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I had previously solved $xy=1,000,000;$ $x,y \in \mathbb Z$, I believe: 1,000,000 has 49 factors so there are 49 pairs since $x$ and $y$ could be both positive or both negative.

Please would you help me find as efficient and systematic a way as possible.

EDIT: since there are many solutions, finding how many there are would be fine. However, unfortunately, (1,1,1000000) is considered distinct from (1000000,1,1).

Thank you

John Smith
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  • What exactly is the question????? – barak manos Nov 24 '16 at 19:22
  • Do you actually want to find all the solutions, or do you want to count the number of solutions (which can be done without listing them out)? Additionally, for your purposes, does $(1,1,1000000)$ count as different from $(1000000,1,1)$? – Aaron Nov 24 '16 at 19:27
  • Counting the number of solutions efficiently would be better: you are right. But I'm afraid those triples would be considered distinct, yes. I will edit accordingly – John Smith Nov 24 '16 at 19:32
  • There are $28^2$ answers. – Thomas Andrews Nov 24 '16 at 19:37
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    @ThomasAndrews You forgot to add a factor of 4 to account for the fact that the numbers can be negative (1 for the positive, 3 for the number of ways of picking 2 things to be negative) – Aaron Nov 24 '16 at 19:39

4 Answers4

5

hint

The only prime factors of $10^6$ are $2$ and $5$. So each of $x,y,z$ (if they are $> 1$) should have the same prime factors. Thus we have $$2^{x_1}\, 5^{x_2} \, 2^{y_1}\, 5^{y_2}\, 2^{z_1}\, 5^{z_2} = 2^{6}\, 5^{6}$$ So you need to count the number of non-negative integer triplets $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ such that \begin{align*} x_1 + y_1 +z_1 & =6\\ x_2+y_2+z_2&=6 \end{align*}

The counting for such equations can be done using $\binom{n+r-1}{r-1}$ stars and bars method.

Once you get the count of positive integer solutions, negatives will be very easy to count.

Anurag A
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$xyz = 10^6 = 2^6 5^6 = 2 \times2 \times2 \times2 \times2 \times2 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5$

Now find the number of possible ways of dividing those 12 elements into 3 sets ($x$, $y$ and $z$).

cubeception
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  • This is what I hoped to do (I should have said, I'm sorry), but I'm not sure how – John Smith Nov 24 '16 at 19:34
  • There are more ways how to do that. My prefered way is to think of it this way. $x_2$ is the number of twos selected to the $x$ set, $y_2$ is the number of twos in $y$ and so on. Then $x_2+y_2+z_2 =6$ and conversely $x_5+y_5+z_5 =6$, where $x_2,y_2,z_2 \in \mathbb{N} \cup 0$. As you see, you just have to say how many solutions there are for one of the equations. The number of all possible $x,y,z$ is then square of that number. – cubeception Nov 24 '16 at 19:51
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If $x$, $y$, and $z$ are positive integers with $xyz=2^6\cdot5^6$, then $x=2^a\cdot5^{a'}$, $y=2^b\cdot5^{b'}$, and $z=2^c\cdot5^{c'}$ with $0\le a,a',b,b',c,c'$ and $a+b+c=a'+b'+c'=6$. There are $1+2+3+4+5+6+7=28$ non-negative solutions to $a+b+c=6$, and likewise for $a'+b'+c'=6$, so there is a total of $28\cdot28$ positive triples with product $1{,}000{,}000$. Each of these triples can be modified in three different ways to give a triple with two negative signs, so the total number of integer solutions to $xyz=1{,}000{,}000$ is $4\cdot28\cdot28=3136$.

To explain the $1+2+3+4+5+6+7$, if $a=6$, then there's only $1$ choice for $b$ and $c$, namely $b=c=0$, while if $a=5$, there are $2$ choices, down to $a=0$, for which there are $7$ solutions to $b+c=6$.

Barry Cipra
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Let $x$ run through the factors of $1,000,000$. For each $x$, find all pairs $(y,z)$ that multiply to $\frac{1,000,000}{x}$.

G Tony Jacobs
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