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

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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: \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} The possible values can now be calculated by hand.

@@ Line 54: / Line 54: @@
 Hence <math>|a-d|=|d|\in\{1,3,5,9\}</math>, and the sum of possible values is <math>1+3+5+9 = \boxed{18}</math>.
+==Solution 3==
+Transforming the absolute values to <math>\pm</math> equations, we can now add:
+\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>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>&&\\
+\hline
+& <math>a</math> & <math>-</math> & <math>d</math> & <math>=</math> & <math>\pm2</math>&<math>\pm3</math>&<math>\pm4</math>\\
+\end{tabular}
+The possible values can now be calculated by hand.
 == 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

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