Absolute Value System

$\begin{aligned} a - d &= (a - b) + (b - c) + (c - d) \\&= \pm 2 + \pm 3 + \pm 4 \\&\in \{ -9, -5, -3, -1, 1, 3, 5, 9 \} \end{aligned}$ So $\lvert a - d \rvert \in \{ 1,3,5,9 \}$ , and $1 + 3 + 5 + 9 = \boxed{18}.$

$a-d=(a-b)+(b-c)+(c-d)\quad \& \\ \\ a-b=\pm 2,\\ \\ b-c=\pm 3,\\ \\ c-d=\pm 4,\& \\ \\ a-d=(a-b)+(b-c)+(c-d)\\ \\ So\quad (a-d)\quad will\quad have\quad 8\quad value\quad \\ \\ 4\quad pair\quad (one\quad +ve\quad \& \quad one\quad -ve)\\ \\ having\quad 4\quad numerically\quad different\quad value,\\ \\ Ans=(2+3+4)+(-2+3+4)+(2-3+4)+(2+3-4)\\ \\ \quad \quad =18\\ \\$

From $\begin{cases} |a-b|=2 & \Rightarrow b = a \pm 2 \\ |b-c|=3 & \Rightarrow c = b \pm 3 \\ |c-d|=4 & \Rightarrow d = c \pm 4 \end{cases}$

Therefore, the possible values of $d$ and $|a-d|$ for each $a$ are as follows:

$\Rightarrow \begin{cases} b = a - 2 \Rightarrow \begin{cases} c = a-5 \Rightarrow \begin{cases} d = a-9 \Rightarrow |a-d| = 9 \\ d = a-1 \Rightarrow |a-d| = 1 \end{cases} \\ c = a+1 \Rightarrow \begin{cases} d = a-3 \Rightarrow |a-d| = 3 \\ d = a+5 \Rightarrow |a-d| = 5 \end{cases} \end{cases} \\ b = a+2 \Rightarrow \begin{cases} c = a-1 \Rightarrow \begin{cases} d = a-5 \Rightarrow |a-d| = 5 \\ d = a+3 \Rightarrow |a-d| = 3 \end{cases} \\ c = a+5 \Rightarrow \begin{cases} d = a+1 \Rightarrow |a-d| = 1 \\ d = a+9 \Rightarrow |a-d| = 9 \end{cases} \end{cases} \end{cases}$

Therefore, the sum of possible values of $|a-d|$ is $1+3+5+9 = \boxed{18}$

hmmm.....There are multiple answers to this since the absolute values are taken. The difference are known to be 2 3 4, so you can add all of them to get 9. But you can also add/subtract them in ways to get answers of 5, 3, and 1.

Obviously, 'a' can be viewed as a reference point, and it doesn't change the value of |a-d|. So set $a$ =0. $|d|=|d-c+c-b+b|=|±4±3±2|\in\{1,3,5,9\}$

Awesome problem :) :)

$| a - d | = \pm 2 + \pm 3 + \pm 4$

by drawing the tree, it's easily found that the possible values of |a-d| are: 1,3,5, 9.

thus, the sum is 18

I did it in a very childish way by supposing a to be 4 and then brute forcing to calculate b and c which gave me 4 distinct values for c. When I reached d I had 8 values out of which 6 were distinct. When I got absolute values for the difference between 4 and those values got 6 values out of which 4 were distinct namely 1 3 5 and 9. Thus making the sum as 18.

There are basically two possibilities for the value of each (a-b), (b-c) and (c-d) as 2 or -2, 3 or -3 and 4 or -4 respectively, hence we have 8 different possible combinations as under;

1. a-b=+2, b-c=+3, c-d=+4 -> a-d=+9
2. a-b=+2, b-c=+3, c-d=-4 -> a-d=+1
3. a-b=+2, b-c=-3, c-d=+4 -> a-d=+3
4. a-b=+2, b-c=-3, c-d=-4 -> a-d=-5
5. a-b=-2, b-c=+3, c-d=+4 -> a-d=+5
6. a-b=-2, b-c=+3, c-d=-4 -> a-d=-3
7. a-b=-2, b-c=-3, c-d=+4 -> a-d=-1
8. a-b=-2, b-c=-3, c-d=-4 -> a-d=-9

From 1 and 8 we get |a-d|=9, From 2 and 7 we get |a-d|=1, From 3 and 6 we get |a-d|=3, and From 4 and 5 we get |a-d|=5.

Hence 9+1+3+5=18 :)

2+3+4, 2-3-4, 2+3-4, 2-3+4

It can be solved by using the number line. First place a and b, and then check for possible locations of c, and the corresponding location of d. c cannot be between a and b. c can either be before a or after b or vice versa. Which give only four possible locations of d. Correspondingly, the distance between a and d has four possible values: 9,3,1,5. So, the sum comes out to be 18.