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?


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

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    $\begingroup$ What's the relevance or interest of this rather special question? $\endgroup$ – 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.


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