Difference between revisions of "2010 AIME II Problems/Problem 13"
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== Solution == | == Solution == | ||
− | + | Once the two cards are drawn, there are <math>\dbinom{50}{2} = 1225</math> ways for the other two people to draw. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below <math>a</math>, which occurs in <math>\dbinom{a-1}{2}</math> ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above <math>a+9</math>, which occurs in <math>\dbinom{43-a}{2}</math> ways. Thus, <cmath>p(a)=\frac{\dbinom{43-a}{2}+\dbinom{a-1}{2}}{1225}.</cmath> Simplifying, we get <math>p(a)=\frac{(43-a)(42-a)+(a-1)(a-2)}{2\cdot1225}</math>, so we need <math>(43-a)(42-a)+(a-1)(a-2)\ge (1225)</math>. If <math>a=22+b</math>, then <cmath>\begin{align*}(43-a)(42-a)+(a-1)(a-2)&=(21-b)(20-b)+(21+b)(20+b)=2b^2+2(21)(20)\ge (1225) \\ b^2\ge \frac{385}{2} &= 192.5 >13^2 \end{align*}</cmath> So <math>b> 13</math> or <math>b< -13</math>, and <math>a=22+b<9</math> or <math>a>35</math>, so <math>a=8</math> or <math>a=36</math>. Thus, <math>p(8) = \frac{616}{1225} = \frac{88}{175}</math>, and the answer is <math>88+175 = \boxed{263}</math>. | |
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− | Once the two cards are drawn, there are <math>\dbinom{50}{2} = 1225</math> ways for the other two people to draw. | ||
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− | Alex and Dylan are the team with higher numbers if Blair and Corey both draw below <math>a</math>, which occurs in <math>\dbinom{a-1}{2}</math> ways. | ||
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− | Alex and Dylan are the team with lower numbers if Blair and Corey both draw | ||
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− | Thus, < | ||
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− | <cmath>\begin{align*}(43-a)(42-a)+(a-1)(a-2)&=(21-b)(20-b)+(21+b)(20+b)=2b^2+2(21)(20)\ge (1225) \\ | ||
− | b^2\ge \frac{385}{2} &= 192.5 >13^2 \end{align*}</cmath> | ||
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− | So <math>b> 13</math> or <math>b< -13</math>, and <math>a=22+b<9</math> or <math>a>35</math>, so <math>a=8</math> or <math>a=36</math>. | ||
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− | Thus, <math>p(8) = \frac{616}{1225} = \frac{88}{175}</math>, and the answer is <math>\boxed{263}</math>. | ||
== See also == | == See also == |
Revision as of 14:37, 26 February 2016
Problem
The cards in a deck are numbered
. Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked, The two persons with lower numbered cards from a team, and the two persons with higher numbered cards form another team. Let
be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards
and
, and Dylan picks the other of these two cards. The minimum value of
for which
can be written as
. where
and
are relatively prime positive integers. Find
.
Solution
Once the two cards are drawn, there are ways for the other two people to draw. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below
, which occurs in
ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above
, which occurs in
ways. Thus,
Simplifying, we get
, so we need
. If
, then
So
or
, and
or
, so
or
. Thus,
, and the answer is
.
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.