two cube numbers a cube number? Is q^k both a fourth power and a sixth power (q prime)?

Step 1: Solve Question 4.

c) किसी संख्या की पाँचवी घात उस संख्या के घन से विभाज्य होती है। (Is the fifth power of a number divisible by the cube of that number?)
Let the number be $x$. Its fifth power is $x^5$ and its cube is $x^3$.
To check divisibility, we look at the ratio:

$$\frac{x^5}{x^3} = x^{5-3} = x^2$$

Since $x^2$ is an integer for any integer $x$, the fifth power of a number is always divisible by the cube of that number (provided $x \neq 0$).
This statement is हमेशा सत्य है (always true).

d) दो घन संख्याओं का गुणनफल एक घन संख्या होती है। (Is the product of two cube numbers a cube number?)
Let the two cube numbers be $a^3$ and $b^3$.
Their product is $a^3 \times b^3$.
Using the property of exponents, $(ab)^3 = a^3 b^3$.
So, $a^3 \times b^3 = (ab)^3$.
Since $(ab)^3$ is a cube number, the product of two cube numbers is always a cube number.
This statement is हमेशा सत्य है (always true).

e) $q^k$ एक चतुर्थ घात और एक छठी घात दोनों है ($q$ एक अभाज्य संख्या है)। ($q^k$ is both a fourth power and a sixth power ($q$ is a prime number).)
If $q^k$ is a fourth power, then $k$ must be a multiple of 4. So, $k = 4m$ for some integer $m$.
If $q^k$ is a sixth power, then $k$ must be a multiple of 6. So, $k = 6n$ for some integer $n$.
For $k$ to be both a multiple of 4 and a multiple of 6, $k$ must be a multiple of the least common multiple (LCM) of 4 and 6.
LCM$(4, 6) = 12$.
So, $k$ must be a multiple of 12. For example, if $k=12$, then $q^{12} = (q^3)^4$ (a fourth power) and $q^{12} = (q^2)^6$ (a sixth power).
Since such a $k$ exists (e.g., $k=12$), this statement is कभी-कभी सत्य है (sometimes true).