Question

The number of proper divisor of $$2^{p}.6^{q}.15^{r}$$ is
A) (p + q + 1)(q + r + 1)( r+ 1)
B) (p + q + 1)(q + r + 1)(r + 1) - 2
C) (p + q)(q + r)r - 2
D) None of these

Answer 1

Hint: In this question it is given that we have to find the number of proper divisors of $$2^{p}.6^{q}.15^{r}$$. So to find the solution we need to know that the divisors of $$a^{n}$$ is $$\left\{ 1,a,a^{2},a^{3},.......,a^{n}\right\} $$, i.e, the total number of divisors of $$a^{n}$$ is k= (n + 1), where a must be the prime number.
But the number of proper divisors is (k-2)=(n+1-2)=(n-1).......(1)
i.e, excluding 1 and the number itself.
Complete step by step answer:
Given term,
$$2^{p}\cdot 6^{q}\cdot 15^{r}$$
=$$2^{p}\cdot \left( 2\times 3\right)^{q}\cdot \left( 3\times 5\right)^{r} $$
=$$2^{p}\cdot 2^{q}\cdot 3^{q}\cdot 3^{r}\cdot 5^{r}$$ [$$\because \left( ab\right)^{n} =a^{n}\cdot b^{n}$$]
=$$2^{\left( p+q\right) }\cdot 3^{\left( q+r\right) }\cdot 5^{r}$$ [$$\because a^{n}\cdot a^{m}=a^{\left( n+m\right) }$$]
Now we can say that the total number of divisors $$2^{\left( p+q\right) }$$ is (p + q + 1),
Similarly the divisors of $$3^{\left( q+r\right) }$$ and $$5^{r}$$ are (q + r + 1) and (r + 1) respectively.
So the total number of divisors of $$2^{p}\cdot 6^{q}\cdot 15^{r}$$ is k = (p + q + 1)(q + r + 1)( r+ 1).
But for proper divisors we have to exclude 1 and the number itself, so by statement (1) we can say that the number of proper divisors are (k - 2) = (p + q + 1)(q + r + 1)(r + 1) - 2.
Hence the correct option is option B.

Note: So for finding the proper divisor you have to know that, the proper divisor of a number is any divisor of that number other than 1 and the number itself. Let us understand by an example, the divisors of 30 is 1, 2, 3, 5, 6, 10, 15, 30, i.e, we have (8=k) divisors but the proper divisors are (6=k-2).