Question

Find the number of odd proper divisors of \[{{3}^{p}}{{6}^{m}}{{21}^{n}}\].
A. $\left( P+1 \right)\left( m+1 \right)\left( n+1 \right)-2$
B. $\left( P+m+n+1 \right)\left( n+1 \right)-1$
C. $\left( P+1 \right)\left( m+1 \right)\left( n+1 \right)-1$
D. None of these

Answer 1

Hint: If any problem where we have to find proper divisors, we first Express that number into its prime factors i.e. we prime factorize the number so here also we prime factorize the number \[{{3}^{p}}\times {{6}^{m}}\times {{21}^{n}}\]. After getting the prime factors we reject even factors as we only want odd proper divisors.
The proper divisor is a divisor of a number excluding the number itself.

Complete step by step solution: Let \[N={{3}^{p}}\times {{6}^{m}}\times {{21}^{n}}\]
first we do the prime factorization of N:
\[\therefore N={{3}^{p}}\times {{\left( 2\times 3 \right)}^{m}}\times {{\left( 3\times 7 \right)}^{n}}\]
\[={{3}^{p}}\times {{2}^{m}}\times {{3}^{m}}\times {{3}^{n}}\times {{7}^{n}}\]
\[={{2}^{m}}\times {{3}^{p+m+n}}\times {{7}^{n}}\]
\[\therefore N={{2}^{m}}\times {{3}^{p+m+n}}\times {{7}^{n}}\]
Since we want only odd proper divisors ,so we don’t want 2 as one of its divisors. Hence, we will reject ${2}^{\text{m}}$ as the proper divisor.
We have:
$\text{N}=2^0 \times {3}^{\text{m+p+n}}\times 7^{\text{n}}$
which simplifies to,
$\text{N}=1 \times 3^{\text{a}} \times 7^{\text{b}}$, where: a = m + p + n and, b = n
Now, since ‘a’ can be any integer between 0 to p+m+n and ‘b’ can be any integer from 0 to n, we have 0≤a≤p+m+n and 0≤b≤n.
This implies ‘a’ can take total p+m+n+1 values and ‘b’ can take n+1 values. But, we cannot include the given number as one of its proper divisors.
Therefore,
                Total number of odd proper divisors=(p+m+n+1)(n+1)-1

Therefore, Correct option is B.

Note: There is one point that you must remember as you can go wrong there that we are rejecting ${2}^{\text{m}}$ as one of its proper divisors because we only want the odd proper divisors. If we include ${2}^{\text{m}}$ then, even proper divisors will also get included in the calculation. So, this is where you must be very careful about this problem.