Difference between revisions of "2005 AIME II Problems/Problem 1"

 
(Solution)
 
(9 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Six circles form a ring with with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle <math>C</math> with radius <math>30</math>. Let <math>K</math> be the area of the region inside circle <math>C</math> and outside of the six circles in the ring. Find <math>\lfloor K \rfloor</math>.
+
 
 +
A game uses a deck of <math> n </math> different cards, where <math> n </math> is an integer and <math> n \geq 6. </math> The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find <math> n. </math>
 +
 
 +
== Video Solution ==
 +
https://youtu.be/IRyWOZQMTV8?t=150
 +
 
 +
~ pi_is_3.14
 +
 
 
== Solution ==
 
== Solution ==
 +
 +
The number of ways to draw six cards from <math>n</math> is given by the [[binomial coefficient]] <math>{n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}</math>. 
 +
 +
The number of ways to choose three cards from <math>n</math> is <math>{n\choose 3} = \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}</math>.
 +
 +
We are given that <math>{n\choose 6} = 6 {n \choose 3}</math>, so <math>\frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1} = 6 \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}</math>.
 +
 +
Cancelling like terms, we get <math>(n - 3)(n - 4)(n - 5) = 720</math>.
 +
 +
We must find a [[factoring|factorization]] of the left-hand side of this equation into three consecutive [[integer]]s. Since 720 is close to <math>9^3=729</math>, we try 8, 9, and 10, which works, so <math>n - 3 = 10</math> and <math>n = \boxed{013}</math>.
 +
 
== See Also ==
 
== See Also ==
*[[2005 AIME II Problems]]
+
{{AIME  box|year=2005|n=II|before=First Question|num-a=2}}
 +
 
 +
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:48, 17 January 2021

Problem

A game uses a deck of $n$ different cards, where $n$ is an integer and $n \geq 6.$ The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find $n.$

Video Solution

https://youtu.be/IRyWOZQMTV8?t=150

~ pi_is_3.14

Solution

The number of ways to draw six cards from $n$ is given by the binomial coefficient ${n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$.

The number of ways to choose three cards from $n$ is ${n\choose 3} = \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}$.

We are given that ${n\choose 6} = 6 {n \choose 3}$, so $\frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1} = 6 \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}$.

Cancelling like terms, we get $(n - 3)(n - 4)(n - 5) = 720$.

We must find a factorization of the left-hand side of this equation into three consecutive integers. Since 720 is close to $9^3=729$, we try 8, 9, and 10, which works, so $n - 3 = 10$ and $n = \boxed{013}$.

See Also

2005 AIME II (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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. AMC logo.png

Invalid username
Login to AoPS