2009 AMC 10A Problems/Problem 16

Revision as of 08:52, 13 February 2009 by Misof (talk | contribs) (New page: == Problem == Let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be real numbers with <math>|a-b|=2</math>, <math>|b-c|=3</math>, and <math>|c-d|=4</math>. What is the...)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

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 $1+3+5+9 = \boxed{18}$.

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem
Followed by
Problem
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
Invalid username
Login to AoPS