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If a×b×c×d is 9, what is a+b+c+d ( a b c d are different integers)

Actually, I know the answer to this problem, (which is 1+(-1)+3+(-3)= 0), but I am not sure whether there is a systematic way to solve it.

I thought of separating the a b c and d to

a= 91/b1/c1/d b=91/a1/c1/d And so on...

However, it didn't work. Should I treat a b c d as a whole?

Edit: Sorry I am on my phone so I can't use the formattings. Thanks for your understanding

Edit: Thanks for all your help :).

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    $\begingroup$ Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ Jul 5 at 7:08
  • $\begingroup$ A similar question: math.stackexchange.com/q/1595118/42969 $\endgroup$
    – Martin R
    Jul 5 at 7:08
  • $\begingroup$ Please use MathJax so that people can see it easily. $\endgroup$
    – RDK
    Jul 5 at 7:12
  • $\begingroup$ Sorry I am on my phone so I can't use MathJax $\endgroup$
    – KongMing
    Jul 5 at 7:15
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    $\begingroup$ I have already written many posts with MathJax on a smartphone. $\endgroup$ Jul 5 at 7:35

1 Answer 1

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If $a, b, c, d$ are distinct integers which satisfies $abcd=9$, find $a+b+c+d$.

$a|9, b|9, c|9, d|9$.

The factors of $9$ is $-9, -3, -1, 1, 3, 9$.

WLOG, $a<b<c<d$.

If $a=-9:$ $bcd=-1$ for all distinct integers $b, c, d$, which has a contradiction.

If $b=9:$ $abc=1$ for all distinct integers $a, b, c$, which has a contradiction.

Therefore, $a=-3, b=-1, c=1, d=3.$

So, $a+b+c+d=0$.

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    $\begingroup$ Please do not post answers to duplicate questions and/or questions which show little or no effort from the OP. This is against the Quality Standards of the site. $\endgroup$ Jul 5 at 7:20

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