Find the number of common, odd positive divisors of 27, 900 & 20, 700. Question as in title, 

Find the number of common, odd positive divisors of $27, 900$ & $20, 700$.

explaining the steps in your method would be much appreciated. 
I prime factorized and got that the common factor is $2^2\cdot3^2\cdot5^2$, then my notes from the relevant lecture simply go on to a solution implied from that step that the number of common odd positive divisors = $(2+1)(2+1)=9$. How the one leads to the other is where I've fallen down.
So how can you tell the number of common, odd positive divisors of $2^2\cdot3^2\cdot5^2$?
 A: Factorize $27{,}900$ and $20{,}700$ :
$$27{,}900=2^2 \times 3^2 \times 5^2 \times 31 ~;~ 20{,}700=2^2 \times 3^2 \times 5^2 \times 23$$
For their common odd divisors ; you need total number of divisors of $3^2 \times 5^2$. 
For counting number of divisors of $3^2 \times 5^2$, you can have $0 , 1$ or $2$ power of $3$ from $3^2$. Hence three ways to chose exponent of $3$ in the common divisor. Similarly for exponent of $5$.
Therefore we have : $3 \times 3 = 9$ total number of common, odd positive divisors of $27{,}900$ & $20{,}700$.
In general, for exponent $\alpha$ of prime $p$ in a number, you'll have $\alpha+1$ ways to select an exponent of $p$ in the divisor.
A: Alternative hint (without factoring the two numbers):


*

*determine $\gcd(27900,20700)=900$ using the Euclidean algorithm for example;

*drop the factors of $2$ from $900$ by halving until you get an odd number, in this case $225\,$.
The common, odd positive divisors of the two original numbers will be precisely the odd, positive divisors of $225=2^2 \cdot 3^2\,$, and there is $(2+1)\cdot(2+1)=9$ of them, as pointed out already.
The advantage of this approach is that you only have to factor the (small) number $225\,$, rather than the two (big) numbers originally given.
