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

Revision as of 21:16, 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 $T$ :

$(5, 5, 5)$

$(5, 5, 4)$

$(5, 5, 3)$

$(5, 5, 2)$

$(5, 5, 1)$

$(5, 4, 4)$

$(5, 4, 3)$

$(5, 4, 2)$

$(5, 3, 3)$

$(4, 4, 4)$

$(4, 4, 3)$

$(4, 4, 2)$

$(4, 4, 1)$

$(4, 3, 3)$

$(4, 3, 2)$

$(3, 3, 3)$

$(3, 3, 2)$

$(3, 3, 1)$

$(3, 2, 2)$

$(2, 2, 2)$

$(2, 2, 1)$

$(1, 1, 1)$

It should be clear that $|S|$ is simply $|T| - t$ , where $t$ is the number of triples $(d, e, f)$ such that there exists at least one triple $(kd, ke, kf)$ where $k \ge 1$ and $k \in \mathbb{N}$ . So, $t$ is... and the answer is ... ...

Revision as of 21:15, 20 February 2014 (view source) Mathgreen (talk \| contribs) (→‎Solution)		Revision as of 21:16, 20 February 2014 (view source) Mathgreen (talk \| contribs) (→‎Solution) Newer edit →
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	<math>(1, 1, 1)</math>		<math>(1, 1, 1)</math>

−	It should be clear that <math>\|S\|</math> is simply <math>\|T\| - t</math>, where <math>x</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>x</math> is	+	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 ... ...

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

Revision as of 21:16, 20 February 2014

Solution