-3

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 :).

KongMing
  • 7
  • 1
    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. – José Carlos Santos Jul 05 '22 at 07:08
  • A similar question: https://math.stackexchange.com/q/1595118/42969 – Martin R Jul 05 '22 at 07:08
  • Please use MathJax so that people can see it easily. – RDK Jul 05 '22 at 07:12
  • Sorry I am on my phone so I can't use MathJax – KongMing Jul 05 '22 at 07:15
  • 1
    I have already written many posts with MathJax on a smartphone. – José Carlos Santos Jul 05 '22 at 07:35
  • Prime factorization is $9=3^2$. Then $9$ has $2\cdot(2+1)=6$ distinct integer divisors. Using one of $\pm9$ multipliers excludes using any $\pm3$ and other $\pm9$ with only remaining multipliers $\pm1$, which makes total number of multipliers not greater than 3, which is not enough. Then one cannot use $\pm9$, then one can use only $\pm1$ and $\pm3$, which count is exactly four. – Ivan Kaznacheyeu Jul 05 '22 at 07:52

1 Answers1

-2

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

RDK
  • 2,623
  • 1
  • 8
  • 30