Find the number of odd proper divisors of `3^pxx6^mxx21^ndot`

Options

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

Explanation

We have,
`3^(m)xx6^(n)xx21^(p)=3^(m)xx2^(n)xx3^(n)xx3^(p)xx7^(p)=2^(n)xx3^(m+n+p)xx7^(p)`
Clearly,
Number of odd proper divisors
= Number of ways of selecting any number of `3^(s)and7^(s)` from (m+n+p) identical `3^(s)andP` identical `7^(s)`.
`=(m+n+p+1)(p+1)-1`.