Question

The number of odd proper divisors of $3^p.6^m.{21}^n$ is

• A. (p + 1) (m + 1) (n + 1) - 2
• B. (p + m + n + 1) (n + 1) - 1
• C. (p + 1) (m + 1) (n + 1) - 1
• D. None of these

Answer 1

The correct option is B (p + m + n + 1) (n + 1) - 1

$3^p{6}^m{21}^n$
$=3^p\cdot 3^m\cdot 2^m\cdot 3^n\cdot 7^n$
$=3^{p+m+n}\cdot 2^m\cdot 7^n$
Since we want proper division so will won't take 2 in any way.
So total no. of odd proper division are $=(p+m+n+1)(n+1)-1$ in end $(-1)$ is for when powers of both 3 & 7 becomes 0
So, we don't want that case as in that case the no. will become even.