https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mrbillybohilly&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T11:06:52ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_16&diff=333322009 AMC 10A Problems/Problem 162010-02-07T21:22:29Z<p>Mrbillybohilly: /* Solution */</p>
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<div>== Problem ==<br />
<br />
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 sum of all possible values of <math>|a-d|</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 9<br />
\qquad<br />
\mathrm{(B)}\ 12<br />
\qquad<br />
\mathrm{(C)}\ 15<br />
\qquad<br />
\mathrm{(D)}\ 18<br />
\qquad<br />
\mathrm{(E)}\ 24<br />
</math><br />
<br />
== Solution ==<br />
<br />
Solution 1:<br />
<br />
From <math>|a-b|=3</math> we get that <math>a=b+-2</math><br />
<br />
Similarly, <math>b=c+-3</math> and <math>c=d+-4</math>.<br />
<br />
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>:<br />
<br />
<math>4+3+2=\boxed{9}</math>, <br />
<math>4+3-2=\boxed{5}</math>, <br />
<math>4-3+2=\boxed{3}</math>, <br />
<math>-4+3+2=\boxed{1}</math>, <br />
<math>4-3-2=\boxed{-1}</math>, <br />
<math>-4+3-2=\boxed{-3}</math>, <br />
<math>-4-3+2=\boxed{-5}</math>, <br />
<math>-4-3-2=\boxed{-9}</math><br />
<br />
Therefore, the only possible values of <math>|a-d|</math> are 9, 5, 3, and 1. Their sum is <math>\boxed{18}</math>. <br />
<br />
Solution 2:<br />
<br />
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>.<br />
<br />
From <math>|a-b|=2</math> we get that <math>|b|=2</math>, hence <math>b\in\{-2,2\}</math>.<br />
<br />
If we multiply all four numbers by <math>-1</math>, we will not change any of the differences. Hence we can assume that <math>b=2</math>.<br />
<br />
From <math>|b-c|=3</math> we get that <math>c\in\{-1,5\}</math>.<br />
<br />
From <math>|c-d|=4</math> we get that <math>d\in\{-5,1,3,9\}</math>.<br />
<br />
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>.<br />
<br />
== See Also ==<br />
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{{AMC10 box|year=2009|ab=A|num-b=15|num-a=17}}</div>Mrbillybohilly