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

(Solution 3)
(Solution 3)
Line 57: Line 57:
 
==Solution 3==
 
==Solution 3==
 
Transforming the absolute values to <math>\pm</math> equations, we can now add:
 
Transforming the absolute values to <math>\pm</math> equations, we can now add:
<math>\begin{tabular}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c}
+
<math></math>\begin{tabular}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c}
& </math>a<math> & </math>-<math> & </math>\cancel{b}<math> & </math>=<math> & </math>\pm2<math> &&\
+
& <math>a</math> & <math>-</math> & <math>\cancel{b}</math> & <math>=</math> & <math>\pm2</math> &&\
& </math>b<math> & </math>-<math> & </math>\cancel{c}<math> & </math>=<math> & </math>\pm3<math>&&\
+
& <math>b</math> & <math>-</math> & <math>\cancel{c}</math> & <math>=</math> & <math>\pm3</math>&&\
</math>+<math>& </math>c<math> & </math>-<math> & </math>\cancel{d}<math> & </math>=<math> & </math>\pm4<math>&&\
+
<math>+</math>& <math>c</math> & <math>-</math> & <math>\cancel{d}</math> & <math>=</math> & <math>\pm4</math>&&\
 
\hline
 
\hline
& </math>a<math> & </math>-<math> & </math>d<math> & </math>=<math> & </math>\pm2<math>&</math>\pm3<math>&</math>\pm4<math>\
+
& <math>a</math> & <math>-</math> & <math>d</math> & <math>=</math> & <math>\pm2</math>&<math>\pm3</math>&<math>\pm4</math>\
\end{tabular}</math>
+
\end{tabular}<math></math>
 
The possible values can now be calculated by hand.
 
The possible values can now be calculated by hand.
  

Revision as of 15:59, 2 August 2020

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

Solution 3

Transforming the absolute values to $\pm$ equations, we can now add: $$ (Error compiling LaTeX. Unknown error_msg)\begin{tabular}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} & $a$ & $-$ & $\cancel{b}$ & $=$ & $\pm2$ &&\ & $b$ & $-$ & $\cancel{c}$ & $=$ & $\pm3$&&\ $+$& $c$ & $-$ & $\cancel{d}$ & $=$ & $\pm4$&&\ \hline & $a$ & $-$ & $d$ & $=$ & $\pm2$&$\pm3$&$\pm4$\ \end{tabular}$$ (Error compiling LaTeX. Unknown error_msg) The possible values can now be calculated by hand.

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

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