# 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 1

From $|a-b|=2$ we get that $a=b\pm 2$

Similarly, $b=c\pm3$ and $c=d\pm4$.

Substitution gives $a=d\pm 4\pm 3\pm 2$. This gives $|a-d|=|\pm 4\pm 3\pm 2|$. There are $2^3=8$ possibilities for the value of $\pm 4\pm 3\pm2$:

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

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

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

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

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

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

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

$-4-3-2=\boxed{-9}$

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. (This is due to the fact that we are calculating |d| at the end ~Williamgolly) Hence we can WLOG 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 $1+3+5+9 = \boxed{18}$.

## Solution 3

Transforming the absolute values to $\pm$ equations, we can now add: $$(Error compiling LaTeX. ! Missing  inserted.)\begin{tabular}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} & $a$ & $-$ & $\cancel{b}$ & $=$ & $\pm2$ &&\\ & $b$ & $-$ & $\cancel{c}$ & $=$ & $\pm3$&&\\ $+$& $c$ & $-$ & $\cancel{d}$ & $=$ & $\pm4$&&\\ \hline & $a$ & $-$ & $d$ & $=$ & $\pm2$&$\pm3$&$\pm4$\\ \end{tabular}$$ (Error compiling LaTeX. ! Missing \$ inserted.) The possible values can now be calculated by hand.

## See Also

 2009 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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