View source for 2010 AMC 12B Problems/Problem 8

{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #8]] and [[2010 AMC 10B Problems|2010 AMC 10B #17]]}}

== Problem ==
Every high school in the city of Euclid sent a team of <math>3</math> students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed <math>37</math><sup>th</sup> and <math>64</math><sup>th</sup>, respectively. How many schools are in the city?

<math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26</math>

== Solution 1 ==
There are <math>x</math> schools. This means that there are <math>3x</math> people. Because no one's score was the same as another person's score, that means that there could only have been <math>1</math> median score. This implies that <math>x</math> is an odd number. <math>x</math> cannot be less than <math>23</math>, because there wouldn't be a <math>64</math>th place if <math>x</math> was. <math>x</math> cannot be greater than <math>23</math> either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is <math>\boxed{\mathbf{(B)}\ 23}</math>.

== Solution 2 ==
Let <math>a</math> be Andrea's place. We know that she was the highest on her team, so <math>a < 37</math>. 

Since <math>a</math> is the median, there are <math>a-1</math> to the left and right of the median, so the total number of people is <math>2a-1</math> and the number of schools is <math>(2a-1)/3</math>. This implies that <math>2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}</math>. 

Also, since <math>2a-1</math> is the rank of the last-place person, and one of Andrea's teammates already got 64th place, <math>2a-1 > 64 \implies a \ge 33</math>.

Putting it all together: <math>33 \le a < 37</math> and <math>a \equiv 2 \pmod{3}</math>, so clearly <math>a = 35</math>, and the number of schools as we got before is <math>(2a-1)/3 = 69/3 = \boxed{\mathbf{(B)}\ 23}</math>.

~adihaya

==Video Solution==
https://youtu.be/FQO-0E2zUVI?t=297

~IceMatrix

== See also ==
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=B}}
{{AMC10 box|year=2010|num-b=16|num-a=18|ab=B}}
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