How to Find the smallest integer with exactly N odd divisors. Hi All I was trying one problem in which is it asking for the smallest number having N odd divisors. As I know the smallest number having n divisors can be find easily.First we need to find the prime factor of the given number for example take $12$. we can write 12 as a product of integers There are only four ways: $12 = 12 = 2 \times 6 = 3 \times 4 = 2 \times 2 \times 3$ and answer is answer is $5^1 \times 3^1 \times 2^2 = 60$. So $60$ is the smallest number with $12$ divisors.
But Here I am stuck how to find smallest number having all odd divisors.
Thanks.

Every divisor of an odd number is odd, su just skip the prime factor $2$. That is, if you want to find a number with $N$ odd divisors, where
$$N=p_1\cdot\ldots\cdot p_s$$
and
$$p_1\ge\ldots\ge p_s$$
are prime numbers, not necessarily different, take
$$3^{p_1-1}\cdot 5^{p_2-1}\cdot\ldots\cdot q_s^{p_s-1}$$
where $q_s$ is the $s$-th odd prime number.
For your example ($N=12$). Since $N=3\cdot2\cdot 2$, the number you re looking for is $3^2\cdot 5\cdot 7=315$.
