Number of divisors of $2^2.3^3.5^3.7^5$ of the form (4n+1)

Question

The number of divisors of $2^2.3^3.5^3.7^5$ of the form (4n+1).

Since $4n +1$ is odd the divisors must be of the form $3^a.5^b.7^c$ Since $a,b,c$ can have $4,4,6$ values respectively, answer should be $4.4.6$
This answer says that answer is 48 but this question came in my test in which the correct answer was 47.
Please explain without the use of modular arithmetic.

Answer 1

Your exponent $b$ can be any of $4$ options, but that's not entirely true for $a$ and $c$. In fact, if $a+c$ is odd, we have a problem. (Note, for example, that $3^17^0=3$ fails to be of the proper form.) Thus, for each choice of $a$, there are only $3$ viable choices for $c$, namely, those with the same parity as $a$.

I don't see any reason why the answer should be $47$ instead of $48$, though, unless we're excluding the number $1$ as a divisor? That would seem strange.

Edit

As noted below in a comment: If we exclude $0$ as a natural number, as some authors do, then the number $1$ is not of the correct form, so we would exclude the case $a=b=c=0$.