SOLUTION: Let a, b, c, and d be real numbers with |a-b|=2,|b-c|=3, and|c-d|=4. What is the sum of all possible values of |a-d|?

We can choose an arbitrary value for "a";

    then find 2 possible values for "b" using  |a-b| = 2;

        then find 4 possible values for "c" using |b-c| = 3;

            then find 8 possible values for "d" using |c-d| = 4.



Let a = 10.

    then 2 possible values for "b" are 8 and 12;

        then 4 possible values for "c" are 5, 11, 9 and 15;

            then 8 possible values for "d"  are 1, 9, 7, 15, 5, 13, 11 and 19.



Now you can calculate the sum of all possible distances |a-d|.


These distances are 9, 1, 3, 5, 5, 3, 1 and 9.


Their sum is  9 + 1 + 3 + 5 + 5 + 3 + 9 + 1 = 36.     ANSWER