Given that a, B, C, D, e and F are six mutually unequal integers and a × B × C × D × e × f = 36, then a + B + C + D + e + F = ()

36 can be divided into 2 × 2 × 3 × 3, and the difference integer can be supplemented by ± 1, so the six numbers are
±3±2±1
So the sum of them is 0
a+b+c+d+e+f=0

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