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Difference between revisions of "2014 AMC 12B Problems/Problem 12"

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<math>(1, 1, 1)</math>
 
<math>(1, 1, 1)</math>
  
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It should be clear that <math>|S|</math> is simply <math>|T| - t</math>, where <math>t</math> is the number of triples <math>(d, e, f)</math> such that there exists at least one triple <math>(kd, ke, kf)</math> where <math>k \ge 1</math> and <math>k \in \mathbb{N}</math>. So, <math>t</math> is... and the answer is ... ...
+
It should be clear that <math>|S|</math> is simply <math>|T|</math> minus the larger "duplicates" (e.g. <math>(2, 2, 2)</math> is a larger duplicate of <math>(1, 1, 1)</math>). Since <math>|T|</math> is 13 and the number of higher duplicates is 4, the answer is <math>13 - 4</math> or <math>9 B</math>

Revision as of 21:28, 20 February 2014

Solution

Define $T$ to be the set of all triples $(a, b, c)$ such that $a \ge b \ge c$, $b+c > a$, and $a, b, c \le 5$. Now we enumerate the elements of $T$:

$(4, 4, 4)$

$(4, 4, 3)$

$(4, 4, 2)$

$(4, 4, 1)$

$(4, 3, 3)$

$(4, 3, 2)$

$(3, 3, 3)$

$(3, 3, 2)$

$(3, 3, 1)$

$(3, 2, 2)$

$(2, 2, 2)$

$(2, 2, 1)$

$(1, 1, 1)$

It should be clear that $|S|$ is simply $|T|$ minus the larger "duplicates" (e.g. $(2, 2, 2)$ is a larger duplicate of $(1, 1, 1)$). Since $|T|$ is 13 and the number of higher duplicates is 4, the answer is $13 - 4$ or $9 B$

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