Question 187
We are given $n$ identical cubes, each of size $1 \times 1 \times 1$. We arrange all of these $n$ cubes to produce one or more congruent rectangular solids, and let $B(n)$ be the number of ways to do this. For example, if $n=12$, then one arrangement is twelve $1 \times 1 \times 1$ cubes, another is one $3 \times 2 \times 2$ solid, another is three $2 \times 2 \times 1$ solids, another is three $4 \times 1 \times 1$ solids, etc. We do not consider, say, $2 \times 2 \times 1$ and $1 \times 2 \times 2$ to be different; these solids are congruent. You may wish to verify, for example, that $B(12)=11$.
Find, with proof, the integer $m$ such that $10^{m}\lt B\left(2015^{100}\right)\lt 10^{m+1}$.
Find, with proof, the integer $m$ such that $10^{m}\lt B\left(2015^{100}\right)\lt 10^{m+1}$.