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

(Solution)
m (Solution)
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== Solution ==
 
== Solution ==
 +
=== Solution 1 ===
  
Solution 1:
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From <math>|a-b|=2</math> we get that <math>a=b\pm 2</math>
  
From <math>|a-b|=3</math> we get that <math>a=b+-2</math>
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Similarly, <math>b=c\pm3</math> and <math>c=d\pm4</math>.
  
Similarly, <math>b=c+-3</math> and <math>c=d+-4</math>.
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Substitution gives <math>a=d\pm 4\pm 3\pm 2</math>. This gives <math>|a-d|=|\pm 4\pm 3\pm 2|</math>. There are <math>2^3=8</math> possibilities for the value of <math>\pm 4\pm 3\pm2</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{9}</math>,
 
 
<math>4+3-2=\boxed{5}</math>,  
 
<math>4+3-2=\boxed{5}</math>,  
 +
 
<math>4-3+2=\boxed{3}</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{-1}</math>,  
 
<math>4-3-2=\boxed{-1}</math>,  
 +
 
<math>-4+3-2=\boxed{-3}</math>,  
 
<math>-4+3-2=\boxed{-3}</math>,  
 +
 
<math>-4-3+2=\boxed{-5}</math>,  
 
<math>-4-3+2=\boxed{-5}</math>,  
 +
 
<math>-4-3-2=\boxed{-9}</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>.  
 
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:
+
=== 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 02:44, 8 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 $|a-b|=2$ 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. 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 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions
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