The number of proper divisors of $2^{p} {.6}^{q} {.15}^{r}$ is

(p + q + 1) (q + r + 1) (r + 1)

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(p + q + 1) (q + r + 1) (r + 1) - 2

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(p + q) (q + r) r - 2

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none of these

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Solution

The correct option is B $(p + q + 1) (q + r + 1) (r + 1) - 2$
$2^{p} {.6}^{q} {.15}^{r}$ We need to find proper division.
Suppose $a^{n}$ is there then $a^{n}$ has factors $(1, a, a^{2} . . . . . . . . . a^{n})$ and $a$ is proper i.e. $a^{n}$ has total division $= (n + 1)$
Now $2^{p} {.6}^{q} {.15}^{r} = 2^{p} {.2}^{q} {.3}^{q} {.3}^{r} {.5}^{r}$
$= 2^{(p + q)} {.3}^{(q + r)} {.5}^{r}$
So, total factors $= (p + q + 1) (q + r + 1) (r + 1)$
However, proper divisiors exclude $1$ and the number itself.
Hence, the answer is $(p + q + 1) (q + r + 1) (r + 1) - 2.$

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