# Integral solutions to $a_1 \times a_2 \times … \times a_k = N$ [closed]

How many integral solutions are possible for the equation $a_1 \times a_2 \times \ldots \times a_k = N$ where each of $a_1, a_2, \ldots, a_k$ satisfy the property $0 \leq a_i \leq 9$?

The question is to find out the number of possible combinations $(a_1, a_2, \dots, a_L)$ such that $(\frac{a_1}{a_2})(\frac{a_3}{a_4}) \dots = N$ with the constraint that $a_1, a_2, \dots ,a_L$ satisfy $0 \leq a_i \leq 9$. So, my approach was to consider $N$ as $\frac{N}{1}, \frac{2N}{2}, \ldots$ till $\frac{mN}{m}$ such that $mN \leq 9^k, k = L - 1 or L$. Now, if I get the possible combinations for both numerator and denominator(for all these fractions), then I could multiply and add these combination numbers to get the final result. Can any other approach be adopted for doing it?

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## closed as off topic by Steven Landsburg, Emil Jeřábek, Andreas Blass, Brendan McKay, Felipe VolochJan 10 '13 at 20:21

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What's the relevance or interest of this rather special question? – Peter Mueller Jan 10 '13 at 16:10

Write $N=2^a3^b5^c7^d$. (If $N$ has not this shape, there are no solutions.) In a solution, let $m_i$ be the number of occurrences of the factor $i$. So $m_1+m_2+\dots+m_9=k$, $m_2+2m_4+m_6+3m_8=a$, $m_3+m_6+2m_9=b$, $m_5=c$, and $m_7=d$. The tuples $m_i$ can be computed in terms of $a$, $b$, $c$ and $d$. For each such tuple, the number of solutions equals the multinomial coefficient $\binom{k}{m_1,m_2,\dots,m_9}$ (as the $a_i$'s are not ordered). I doubt that a more precise answer or a closed formula can be obtained.