How many triplets of non-negative integers $(x, y, z)$ satisfy the equation $x y z + x y + y z + z x + x + y + z = 2012$ ?
(correct answer + 3, wrong answer 0)

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Solution

$(x + 1) (y + 1) (z + 1) = 2013 = 3 \times 11 \times 61$
If all $x, y, z$ are positive, there are $3! = 6$ solutions.
If exactly one of $x, y, z$ is $0,$ there are $3 \times 6 = 18$ solutions.
If exactly two of $x, y, z$ are $0,$ there are $3$ solutions.
Total solutions $= 6 + 18 + 3 = 27$

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