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I can't solve the following:

Find the number of positive-integer solutions that satisfy $x_1\cdot x_2\cdot x_3\cdot x_4=1,000,000$.

Thanks.

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Hi, I am just curious. What if another condition is imposed? $x_1\le x_2\le x_3\le x_4$ – freak_warrior Feb 26 '15 at 8:50
up vote 8 down vote accepted

HINT: Start with the prime factorization of the righthand side: $1,000,000=10^6=2^6\cdot5^6$. How many ways are there to split the $12$ prime factors of $1,000,000$ amongst $x_1,x_2,x_3$, and $x_4$? Remember, it’s possible to have $x_k=1$, with none of the factors.

I suggest writing $x_k=2^{a_k}\cdot5^{b_k}$, so that

$$\begin{align*} x_1x_2x_3x_4&=\left(2^{a_1}\cdot5^{b_1}\right)\left(2^{a_2}\cdot5^{b_2}\right)\left(2^{a_3}\cdot5^{b_3}\right)\left(2^{a_4}\cdot5^{b_4}\right)\\ &=2^{a_1+a_2+a_3+a_4}\cdot5^{b_1+b_2+b_3+b_4}\;; \end{align*}$$

In how many ways can you choose the $a_k$’s and $b_k$’s so that $a_1+a_2+a_3+a_4=6$ and $b_1+b_2+b_3+b_4=6$? (You’ve probably seen problems like these before, but if not, take a look at this article.

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Hi, I am just curious. What if another condition is imposed? $x_1\le x_2\le x_3\le x_4$ – freak_warrior Feb 26 '15 at 8:50
    
@freak_warrior, that is another (much harder) question. Please post it as such. – vonbrand Aug 12 '15 at 19:23

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