Odd Numbers 3

Odd divisors are obtained only from powers of $5$ and $7$ .

$Reason$ :-
In general, we can just use:

odd * odd = odd and even * even = even .

However, for a more detailed proof for odd numbers, consider any odd number of the form $2k+1$ . Thus, from the binomial expansion of the $n^{th}$ power of this odd number, we get:

$(2k+1)^{n} = \displaystyle \sum_{r=0}^n {n\choose r} \times (2k)^{r} \times (1)^{n-r}$

In this expansion, all the terms except the first are even .The first term is $1$ .

Thus, $(2k+1)^{n}$ is of of the form $2m+1$ which is an odd number . Thus, powers of odd numbers are also odd .

Coming back to the question, in the given number, there are $7$ powers of $5$ available which are:

$5^{0}, 5^{1}, 5^{2}, \ldots, 5^{6}.$

Similarly, for $7$ we have $9$ powers available:

$7^{0}, 7^{1}, 7^{2}, \ldots, 7^{8}.$

Thus, the total number of odd divisors is the number of products of these powers of $5$ and $7$ taking one power from each set at a time.

Hence, total number of odd divisors is given by the number of combinations of powers of $5$ and $7$ :

$\Rightarrow$ $7 \choose 1$ $\times$ $9 \choose 1$

$\Rightarrow$ $7 \times 9$

$\Rightarrow \boxed{63}$