# Difference between revisions of "2009 AMC 10A Problems/Problem 16"

## Problem

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 $|a-d|$?

$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$

## Solution

Solution 1:

From $|a-b|=3$ we get that $a=b+-2$

Similarly, $b=c+-3$ and $c=d+-4$.

Substitution gives $a=d+-4+-3+-2$. This gives $|a-d|=|+-4+-3+-2|$. There are $2^3=8$ possibilities for the value of $|+-4+-3+-2|$:

$4+3+2=\boxed{9}$, $4+3-2=\boxed{5}$, $4-3+2=\boxed{3}$, $-4+3+2=\boxed{1}$, $4-3-2=\boxed{-1}$, $-4+3-2=\boxed{-3}$, $-4-3+2=\boxed{-5}$, $-4-3-2=\boxed{-9}$

Therefore, the only possible values of $|a-d|$ are 9, 5, 3, and 1. Their sum is $\boxed{18}$.

Solution 2:

If we add the same constant to all of $a$, $b$, $c$, and $d$, we will not change any of the differences. Hence we can assume that $a=0$.

From $|a-b|=2$ we get that $|b|=2$, hence $b\in\{-2,2\}$.

If we multiply all four numbers by $-1$, we will not change any of the differences. Hence we can assume that $b=2$.

From $|b-c|=3$ we get that $c\in\{-1,5\}$.

From $|c-d|=4$ we get that $d\in\{-5,1,3,9\}$.

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