Big circle, little circles

The statement that $N$ circles of radius $r$ are together equal in area to the one circle of radius $R$ is equivalent to this mathematical statement:

$\pi R^2 = N \pi r^2$

We can solve that for N:

$\frac{R^2}{r^2} = N$

Since we are requiring that $N$ be an integer, the prime factorization of $r$ must be a subset of the prime factorization of $R$ , excluding the whole set, since $N > 1$ .

This problem is now reduced to finding how many unique subsets there are of the prime factorization of $R$ . If $R$ has $P$ different prime factors, and the kth unique prime factor is repeated $n_k$ times, then the number of unique subsets of the prime factorization is

$\left(\displaystyle \prod_{k=1}^{P} \left(n_k + 1\right) \right) - 1$

The " $+ 1$ " is there because for any given subset, we may choose zero of a particular prime factor, and the " $- 1$ " is there because we want to exclude the case where the subset is the whole set.

In this case, $R = 90\,000 = 2 \cdot 2\cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 5 \cdot 5$

There are three prime factors, 2, 3, and 5, repeated 4, 2, and 4 times respectively. Thus the number of permissible $r$ is:

$\left(\displaystyle \prod_{k=1}^{3} \left(n_k + 1\right)\right) - 1 = ((4 + 1) \cdot (2 + 1) \cdot (4 + 1)) - 1 = 74$

Thus, the number of different $r$ for $R = 90\,000$ is 74.