Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎See Also: Fixed the box)
(→‎Solution)
Line 16: Line 16:
  
 
== Solution ==
 
== Solution ==
 +
 +
Solution 1:
 +
 +
From <math>|a-b|=3</math> we get that <math>a=b+-2</math>
 +
 +
Similarly, <math>b=c+-3</math> and <math>c=d+-4</math>.
 +
 +
Substitution gives <math>a=d+-4+-3+-2</math>. This gives <math>|a-d|=|+-4+-3+-2|</math>. There are <math>2^3=8</math> possibilities for the value of <math>|+-4+-3+-2|</math>:
 +
 +
<math>4+3+2=\boxed{9}</math>,
 +
<math>4+3-2=\boxed{5}</math>,
 +
<math>4-3+2=\boxed{3}</math>,
 +
<math>-4+3+2=\boxed{1}</math>,
 +
<math>4-3-2=\boxed{-1}</math>,
 +
<math>-4+3-2=\boxed{-3}</math>,
 +
<math>-4-3+2=\boxed{-5}</math>,
 +
<math>-4-3-2=\boxed{-9}</math>
 +
 +
Therefore, the only possible values of <math>|a-d|</math> are 9, 5, 3, and 1. Their sum is <math>\boxed{18}</math>.
 +
 +
Solution 2:
  
 
If we add the same constant to all of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, we will not change any of the differences. Hence we can assume that <math>a=0</math>.
 
If we add the same constant to all of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, we will not change any of the differences. Hence we can assume that <math>a=0</math>.

Revision as of 17:22, 7 February 2010

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

See Also

2009 AMC 10A (Problems • Answer Key • Resources)
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_16&oldid=33332"