Multiplicative Partitions $abcd=2^8\cdot 3^8$ Suppose $a,b,c,d$ are positive integers and $abcd=2^8\cdot 3^8$.
We consider permutations of an $(a,b,c,d)$ solution identical. 
For example, solutions $(1,1,1,2^8\cdot 3^8),(1,1,2^8\cdot 3^8,1), (1,2^8\cdot 3^8,1,1), (2^8\cdot 3^8,1,1,1)$ are considered equivalent.
How many solutions $(a,b,c,d)$ are there?
 A: If $a_i=2^{x_i}\cdot 3^{y_i}$ then the $x_i$ as well as the $y_i$ form a partition of $8$ into four parts $\geq0$. There are $15$ such partitions. We can sort these $15$ partitions into five  types according to the multiplicities of the occurring sizes as follows:
$$\eqalign{1{\rm\  of\ type\ 1}:\quad &2222\cr
2{\rm\  of\ type\ 2}:\quad &8000, 5111\cr
2{\rm\  of\ type\ 3}:\quad &4400, 3311\cr
8{\rm\  of\ type\ 4}:\quad &7100, 6200,  5300,  6011, 4211, 4022, 3122, 2033\cr
2{\rm\  of\ type\ 5}:\quad &5210, 4310\cr}\tag{1}$$
We now have to determine for each choice of two types $j$ and $k$ the number $q_{jk}$ of different quadruples $(a_1,a_2,a_3,a_4)$ that can be formed by a partition of type $j$ used with base $2$ and a partition of type $k$ used with base $3$. This has to be done "by hand". The symmetric matrix $Q=\bigl[q_{jk}\bigr]$ defined in this way looks as follows:
$$Q=\left[\matrix{
1&1&1&1&1 \cr
1&2&2&3&4 \cr
1&2&3&4&6 \cr
1&3&4&7&12 \cr
1&4&6&12&24 \cr}\right]\ .$$
Let $s=[1\ 2\ 2\ 8\ 2]$ be the vector listing the cardinalities of the five types in $(1)$. Then the total number $N$ of quadruples of the described kind is given by
$$N=s\>Q\>s'=1297\ .$$
