Question
The number of proper divisors of $${2^p} \cdot {6^q} \cdot {15^r}$$ is
A.
$$\left( {p + q + 1} \right)\left( {q + r + 1} \right)\left( {r + 1} \right)$$
B.
$$\left( {p + q + 1} \right)\left( {q + r + 1} \right)\left( {r + 1} \right) - 2$$
C.
$$\left( {p + q} \right)\left( {q + r} \right)r - 2$$
D.
None of these
Answer :
$$\left( {p + q + 1} \right)\left( {q + r + 1} \right)\left( {r + 1} \right) - 2$$
Solution :
$${2^p} \cdot {6^q} \cdot {15^r} = {2^{p + q}} \cdot {3^{q + r}} \cdot {5^r}$$
∴ the number of proper divisors
= {total number of selections from $$\left( {p + q} \right)$$ twos, $$\left( {q + r} \right)$$ threes and $$r$$ fives} $$- 2$$
$$ = \left( {p + q + 1} \right)\left( {q + r + 1} \right)\left( {r + 1} \right) - 2.$$