Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.1 AMC 10/12 Find the answer

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 $|a-d|$ ?

Pick one

Solution

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 $\pm 4\pm 3\pm2$ :
$4+3+2=9$ ,
$4+3-2=5$ ,
$4-3+2=3$ ,
$-4+3+2=1$ ,
$4-3-2=-1$ ,
$-4+3-2=-3$ ,
$-4-3+2=-5$ ,
$-4-3-2=-9$
Therefore, the only possible values of $|a-d|$ are $9$ , $5$ , $3$ , and $1$ . Their sum is $\boxed{\textbf{(D) } 18}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.