If a, B and C are three digits of a three digit number, and a ≤ B ≤ C, what is the maximum value of | A-B | + | B-C | + | C-A |?
Because a ≤ B ≤ C
therefore
|a-b|+|b-c|+|c-a|=(b-a)+(c-b)+(c-a)=2(c-a)
The maximum probability of a three digit number minus a hundred is 9-1 = 8
So the answer is 8 * 2 = 16
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