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Note: Someone else has posted this proof as a question; however, that post is largely inactive. I am interested in exploring this proof, but don't know how to "bump" someone's post, so I'm re-posting it here. If there is another way I should go about this, I'm happy to do so. <https://math.stackexchange.com/questions/1988393/not-sure-about-how-to-prove-this-primal-triple>

For the integers $(x, y, z)$ such that $2x = 3y = 5z$

Show that $xyz=n900$.

Any help to get me moving in the right direction?

Community
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Math1
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  • Providing a link to the post you are talking about would be helpful. – Mike Pierce Oct 29 '16 at 17:28
  • Here is a link to that question. http://math.stackexchange.com/questions/1988393/not-sure-about-how-to-prove-this-primal-triple I voted to close the other one, as it lack context. – quid Oct 29 '16 at 17:30
  • I've updated the post to include the link, thank-you. – Math1 Oct 29 '16 at 17:34

3 Answers3

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Since 2, 3 and 5 are all prime and pairwise relatively prime: $$2x=3y=5z\implies15\mid x,\ 10\mid y,\ 6\mid z$$ (This may be seen from prime factorisations.) Therefore $$15\cdot10\cdot6\mid xyz$$ $$900\mid xyz$$

Parcly Taxel
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$$ 2x=3y = 5z $$

If $2x = 3y$ then $2x$ is a multiple of $3$.

Since $2$ is prime and not a multiple of $3$, we must conclude $x$ is a multiple of $3$.

For the same reason, $x$ must be a multiple of $5$.

So $x = 3\cdot5\cdot j$ for some integer $j$.

Similarly, $y =2\cdot5\cdot k$ and $z = 2\cdot3\cdot \ell$.

So $xyz = (3\cdot5\cdot j)(2\cdot5\cdot k)(2\cdot 3\cdot\ell) = (2\cdot2\cdot3\cdot3 \cdot5\cdot5)jk\ell = 900 jk\ell.$

Michael Hardy
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Denote $$2x=3y=5z=k$$

Then $x=\frac{k}{2}$ is an integer, thus $2|k$. Similarly $3|k$ and $5|k$.

As $2,3,5$ are pairwise relatively prime, it follows that their product, 30, divides $k$.

Thus, we can write $k=30n$.

Then $$2x=3y=5z=30n\\ x=15n \\ y=10n \\ z=6 n \\ xyz=900 n^3$$

N. S.
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