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

Revision as of 18:17, 17 April 2009

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 $|a-d|$ ?

$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$

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

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 10 Problems and Solutions

Revision as of 07:52, 13 February 2009 (view source) Misof (talk \| contribs) (New page: == Problem == Let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be real numbers with <math>\|a-b\|=2</math>, <math>\|b-c\|=3</math>, and <math>\|c-d\|=4</math>. What is the...)		Revision as of 18:17, 17 April 2009 (view source) 5849206328x (talk \| contribs) m (→‎See Also: Fixed the box) Newer edit →
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	== See Also ==		== See Also ==

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