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$a,b,c,d$ and $e$ are five different integers. If $(4-a)(4-b)(4-c)(4-d)(4-e)=12$, then calculate $a+b+c+d+e$.

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    $\begingroup$ Even if you restrict the numbers to integers ${2,2,3,3,1}$ and ${0,3,3,3,1}$ gives two different results. I guess maybe the question means "positive integers"? $\endgroup$
    – cr001
    Sep 16, 2017 at 21:40
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    $\begingroup$ what has been tried ? $\endgroup$
    – user451844
    Sep 16, 2017 at 21:53
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    $\begingroup$ Are $a$,$b$,$c$,$d$ and $e$ real or integers? Are they positive or negative? You should always post the full princlement statement along with what you've tried so far. $\endgroup$
    – user472341
    Sep 16, 2017 at 21:54
  • $\begingroup$ the numbers are different integers $\endgroup$
    – hachemy
    Sep 16, 2017 at 22:13

1 Answer 1

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I found the solution: $12$ can be expressed as a product of five integers only as $1\times(-1)\times2\times(-2)\times3 =12$, the $a-4=1, b-4=-1, c-4=2, d-4=-2$ and $e-4=3$. we deduce from that $a+b+c+d+e=23$.

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