### Instruction: 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 $|a-d|$?### Response: I notice that the problem involves absolute values, which can change sign depending on the order of the numbers inside.
This means that there are two possible cases for each equation: either the left-hand side is positive or negative.
For example, $|a-b|=2$ implies either $a-b=2$ or $a-b=-2$.
Similarly, $|b-c|=3$ implies either $b-c=3$ or $b-c=-3$, and $|c-d|=4$ implies either $c-d=4$ or $c-d=-4$.
To find the possible values of $|a-d|$, I need to consider all the combinations of these cases and see what they imply for $a-d$.
I can use a tree diagram to organize the cases and keep track of the results.
Starting from $|a-b|=2$, I have two branches: $a-b=2$ and $a-b=-2$.
From each branch, I can further branch out based on the value of $|b-c|=3$.
If $a-b=2$, then $b=a-2$.
This means that either $b-c=3$ or $b-c=-3$.
If $b-c=3$, then $c=b-3=a-5$.
This means that either $c-d=4$ or $c-d=-4$.
If $c-d=4$, then $d=c-4=a-9$.
This implies that $a-d=9$.
If $c-d=-4$, then $d=c+4=a-1$.
This implies that $a-d=1$.
Similarly, if $b-c=-3$, then $c=b+3=a+1$.
This means that either $c-d=4$ or $c-d=-4$.
If $c-d=4$, then $d=c-4=a-3$.
This implies that $a-d=3$.
If $c-d=-4$, then $d=c+4=a+5$.
This implies that $a-d=-5$.
Therefore, from the branch where $a-b=2$, I get four possible values of $a-d$: $9$, $1$, $3$, and $-5$.
Similarly, from the branch where $a-b=-2$, I get four possible values of $a-d$: $-9$, $-1$, $-3$, and $5$.
These are all the possible values of $a-d$ that can result from the given equations.
To find the sum of all possible values of $|a-d|$, I need to take the absolute value of each term and add them up.
I get $|9|+|1|+|3|+|-5|+|-9|+|-1|+|-3|+|5|=9+1+3+5+9+1+3+5=36$.