2009 AMC 10A Problems/Problem 16

Revision as of 17:00, 2 August 2020 by Coltsfan10 (talk | contribs)


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 $\pm 4\pm 3\pm2$:









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

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 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. AMC logo.png

Invalid username
Login to AoPS