most recent 30 from http://mathoverflow.net 2013-05-24T22:11:19Z http://mathoverflow.net/feeds/question/118522 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118522/integral-solutions-to-a-1-times-a-2-times-times-a-k-n jigsawmnc 2013-01-10T11:05:48Z 2013-01-11T14:55:38Z

<p>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 $?</p> <p>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?</p>

http://mathoverflow.net/questions/118522/integral-solutions-to-a-1-times-a-2-times-times-a-k-n/118561#118561 Peter Mueller 2013-01-10T18:50:29Z 2013-01-10T18:50:29Z

<p>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. </p>